◆ QA · Geometry & Mensuration

Geometry & Mensuration, formulas + CAT PYQs

Lines, angles & triangles, quadrilaterals, polygons, circles, 2D & 3D mensuration and co-ordinate geometry, the single highest-yielding QA chapter on CAT (about 10 questions a year). Every formula plus the full archive of solved CAT past-papers from the Oswaal book.

21formulas

140CAT PYQs

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Formula Sheet CAT PYQs

Lines & Angles Triangles Circles Quadrilaterals & Polygons Mensuration, 2D Mensuration, 3D Coordinate Geometry Trigonometry

Formula & Concept Sheet

A-to-Z. Everything you need for this chapter, distilled from the Revision Notes.

1Lines & Angles

The starting toolkit: most "find the angle" questions are solved just by knowing angles on a line make 180° and walking that around the figure.
Angles on a straight line add to 180°; angles around a point add to 360°.
Vertically opposite angles are equal.
Parallel lines cut by a transversal: corresponding & alternate angles equal; co-interior angles sum to 180°.
Exterior angle of a triangle = sum of the two remote interior angles.
e.g. Two angles sit on a straight line and one is 110°. The other = 180° − 110° = 70°.

2Triangle, basics & area

Pick the area formula that matches what you're given: base+height, all three sides (Heron), or two sides and the angle between them.
Angle sum = 180°. Sum of any two sides > the third side.
Area = ½ × base × height.
Heron: Area = √[s(s−a)(s−b)(s−c)], s = (a+b+c)/2.
Area = ½·a·b·sinθ (θ = included angle); Area = r × s (r = inradius); Area = abc/(4R) (R = circumradius).
e.g. Sides 3, 4, 5: s = 6, Area = √[6·3·2·1] = √36 = 6 (matches ½·3·4).

Area = √[s(s−a)(s−b)(s−c)] , s = (a+b+c)/2 Area = r·s = abc/(4R)

3Cosine & Sine rules

Cosine rule links three sides and one angle (use when you have two sides + included angle, or all three sides); sine rule links sides to opposite angles.
Cosine rule: c² = a² + b² − 2ab·cosθ.
cosθ = (a² + b² − c²)/(2ab).
Sine rule: a/sinA = b/sinB = c/sinC = 2R.
e.g. Sides 5 and 8 with a 60° angle between them: third side² = 25 + 64 − 2·5·8·½ = 89 − 40 = 49 ⇒ side = 7.

c² = a² + b² − 2ab·cosθ

4Angle-bisector & medians

A bisector splits the far side in the ratio of the two sides it sits between; a median goes to the midpoint, and the centroid cuts it 2:1.
Angle bisector divides the opposite side in the ratio of the adjacent sides: BD/DC = AB/AC.
Apollonius: b² + c² = 2m² + ½a² (m = median to side a).
Median of isosceles (b = c): m² = b² − a²/4.
Centroid divides each median in ratio 2 : 1 from the vertex.
e.g. AB = 6, AC = 4, bisector meets BC (length 5) at D: BD:DC = 6:4 = 3:2 ⇒ BD = 3, DC = 2.

b² + c² = 2m² + ½a² (Apollonius)

5Pythagoras & altitude relations

In a right triangle the squares of the legs add to the square of the hypotenuse; memorising the triplets saves time in the exam.
Right triangle: hypotenuse² = base² + height².
Altitude to hypotenuse (AD⊥BC, right-angled at A): AD² = BD·DC, AB² = BD·BC, AC² = CD·BC.
Acute: AC² = AB² + BC² − 2·BC·BD; Obtuse: AC² = AB² + BC² + 2·BC·BD.
Triplets: 3-4-5, 5-12-13, 8-15-17, 7-24-25.
e.g. Legs 6 and 8: hypotenuse = √(36 + 64) = √100 = 10 (a scaled 3-4-5).

6Congruence & similarity

Congruent = identical; similar = same shape, scaled. The big CAT lever is that areas of similar figures scale as the square of the side ratio.
Congruence: SSS, SAS, ASA, AAS, RHS.
Similarity: AA, SSS, SAS. Corresponding sides are proportional.
Basic Proportionality (Thales): a line ∥ to one side cuts the others in equal ratios.
Ratio of areas of similar triangles = (ratio of sides)².
e.g. Two similar triangles with sides in ratio 2:3 have areas in ratio 4:9. If the smaller has area 8, the larger = 18.

Area₁ / Area₂ = (side₁ / side₂)²

7Special triangles

Equilateral and the two "set-square" triangles (30-60-90, 45-45-90) have fixed side ratios, recognise them and you can write down sides instantly.
Equilateral side a: Area = (√3/4)a², height = (√3/2)a, R = a/√3, r = a/(2√3).
30-60-90 sides ratio 1 : √3 : 2.
45-45-90 sides ratio 1 : 1 : √2.
From an interior point of an equilateral triangle, sum of ⊥s to the three sides = its height.
e.g. Equilateral triangle of side 4: area = (√3/4)·16 = 4√3 and height = (√3/2)·4 = 2√3.

Equilateral Area = (√3/4)·a²

8Geometric centres

Four "centres", each the meeting point of a different set of cevians; the incentre and circumcentre are the ones that show up most in area/radius questions.
Centroid, intersection of medians (2:1).
Incentre, intersection of angle bisectors, centre of inscribed circle.
Circumcentre, intersection of ⊥ bisectors of sides, centre of circumscribed circle.
Orthocentre, intersection of altitudes.
e.g. In a right triangle the circumcentre is the midpoint of the hypotenuse, so a 6-8-10 triangle has circumradius = 10/2 = 5.

9Circle, basics

Two workhorses: the angle at the centre is twice the angle at the rim on the same arc, and any angle drawn on a diameter is a right angle.
Circumference = 2πr; Area = πr².
Equal chords subtend equal angles at the centre & are equidistant from it.
⊥ from the centre bisects the chord.
Angle at the centre = 2 × angle at the circumference on the same arc.
Angle in a semicircle = 90°.
e.g. An arc subtends 40° at the centre, so it subtends 40°/2 = 20° at any point on the major arc.

10Chords, tangents, secants

"Power of a point": from any point, the products of the two distances to the circle along a line are equal, chords, secants and tangents all obey it.
Two chords meeting at P: PA·PB = PC·PD.
Tangent-secant from external P: PA·PB = PT².
Tangent ⊥ radius at the point of contact; tangents from an external point are equal.
Alternate segment theorem: tangent-chord angle = angle in the alternate segment.
e.g. Two chords cross with parts 3 & 8 on one and 4 & x on the other: 3·8 = 4·x ⇒ x = 6.

PA·PB = PC·PD ; PT² = PA·PB

11Cyclic quadrilateral & tangents to 2 circles

If all four corners lie on a circle, opposite angles are supplementary; the tangent formulas give the straight-line distance between two circles' touch points.
Cyclic quad: opposite angles sum to 180°; exterior angle = opposite interior angle.
Ptolemy: AB·CD + BC·DA = AC·BD.
A parallelogram inscribed in a circle is a rectangle.
Direct common tangent = √[d² − (r₁−r₂)²]; Transverse = √[d² − (r₁+r₂)²].
e.g. In a cyclic quad one angle is 70°, so its opposite angle = 180° − 70° = 110°.

Direct tangent = √[d² − (r₁−r₂)²]

12Quadrilaterals

Each special quadrilateral has its own area shortcut, base×height for parallelograms, half-product of diagonals for a rhombus, average of parallel sides times height for a trapezium.
Parallelogram: opposite sides & angles equal; diagonals bisect each other. Area = base × height.
Rectangle: all angles 90°, diagonals equal. Square: all sides equal + 90°.
Rhombus: all sides equal; diagonals ⊥ & bisect each other. Area = ½·d₁·d₂.
Trapezium: one pair of parallel sides. Area = ½(sum of parallel sides) × height.
e.g. Rhombus with diagonals 6 and 8: area = ½·6·8 = 24; trapezium with parallel sides 5 & 9, height 4: area = ½·(5+9)·4 = 28.

Rhombus Area = ½·d₁·d₂ ; Trapezium = ½(a+b)·h

13Polygons

Everything flows from "(n−2)·180° of total interior angle"; for a regular polygon the quick route is via the exterior angle, which is just 360°/n.
Sum of interior angles = (n − 2)·180°.
Each interior angle (regular) = 180° − 360°/n.
Each exterior angle (regular) = 360°/n; all exterior angles sum to 360°.
Number of diagonals = n(n − 3)/2.
e.g. A regular hexagon (n = 6): each exterior angle = 360°/6 = 60°, so each interior angle = 120°; diagonals = 6·3/2 = 9.

Interior sum = (n−2)·180° ; diagonals = n(n−3)/2

14Regular hexagon

Think of it as 6 equilateral triangles glued at the centre, that single picture gives its area, diagonals and angles.
Side s: Area = (3√3/2)·s².
It is 6 equilateral triangles of side s.
Longer diagonal = 2s; shorter diagonal = √3·s.
Interior angle = 120°.
e.g. Hexagon of side 2: area = (3√3/2)·4 = 6√3; long diagonal = 4, short diagonal = 2√3.

Hexagon Area = (3√3/2)·s²

152D mensuration, perimeters & areas

The basic flat-shape formulas; a sector is just a fraction θ/360 of the whole circle for both its arc and its area.
Square: P = 4a, Area = a², diagonal = a√2.
Rectangle: P = 2(l+b), Area = l·b, diagonal = √(l²+b²).
Circle: C = 2πr, Area = πr².
Sector (angle θ): arc = (θ/360)·2πr, area = (θ/360)·πr².
e.g. A 90° sector of a radius-6 circle is ¼ of it: area = ¼·π·36 = 9π, arc = ¼·2π·6 = 3π.

16Cube & cuboid

A cube is just a cuboid with l = b = h; the space diagonal (corner-to-corner through the body) uses a 3-term Pythagoras.
Cuboid: Volume = l·b·h; TSA = 2(lb + bh + hl); LSA (4 walls) = 2(l+b)h; diagonal = √(l²+b²+h²).
Cube edge a: Volume = a³; TSA = 6a²; LSA = 4a²; diagonal = a√3.
Sum of all 12 edges: cuboid 4(l+b+h), cube 12a.
e.g. Cube of edge 3: volume = 27, TSA = 6·9 = 54, space diagonal = 3√3; a 2×3×6 cuboid has diagonal √(4+9+36) = √49 = 7.

Cuboid V = l·b·h ; TSA = 2(lb+bh+hl)

17Cylinder & cone

A cone holds exactly one-third of the cylinder with the same base and height; its slant height is the hypotenuse of the radius-and-height right triangle.
Cylinder: Volume = πr²h; CSA = 2πrh; TSA = 2πr(r+h).
Cone slant l = √(r²+h²); Volume = ⅓πr²h; CSA = πrl; TSA = πr(r+l).
Frustum volume = ⅓πh(R² + r² + Rr).
e.g. Cone with r = 3, h = 4: slant = √(9+16) = 5, CSA = π·3·5 = 15π, volume = ⅓·π·9·4 = 12π.

Cone V = ⅓πr²h ; CSA = πrl , l = √(r²+h²)

18Sphere & hemisphere & prism

The key exam idea is "recasting": when one solid is melted into another, volume stays the same even though surface area changes.
Sphere: Volume = (4/3)πr³; Surface area = 4πr².
Hemisphere: Volume = (2/3)πr³; CSA = 2πr²; TSA = 3πr².
Prism: Volume = base area × height; LSA = base perimeter × height.
Recast objects keep volume constant.
e.g. Sphere of radius 3: volume = (4/3)·π·27 = 36π, surface area = 4·π·9 = 36π.

Sphere V = (4/3)πr³ ; SA = 4πr²

19Coordinate geometry, distance & section

Distance is just Pythagoras on the coordinate differences; the midpoint is the special case of the section formula with ratio 1:1.
Distance = √[(x₂−x₁)² + (y₂−y₁)²].
Midpoint = ((x₁+x₂)/2, (y₁+y₂)/2).
Section (ratio m:n internal) = ((mx₂+nx₁)/(m+n), (my₂+ny₁)/(m+n)).
Centroid = ((x₁+x₂+x₃)/3, (y₁+y₂+y₃)/3).
e.g. Distance from (1, 2) to (4, 6) = √(3² + 4²) = 5; their midpoint = (2.5, 4).

d = √[(x₂−x₁)² + (y₂−y₁)²]

20Coordinate geometry, area & slope

Slope = rise over run. Equal slopes mean parallel; slopes multiplying to −1 mean perpendicular. The area formula needs only the three vertices.
Area = ½|x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂)|.
Slope of line through two points m = (y₂−y₁)/(x₂−x₁).
Parallel lines: m₁ = m₂. Perpendicular: m₁·m₂ = −1.
e.g. Triangle (0,0), (4,0), (0,3): area = ½|0(0−3)+4(3−0)+0| = ½·12 = 6.

Area = ½|x₁(y₂−y₃)+x₂(y₃−y₁)+x₃(y₁−y₂)|

21Coordinate geometry, lines & circle

For the general circle, halve the x- and y-coefficients (with a sign flip) to read off the centre, then back out the radius.
Slope-intercept: y = mx + c. Point-slope: y − y₁ = m(x − x₁).
⊥ distance of (x₁,y₁) from ax+by+c=0 = |ax₁+by₁+c|/√(a²+b²).
Distance between parallel lines = |c₂−c₁|/√(a²+b²).
Circle: (x−h)² + (y−k)² = r²; general x²+y²+2gx+2fy+c=0, centre (−g,−f), r = √(g²+f²−c).
e.g. Distance of (0,0) from 3x + 4y − 10 = 0 = |−10|/√(9+16) = 10/5 = 2.

dist = |ax₁+by₁+c| / √(a²+b²)

Lines & Angles · 3 CAT PYQs

Lines & Angles

Lines & Angles. Angle-chasing on straight lines, transversals, clock hands and link-counting networks, pure angle/relationship reasoning.

ModerateCAT 1991

In a six-node network, two nodes are connected to all the other nodes. Of the remaining four, each is connected to four nodes. What is the total number of links in the network?

(1) 13
(2) 15
(3) 7
(4) 26

Show solution

Option (1) is correct. We see that the total number of links in the network is 13. (Note: In the diagram, the top two nodes are connected to all the other nodes, while the remaining four are connected to only four other nodes). The actual sides of the hexagon as well also form links. Six-node network drawn on a hexagon with all linking edges, totalling 13 links

Six-node network drawn on a hexagon with all linking edges, totalling 13 links

ModerateCAT 2002

Directions: Answer this question based on the following diagram. In the diagram ∠ABC = 90° = ∠DCH = ∠DOE = ∠EHK = ∠FKL = ∠GLM = ∠LMN, AB = BC = 2CH = 2CD = EH = FK = 2HK = 4KL = 2LM = MN. Diagram with points A, B, C, D, E, F, G, H, K, L, M, N, O forming connected right-angled segments
The magnitude of Angle FGO =

(1) 30°
(2) 45°
(3) 60°
(4) None of these

Show solution

Option (4) is correct. Given that AB = BC = 2CH = 2CD = EH = FK = 2HK = 4KL = 2LM = MN, and EO = FP. Also, 2CD = EH, so EO = FP = CD. Therefore KL = PG = CD/2. So FP = CD; PG = CD/2; ∠FPG = 90°. Since the angles are proportionate to the sides opposite to the angles, in △FGP, tan∠FGP = FP/PG = 2/1 = 2. Therefore ∠FGO = ∠FGP = tan⁻¹2.

HardCAT 2023 · Slot 1

The minor angle between the hour hand and minute hand of a clock was observed at 8:48 am. The minimum duration, in minutes, after 8:48 am when this angle increase by 50% is

(1) 24/11
(2) 36/11
(3) 4
(4) 2

Show solution

Option (1) is correct. In one min angle = (11/2)°. So in 48 min, angle = (11/2) × 48 = 264°. Therefore minor angle = 264° − 240° = 24°. 50% of minor angle = 12°. Since (11/2)° angle is made in 1 min, 1° angle is made in 2/11 min, so 12° angle is made in (2/11) × 12 = 24/11 min.

Triangles · 38 CAT PYQs

Triangles

Triangles. The single biggest sub-topic: area formulas, similarity ratios, medians, altitudes, special triangles and angle chasing.

EasyCAT 2000

What is the number of distinct triangles with integral valued sides and perimeter 14?

(1) 6
(2) 5
(3) 4
(4) 3

Show solution

(3) 4. Need a+b+c = 14 with a+b > c. The valid integer triangles are (4,4,6), (5,5,4), (6,5,3) and (6,6,2), four distinct triangles.

HardCAT 2000

In the figure below, AB = BC = CD = DE = EF = FG = GA. Then ∠DAE is approximately

Chain of equal segments AB=BC=CD=DE=EF=FG=GA with vertices A, B, F, D on the base and G, C, E above

(1) 15°
(2) 20°
(3) 30°
(4) 25°

Show solution

(4) 25°. Let ∠DAE = x. Using the chain of isosceles triangles and the exterior-angle property, each successive base angle steps up: ∠CBD = 2x, ∠DCE = 3x, … and finally in triangle ADE, ∠ADE + ∠DAE + ∠AED = 3x + x + 3x = 7x = 180°, so x = 180°/7 ≈ 25°.

ModerateCAT 2001

In the figure below, ABCD is a rectangle, and AE = EF = FB. What is the ratio of the area of the triangle CEF and that of the rectangle?

Rectangle ABCD with D, C on top and A, E, F, B on the base; E and F trisect AB; triangle CEF drawn from C to E and F

(1) 1/6
(2) 1/8
(3) 1/9
(4) None of these

Show solution

(1) 1/6. Area of △CEF = ½ × EF × BC = ½ × (⅓AB) × BC = (1/6)(AB·BC) = 1/6 of the rectangle's area.

ModerateCAT 2001

Euclid has a triangle in mind, its longest side has length 20 and another of its sides has length 10. Its area is 80. What is the exact length of its third side?

(1) √260
(2) √250
(3) √240
(4) √270

Show solution

(1) √260. With base 20 (= AB) and AC = 10, Area = ½·AB·CD = 80 ⇒ height CD = 8. Foot D: AD = √(10²−8²) = 6, so DB = 20−6 = 14. Third side CB = √(14²+8²) = √260.

HardCAT 2001

In ΔDEF shown below, points A, B, and C are taken on DE, DF and EF respectively such that EC = AC and CF = BC. If ∠D = 40°, then what is ∠ACB in degrees?

Triangle DEF with A on DE, B on DF, C on EF; EC=AC and CF=BC marked with tick marks

(1) 140
(2) 70
(3) 100
(4) None of these

Show solution

(3) 100. Let base angles at E and F be x and y; x + y + 40 = 180 ⇒ x + y = 140. ∠ACE = 180−2x, ∠BCF = 180−2y. So ∠ACB = 180 − (180−2x) − (180−2y) = 2(x+y) − 180 = 280 − 180 = 100°.

HardCAT 2002

The internal bisector of an angle A in a triangle ABC meets the side BC at point D. AB = 4, AC = 3 and angle A = 60°. Then what is the length of the bisector AD?

(1) 12√3/7
(2) 12√13/7
(3) 4√13/7
(4) 4√3/7

Show solution

(1) 12√3/7. Area(ABC) = Area(ABD) + Area(ACD): ½·4·3·sin60° = ½·4·AD·sin30° + ½·3·AD·sin30° ⇒ 12·sin60° = 7·AD·sin30° ⇒ AD = 12·sin60°/(7·sin30°) = 12√3/7.

ModerateCAT 2002

In the figure given below, find the distance PQ.

Right triangle BAC with right angle at apex A, AB = 15, AC = 20; altitude AD to base BC; two circles inscribed in triangles ABD and ADC with centres P and Q

(1) 7 m
(2) 4.5 m
(3) 10.5 m
(4) 6 m

Show solution

(1) 7 m. △ABC is right-angled at A with AB = 15, AC = 20, so BC = √(15²+20²) = 25. The altitude AD splits BC into BD = 9 and DC = 16. The two circles are inscribed in △ABD and △ADC; using A = r·s their inradii are 3 m and 4 m. PQ = (inradius of △ABD) + (inradius of △ADC) = 3 + 4 = 7 m.

ModerateCAT 2003

A vertical tower OP stands at the centre O of a square ABCD. Let h and b denote the lengths OP and AB respectively. Suppose ∠APB = 60°. Then the relationship between h and b can be expressed as ___.

(1) 2b² = h²
(2) 2h² = b²
(3) 3b² = 2h²
(4) 3h² = 2b²

Show solution

(2) 2h² = b². O is the centre, so AO = b/√2 and ∠AOP = 90°. With AP = PB and ∠APB = 60°, △APB is equilateral so AP = AB = b. In right △AOP: AO² + OP² = AP² ⇒ (b/√2)² + h² = b² ⇒ 2h² = b².

ModerateCAT 2003

In a triangle ABC, AB = 6, BC = 8 and AC = 10. A perpendicular dropped from B, meets the side AC at D. A circle of radius BD (with centre B) is drawn. If the circle cuts AB and BC at P and Q respectively, then AP : QC is equal to ___.

(1) 1 : 1
(2) 3 : 2
(3) 4 : 1
(4) 3 : 8

Show solution

(4) 3 : 8. 6-8-10 is right-angled at B. BD = (6·8)/10 = 4.8 = radius. So BP = BQ = 4.8. AP = AB − BP = 6 − 4.8 = 1.2; QC = BC − BQ = 8 − 4.8 = 3.2. AP : QC = 1.2 : 3.2 = 3 : 8.

ModerateCAT 2003

In the diagram given below, ∠ABD = ∠CDB = ∠PQD = 90°. If AB : CD = 3 : 1, the ratio of CD : PQ is ___.

AB and CD perpendicular to base BD with AB on the left and CD on the right; segments AD and BC cross at P; PQ is perpendicular to BD meeting it at Q

(1) 1 : 0.69
(2) 1 : 0.75
(3) 1 : 0.72
(4) None of these

Show solution

(2) 1 : 0.75. AB ∥ CD ∥ PQ. △CPD ∼ △BPA gives CP/PB = CD/AB = 1/3. Then △CBD ∼ △PBQ gives CD/PQ = CB/PB = (y+3y)/3y = 4/3 = 1 : 0.75.

ModerateCAT 2003

In the figure (not drawn to scale), P is a point on AB such that AP : PB = 4 : 3. PQ is parallel to AC and QD is parallel to CP. In ∠ARC, ∠ARC = 90°, and in ∆PQS, ∠PSQ = 90°. The length of QS is 6 cm. What is ratio AP : PD?

Triangle CAB with C at apex; P on AB, D on AB; PQ parallel to AC and QD parallel to CP; R on CA, S inside near P

(1) 10 : 3
(2) 2 : 1
(3) 7 : 3
(4) 8 : 3

Show solution

(3) 7 : 3. PQ ∥ AC ⇒ BP:AP = BQ:QC = 3:4. QD ∥ CP ⇒ BD:DP = BQ:QC = 3:4. Take AP = 4x, PB = 3x. Then PD = (4/7)·PB = 12x/7. So AP : PD = 4x : 12x/7 = 7 : 3.

HardCAT 2003

Direction: Consider three circular parks of equal size with centres at A₁, A₂, and A₃ respectively. The parks touch each other at the edge as shown in the figure (not drawn to scale). There are three paths formed by the triangles A₁, A₂, A₃, B₁, B₂, B₃, and C₁, C₂, C₃, as shown. Three sprinters A, B, and C begin running from points A₁, B₁ and C₁ respectively. Each sprinter traverses her respective triangular path clockwise and returns to her starting point.

Sprinter A traverses distances A₁A₂, A₂A₃, and A₃A₁ at average speeds of 20, 30 and 15, respectively. B traverses her entire path at a uniform speed of (10√3 + 20). C traverses distances C₁C₂, C₂C₃, and C₃C₁ at average speeds of (40/3)(√3 +1), (40/3)(√3 +1), and 120 respectively. All speeds are in the same unit. Where would B and C be respectively when A finishes her sprint?

Three equal circles touching each other, inscribed in a large triangle C₁C₂C₃; the centres A₁A₂A₃ form the inner triangle and B₁B₂B₃ the middle triangle

(1) B₁, C₁
(2) B₃, C₃
(3) B₁, C₃
(4) B₁, Somewhere between C₃ and C₁

Show solution

(3) B₁, C₃. A's total time = 2r/20 + 2r/30 + 2r/15 = 3r/10. In 3r/10, B covers (3r/10)(10√3+20) = 3r(√3+2), which is exactly B's full perimeter, so B returns to B₁. C covers C₁C₂ in [r(2√3+2)]/[(40/3)(√3+1)] = 3r/20, C₂C₃ in another 3r/20, total 3r/10, so C reaches C₃ exactly as A finishes. Hence B is at B₁ and C is at C₃.

ModerateCAT 2004

A father and his son are waiting at a bus stop in the evening. There is a lamp post behind them. The lamp post, the father and his son stand on the same straight line. The father observes that the shadows of his head and his son's head are incident at the same point on the ground. If the heights of the lamp post, the father and his son are 6 metres, 1.8 metres and 0.9 metres respectively, and the father is standing 2.1 metres away from the post, then how far (in metres) is the son standing from his father?

(1) 0.9
(2) 0.75
(3) 0.6
(4) 0.45

Show solution

(4) 0.45. Using similar triangles for both shadows with shadow length y for son and son-to-father distance n: ny/0.9 relation gives n = 2y, and (2.1 + n + y)/1.8 = 6/(n+y). Solving gives y = n/2 and n = 0.45 m.

ModerateCAT 2005

Consider the triangle ABC shown in the following figure where BC = 12 cm, DB = 9 cm, CD = 6 cm and ∠BCD = ∠BAC. What is the ratio of the perimeter of the triangle ADC to that of the triangle BDC?

Triangle ABC with D on AB; BD=9, DC=6, BC=12; angle BCD equals angle BAC

(1) 7/9
(2) 8/9
(3) 6/9
(4) 5/9

Show solution

(1) 7/9. ∠BCD = ∠BAC and ∠B common ⇒ △ABC ∼ △CBD. So AB/CB = BC/BD = AC/CD ⇒ AB = 16, AC = 8, AD = AB − BD = 7. Perimeter △ADC = 7+6+8 = 21; △BDC = 9+6+12 = 27. Ratio = 21/27 = 7/9.

ModerateCAT 2006

An equilateral triangle BPC is drawn inside a square ABCD. What is the value of the angle APD in degrees?

(1) 75
(2) 90
(3) 120
(4) 150

Show solution

(4) 150. △ABP and △DCP are isosceles with ∠ABP = ∠DCP = 90° − 60° = 30°, so ∠APB = ∠DPC = (180−30)/2 = 75°. ∠BPC = 60°. ∠APD = 360 − (75+75+60) = 150°.

ModerateCAT 2008

In a triangle ABC, the lengths of the sides AB and AC equal 17.5 cm and 9 cm respectively. Let D be a point on the line segment BC such that AD is perpendicular to BC. If AD = 3 cm, then what is the radius (in cm) of the circle circumscribing the triangle ABC?

(1) 17.05
(2) 27.85
(3) 22.45
(4) 26.25

Show solution

(4) 26.25. R = abc/(4·Area). Area = ½·AD·BC = ½·a·3. So R = (a·17.5·9)/(4·½·a·3) = (17.5·9)/(2·3) = 26.25 cm.

ModerateCAT 2017TITA

Let ABC be a right-angled triangle with BC as the hypotenuse. Lengths of AB and AC are 15 km and 20 km, respectively. The minimum possible time, in minutes, required to reach the hypotenuse from A at a speed of 30 km per hour is:

Show solution

24. Shortest distance from A to BC is the altitude. Hypotenuse = √(15²+20²) = 25. Altitude = (15·20)/25 = 12 km. Time = 12/30 × 60 = 24 minutes.

HardCAT 2017TITA

Let P be an interior point of a right-angled isosceles triangle ABC with hypotenuse AB. If the perpendicular distance of P from each of AB, BC, and CA is 4(√2 − 1) cm, then the area, in sq cm, of the triangle ABC is:

Show solution

16. P equidistant from all sides ⇒ P is the incentre, distance = inradius r = 4(√2−1). For a right isosceles triangle with legs a, r = a(√2−1)/… giving a = 8/√2; Area = ½·a² = ½·(8/√2)² = 16 sq cm.

ModerateCAT 2017

From a triangle ABC with sides of lengths 40 ft, 25 ft and 35 ft, a triangular portion GBC is cut off where G is the centroid of ABC. The area, in sq ft, of the remaining portion of triangle ABC is:

(1) 225√3
(2) 500/√3
(3) 275/√3
(4) 250/√3

Show solution

(2) 500/√3. s = 50, Area = √(50·10·25·15) = 250√3. The centroid triangle GBC is ⅓ of the whole, so remaining = ⅔·250√3 = 500√3/3 = 500/√3.

ModerateCAT 2018

Given an equilateral triangle T₁ with side 24 cm, a second triangle T₂ is formed by joining the midpoints of the sides of T₁. Then a third triangle T₃ is formed by joining the midpoints of the sides of T₂. If this process of forming triangles is continued, the sum of the areas, in sq cm, of infinitely many such triangles T₁, T₂, T₃, … will be

(1) 188√3
(2) 248√3
(3) 164√3
(4) 192√3

Show solution

(4) 192√3. Area T₁ = (√3/4)·24² = 144√3. Each next triangle is ¼ the previous. Sum = T₁/(1 − ¼) = (4/3)·144√3 = 192√3 sq cm.

ModerateCAT 2019

Let ABC be a right-angled triangle with hypotenuse BC of length 20 cm. If AP is perpendicular on BC, then the maximum possible length of AP, in cm, is

(1) 10
(2) 5
(3) 8√2
(4) 6√2

Show solution

(1) 10. AP is maximised when △ABC is isosceles. Then AB = AC = a with 20 = a√2 ⇒ a = 10√2. Area = ½·a² = ½·AP·BC ⇒ 10√2·10√2 = AP·20 ⇒ AP = 10.

ModerateCAT 2019

In a triangle ABC, medians AD and BE are perpendicular to each other, and have lengths 12 cm and 9 cm, respectively. Then, the area of triangle ABC, in sq cm, is

(1) 78
(2) 80
(3) 72
(4) 68

Show solution

(3) 72. Centroid divides medians 2:1: AG = 8, GD = 4, BG = 6, GE = 3. Area(ABE) = ½·BE·AG = ½·9·8 = 36. Area(ABC) = 2·Area(ABE) = 72 sq cm. (Alternatively, area = ⅔·product of perpendicular medians = ⅔·12·9 = 72.)

ModerateCAT 2020

From an interior point of an equilateral triangle, perpendiculars are drawn on all three sides. The sum of the lengths of the three perpendiculars is s. Then the area of the triangle is:

(1) s²/(2√3)
(2) 2s²/√3
(3) √3s²/2
(4) s²/√3

Show solution

(4) s²/√3. Sum of perpendiculars = height = s, so side a satisfies (√3/2)a = s ⇒ a = 2s/√3. Area = (√3/4)a² = (√3/4)(4s²/3) = s²/√3.

ModerateCAT 2021 · Slot 2TITA

Let D and E be points on sides AB and AC, respectively, of a triangle ABC, such that AD : BD = 2 : 1 and AE : CE = 2 : 3. If the area of the triangle ADE is 8 sq cm, then the area of the triangle ABC, in sq cm, is

Show solution

30. Draw BF ∥ DE; △ADE ∼ △ABF with AD:AB = 2:3 ⇒ Area(ABF) = 8·(3/2)² = 18. △ABF and △ABC share height from B: ratio = AF:AC = (AE/AF... ) gives Area(ABF)/Area(ABC) = 3/5 ⇒ Area(ABC) = 18·5/3 = 30 sq cm.

HardCAT 2021 · Slot 3

In a triangle ABC, ∠BCA = 50°. D and E are points on AB and AC, respectively, such that AD = DE. If F is a point on BC such that BD = DF, then ∠FDE, in degrees, is equal to

(1) 96
(2) 100
(3) 80
(4) 72

Show solution

(3) 80. Let ∠DAE = x, ∠DBF = y. AD = DE ⇒ ∠ADE = 180−2x; BD = DF ⇒ ∠BDF = 180−2y. In △ABC: x + y + 50 = 180 ⇒ x+y = 130. On line AB: (180−2x) + ∠FDE + (180−2y) = 180 ⇒ ∠FDE = 2(x+y) − 180 = 260 − 180 = 80°.

ModerateCAT 2022 · Slot 1

The length of each side of an equilateral triangle ABC is 3 cm. Let D be a point on BC such that the area of triangle ADC is half the area of triangle ABD. Then, the length of AD, in cm, is:

(1) √7
(2) √6
(3) √8
(4) √5

Show solution

(1) √7. Areas are in ratio of bases (same height), so DC = ½·BD ⇒ BC = 3·BD = 3 ⇒ BD = 1. In △ABD with ∠B = 60°: cos60° = (3² + 1² − AD²)/(2·3·1) ⇒ ½·6 = 10 − AD² ⇒ AD² = 7 ⇒ AD = √7.

HardCAT 2022 · Slot 2

In triangle ABC, altitudes AD and BE are drawn to the corresponding bases. If ∠BAC = 45° and ∠ABC = θ, then AD/BE equals:

(1) √2 cosθ
(2) 1
(3) √2 sinθ
(4) (sinθ + cosθ)/√2

Show solution

(3) √2·sinθ. In right △ABE, ∠AEB = 90°, ∠BAC = 45° ⇒ AE = BE = x and AB = x√2. In △ABD, sinθ = AD/AB = AD/(x√2). So AD = x√2·sinθ, giving AD/BE = AD/x = √2·sinθ.

HardCAT 2022 · Slot 3TITA

Suppose the medians BD and CE of a triangle ABC intersect at a point O. If the area of triangle ABC is 108 sq cm., then, the area of the triangle EOD, in sq cm., is:

Show solution

9. Median CE halves ABC: Area(ABD) on median split... using centroid 2:1, Area(ABD) = 54, △EBD = ½·54 = 27, and △EOD = ⅓·27 = 9 sq cm.

ModerateCAT 2023 · Slot 2TITA

In a right-angled triangle ΔABC, the altitude AB is 5 cm, and the base BC is 12 cm. P and Q are two points on BC such that the areas of ΔABP, ΔABQ and ΔABC are in arithmetic progression. If the area of ΔABC is 1.5 times the area of ΔABP, the length of PQ, in cm, is

Show solution

2. Area(ABC) = ½·5·12 = 30. Area(ABP) = 30/1.5 = 20 ⇒ BP = (20·2)/5 = 8. The AP middle term = (20+30)/2 = 25 ⇒ BQ = (25·2)/5 = 10. PQ = BQ − BP = 10 − 8 = 2 cm.

HardCAT 2023 · Slot 2

A triangle is drawn with its vertices on the circle C such that one of its sides is a diameter of C and the other two sides have their lengths in the ratio a : b. If the radius of the circle is r, then the area of the triangle is

(1) 2abr²/(a²+b²)
(2) 4abr²/(a²+b²)
(3) abr²/(a²+b²)
(4) abr²/[2(a²+b²)]

Show solution

(1) 2abr²/(a²+b²). Angle in semicircle = 90°. Legs PQ:QR = a:b with PQ²+QR² = (2r)². So PQ = 2rb/√(a²+b²), QR = 2ra/√(a²+b²). Area = ½·PQ·QR = ½·(4r²ab)/(a²+b²) = 2abr²/(a²+b²).

HardCAT 2023 · Slot 3

Let ΔABC be an isosceles triangle such that AB and AC are of equal length. AD is the altitude from A on BC and BE is the altitude from B on AC. If AD and BE intersect at O such that ∠AOB = 105°, then AD/BE equals

(1) 2cos15°
(2) sin15°
(3) 2sin15°
(4) cos15°

Show solution

(1) 2cos15°. Angle chasing gives ∠BAC = 30°, base angles 75°. In △ABE: cos75°... AE/AB = ½ so AB = 2BE. In △ADC: cos15° = AD/AC = AD/AB ⇒ AD = AB·cos15° = 2BE·cos15° ⇒ AD/BE = 2cos15°.

CAT 2024 & 2025, recent

ModerateCAT 2024 · Slot 1 TITA

ABCD is a rectangle with sides AB = 56 cm and BC = 45 cm, and E is the midpoint of side CD. Then, the length, in cm, of radius of incircle of △ADE is

Show solution

10. E is the midpoint of CD, so DE = ½·56 = 28 and AD = 45. △ADE is right-angled at D, hypotenuse AE = √(28² + 45²) = √2809 = 53. Inradius of a right triangle = (leg + leg − hyp)/2 = (28 + 45 − 53)/2 = 10.

EasyCAT 2024 · Slot 2 TITA

The coordinates of the three vertices of a triangle are: (1, 2), (7, 2), and (1, 10). Then the radius of the incircle of the triangle is

Show solution

2. Sides: (1,2)-(7,2) = 6, (1,2)-(1,10) = 8, (7,2)-(1,10) = 10. Since 6² + 8² = 10², it is right-angled. Inradius = (6 + 8 − 10)/2 = 2.

ModerateCAT 2024 · Slot 2

ABCD is a trapezium in which AB is parallel to CD. The sides AD and BC when extended, intersect at point E. If AB = 2 cm, CD = 1 cm, and perimeter of ABCD is 6 cm, then the perimeter, in cm, of △AEB is

(A) 8
(B) 10
(C) 9
(D) 7

Show solution

(A) 8. △ECD ∼ △EAB with ratio CD : AB = 1 : 2, so D and C are midpoints of EA and EB. Perimeter of ABCD = AB + CD + AD + BC = 6, so AD + BC = 6 − 2 − 1 = 3. Since AD = DE and BC = CE, EA + EB = 2(AD + BC) = 6. Perimeter of △AEB = EA + EB + AB = 6 + 2 = 8.

ModerateCAT 2024 · Slot 3 TITA

The midpoints of sides AB, BC, and AC in ΔABC are M, N, and P, respectively. The medians drawn from A, B, and C intersect the line segments MP, MN and NP at X, Y, and Z, respectively. If the area of ΔABC is 1440 sq cm, then the area, in sq cm, of △XYZ is

Show solution

90. The medial triangle MNP has ¼ the area of ABC, so [MNP] = 360. The medians of ABC meet MP, MN, NP at the midpoints of those sides, so X, Y, Z are the midpoints of the sides of △MNP. Hence △XYZ is the medial triangle of MNP and [XYZ] = ¼·[MNP] = ¼·360 = 90 (= 1440/16).

HardCAT 2025 · Slot 2

In a △ABC, points D and E are on the sides BC and AC, respectively. BE and AD intersect at point T such that AD : AT = 4 : 3, and BE : BT = 5 : 4. Point F lies on AC such that DF is parallel to BE. Then, BD : CD is

(A) 9 : 4
(B) 15 : 4
(C) 11 : 4
(D) 7 : 4

Show solution

(C) 11 : 4. From AD : AT = 4 : 3, TD : AT = 1 : 3; from BE : BT = 5 : 4, TE : BT = 1 : 4. DF ∥ BE in △ between the cevians sets up the section ratios. Applying mass points: on AD, AT : TD = 3 : 1; on BE, BT : TE = 4 : 1. Working the masses through F (DF ∥ BE) gives BD : CD = 11 : 4.

ModerateCAT 2025 · Slot 3 TITA

A triangle ABC is formed with AB = AC = 50 cm and BC = 80 cm. Then, the sum of the lengths, in cm, of all three altitudes of the triangle ABC is

Show solution

126. Altitude from A to BC: √(50² − 40²) = √900 = 30, so area = ½·80·30 = 1200. Altitude to BC = 2·1200/80 = 30; altitude to AB = altitude to AC = 2·1200/50 = 48 each. Sum = 30 + 48 + 48 = 126.

HardCAT 2025 · Slot 3

In a triangle ABC, AB = AC = 12 cm and D is a point on side BC such that AD = 8 cm. If AD is extended to point E such that ∠ACB = ∠AEB, then length in cm of AE is:

(A) 16
(B) 18
(C) 14
(D) 20

Show solution

(B) 18. Since AB = AC, ∠ABC = ∠ACB; with ∠ACB = ∠AEB, points A, B, C, E are concyclic. By power of a point (and the resulting similar triangles), AB² = AD·AE, so 12² = 8·AE → AE = 144/8 = 18.

Circles · 38 CAT PYQs

Circles

Circles. Chord/tangent power relations, angles in a circle, cyclic quadrilaterals, inscribed/intersecting/touching circles and segment areas.

ModerateCAT 1998

Three circles, each of radius 20, have centres at P, Q and R. Further, AB = 5, CD = 10 and EF = 12. What is the perimeter of ∆PQR?

Three equal circles with centres P, Q, R and triangle PQR; overlap segments AB, CD, EF

(1) 120
(2) 66
(3) 93
(4) 87

Show solution

(3) 93. PR = (20−5)+5+(20−5) = 35; QR = (20−10)+10+(20−10) = 30; PQ = (20−12)+12+(20−12) = 28. Perimeter = 35+30+28 = 93. (Shortcut: 6·20 − (5+10+12) = 120 − 27 = 93.)

ModerateCAT 1999

The figure below shows two concentric circles with centre O. PQRS is a square inscribed in the outer circle. It also circumscribes the inner circle, touching it at points B, C, D and A. What is the ratio of the perimeter of the outer circle to that of polygon ABCD?

Square PQRS inscribed in outer circle, circumscribing inner circle, touching it at A, B, C, D

(1) π/4
(2) 3π/2
(3) π/2
(4) π

Show solution

(3) π/2. Outer radius = x = OQ; perimeter = 2πx. ABCD is a square of side x (its diagonal equals the square's side relationship), perimeter 4x. Ratio = 2πx/4x = π/2.

HardCAT 2000

Consider a circle with unit radius. There are seven adjacent sectors, S₁, S₂, S₃, …, S₇, in the circle such that their total area is 1/8 of the area of the circle. Further, the area of the j^th sector is twice that of the (j − 1)^th sector, for j = 2, …, 7. What is the angle, in radians, subtended by the arc of S₁ at the centre of the circle?

(1) π/508
(2) π/2040
(3) π/1016
(4) π/1524

Show solution

(3) π/1016. Sector areas x, 2x, …, 64x sum to 127x = (1/8)·(πr²). So total circle = 127x·8 = 1016x. Hence angle of S₁ = π/1016 rad.

HardCAT 2001

A certain city has a circular wall around it, and this wall has four gates pointing north, south, east and west. A house stands outside the city, three km north of the north gate, and it can just be seen from a point nine km east of the south gate. What is the diameter of the wall what surrounds the city?

(1) 6 km
(2) 9 km
(3) 12 km
(4) None of these

Show solution

(2) 9 km. With radius r, set up Pythagoras for the sightline tangent to the wall: solving the two relations gives r = 4.5 km, so diameter = 2r = 9 km.

ModerateCAT 2002

There is a common chord of 2 circles with radius 15 and 20. The distance between the two centres is 25. The length of the chord is

(1) 48
(2) 24
(3) 36
(4) 28

Show solution

(2) 24. Let foot of chord split the line of centres at distance x from the radius-15 centre: 20² = x² + AP², 15² = (25−x)² + AP². Solving gives x = 16, AP = 12. Chord = 2·AP = 24.

HardCAT 2003

There are two concentric circles such that the area of the outer circle is four times the area of the inner circle. Let A, B and C be three distinct points on the perimeter of the outer circle such that AB and AC are tangents to the inner circle. If the area of the outer circle is 12 square centimetres then the area (in square centimetres) of the triangle ABC would be___.

(1) π√12
(2) 9/π
(3) 9√3/π
(4) 6√3/π

Show solution

(3) 9√3/π. R = 2r. The tangents make △ABC equilateral. AB = 2√(R²−r²) = 2√3·r, Area = (√3/4)·(2√3 r)² = 3√3·r². With 4πr² = 12 ⇒ r² = 3/π. Area = 3√3·3/π = 9√3/π.

ModerateCAT 2003

In the figure given below, AB is the chord of a circle with centre O. AB is extended to C such that BC = OB. The straight line CO is produced to meet the circle at D. If ∠ACD = y° and ∠AOD = x° such that x = ky, then the value of k is___.

Chord AB of circle centre O extended to C with BC = OB; CO produced to D on the circle

(1) 3
(2) 2
(3) 1
(4) None of these

Show solution

(1) k = 3. BC = OB ⇒ ∠BOC = ∠BCO = y. Exterior angle ∠OBA = 2y, and OB = OA ⇒ ∠OAB = 2y. Chasing the angles at O: x = 3y, so k = 3.

HardCAT 2003

In the figure below (not drawn to scale), rectangle ABCD is inscribed in the circle with centre at O. The length of side AB is greater than that of side BC. The ratio of the area of the circle to the area of the rectangle ABCD is π : √3. The line segment DE intersects AB at E such that ∠ODC = ∠ADE. What is the ratio of AE : AD?

Rectangle ABCD inscribed in a circle centre O; segment DE meets AB at E

(1) 1 : √3
(2) 1 : √2
(3) 1 : 2√3
(4) 1 : 2

Show solution

(1) 1 : √3. Let sides a, b. πr²/(ab) = π/√3 ⇒ ab = √3·r² with 4r² = a²+b². tan in △DAE and △DBC: AE/AD = tan(∠ADE) = b/a relations resolve to AE : AD = 1 : √3.

HardCAT 2003

In the figure given below (not drawn to scale), A, B and C are three points on a circle with centre O. The chord BA is extended to a point T such that CT becomes a tangent to the circle at point C. If ∠ATC = 30° and ∠ACT = 50°, then the angle ∠BOA is

Circle centre O with points A, B, C; chord BA extended to T where CT is tangent at C; angles 50° at C and 30° at T

(1) 100°
(2) 150°
(3) 80°
(4) Cannot be determined

Show solution

(1) 100°. Alternate segment: ∠ABC = ∠ACT = 50°. ∠CAT exterior to △ABC = ∠ABC + ∠BCA = 100° ⇒ ∠BCA = 50°. ∠BOA = 2·∠BCA = 100° (central = 2× inscribed).

ModerateCAT 2004

In the adjoining figure, chord ED is parallel to the diameter AC of the circle. If ∠CBE = 65°, then what is the value of ∠DEC?

Circle with diameter AC, chord ED parallel to AC, point B on the circle

(1) 35°
(2) 55°
(3) 45°
(4) 25°

Show solution

(4) 25°. ∠AEC = 90° (angle in semicircle). ∠EAC = ∠EBC = 65° (same arc). In △AEC: 65 + 90 + ∠DEC = 180 ⇒ ∠DEC = 25°.

HardCAT 2004

On a semicircle with diameter AD, chord BC is parallel to the diameter. Further, each of the chords AB and CD has length 2, while AD has length 8. What is the length of BC?

Semicircle with diameter AD; chord BC parallel to AD, with AB and CD drawn

(1) 7.5
(2) 7
(3) 7.75
(4) None of these

Show solution

(2) 7. Centre O, OB = OC = 4. With perpendiculars BE, CF and EO = OF = x: from 4 = a² + (4−x)² and 16 = a² + x², solving gives x = 3.5, so BC = 8 − 2x = 7.

HardCAT 2004

Let C be a circle with centre P₀ and AB be a diameter of C. Suppose P₁ is the mid-point of the line segment P₀B, P₂ is the mid-point of the line segment P₁B and so on. Let C₁, C₂, C₃, …… be circles with diameters P₀P₁, P₁P₂, P₂P₃ … respectively. Suppose the circles C₁, C₂, C₃, …. are all shaded. The ratio of the area of the unshaded portion of C to that of the original circle C is

(1) 8 : 9
(2) 9 : 10
(3) 10 : 11
(4) 11 : 12

Show solution

(4) 11 : 12. Shaded areas form a geometric series in (1/4): shaded fraction = (1/4)/(1−1/4) of the appropriately scaled radius² ⇒ unshaded = 11/12 of C.

HardCAT 2004

A circle with radius 2 is placed against a right angle. Another smaller circle is also placed as shown in the adjoining figure. What is the radius of the smaller circle?

Circle of radius 2 in a right-angle corner with a smaller circle tucked into the corner

(1) 3 − 2√2
(2) 4 − 2√2
(3) 7 − 4√2
(4) 6 − 4√2

Show solution

(4) 6 − 4√2. Diagonal of corner square for the big circle = 2√2 + 2. For small circle radius r: 2√2 = 2 + r + r√2 ⇒ r = 2(√2−1)/(√2+1) = 6 − 4√2.

ModerateCAT 2005

What is the distance in cm between two parallel chords of lengths 32 cm and 24 cm in a circle of radius 20 cm?

(1) 1 or 7
(2) 2 or 14
(3) 3 or 21
(4) 4 or 28

Show solution

(4) 4 or 28. Distances from centre: for 32-chord, √(20²−16²) = 12; for 24-chord, √(20²−12²) = 16. Same side: 16 − 12 = 4; opposite sides: 16 + 12 = 28.

ModerateCAT 2005

Four points A, B, C and D lie on a straight line in the X-Y plane, such that AB = BC = CD, and the length of AB is 1 metre. An ant at A wants to reach a sugar particle at D. But there are insect repellents kept at points B and C. The ant would not go within one metre of any insect repellent. The minimum distance in metres the ant must traverse to reach the sugar particle is

(1) 3√2
(2) 1 + π
(3) 4π/3
(4) 5

Show solution

(2) 1 + π. The ant skirts each repellent along a quarter-circle of radius 1. Path = straight HG (= 1) + two quarter arcs (each π/2). Total = 1 + π.

HardCAT 2005

In the following figure, the diameter of the circle is 3 cm. AB and MN are two diameters such that MN is perpendicular to AB. In addition, CG is perpendicular to AB such that AE : EB = 1 : 2, and DF is perpendicular to MN such that NL : LM = 1 : 2. The length of DH in cm is

Circle with perpendicular diameters AB and MN; CG ⊥ AB and DF ⊥ MN intersecting at E, O, H, L

(1) 2√2 − 1
(2) (2√2 − 1)/2
(3) (3√2 − 1)/2
(4) (2√2 − 1)/3

Show solution

(2) (2√2 − 1)/2. Radius 1.5; AE = 1, OE = 0.5. EOLG forms a square so HL = 0.5. In △OLD: DL = √(1.5² − 0.5²) = √2. DH = DL − HL = √2 − 0.5 = (2√2 − 1)/2.

ModerateCAT 2005

Two identical circles intersect so that their centres, and the points at which they intersect, form a square of side 1 cm. The area in sq. cm of the portion that is common to the two circles is:

(1) π/4
(2) π/2 − 1
(3) π/5
(4) √2 − 1

Show solution

(2) π/2 − 1. Each radius = 1; ∠MPN = 90°. Common area = 2(sector PMN − △PMN) = 2(¼·π·1² − ½·1·1) = 2(π/4 − ½) = π/2 − 1.

ModerateCAT 2006

A semi-circle is drawn with AB as its diameter. From C, a point on AB, a line perpendicular to AB is drawn meeting the circumference of the semi-circle at D. Given that AC = 2 cm and CD = 6 cm, the area of the semi-circle (in sq. cm.) will be:

(1) 32π
(2) 50π
(3) 40.5π
(4) 81π

Show solution

(2) 50π. OC = r − 2, CD = 6, OD = r. (r−2)² + 6² = r² ⇒ r² + 4 − 4r + 36 = r² ⇒ 4r = 40 ⇒ r = 10. Semicircle area = ½·π·10² = 50π.

HardCAT 2005

P, Q, S, and R are points on the circumference of a circle of radius r, such that PQR is an equilateral triangle and PS is a diameter of the circle. What is the perimeter of the quadrilateral PQSR?

(1) 2r(1 + √3)
(2) 2r(2 + √3)
(3) r(1 + √5)
(4) 2r + √3

Show solution

(1) 2r(1 + √3). Side of equilateral PQR = r√3. PS = 2r. QS = RS = r (angle in semicircle). Perimeter = PQ + QS + RS + PR = r√3 + r + r + r√3 = 2r√3 + 2r = 2r(1+√3).

HardCAT 2007

Two circles with centres P and Q cut each other at two distinct points A and B. The circles have the same radii and neither P nor Q falls within the intersection of the circles. What is the smallest range that includes all possible values of the angle AQP in degrees?

(1) Between 0 and 90
(2) Between 0 and 30
(3) Between 0 and 60
(4) Between 0 and 75

Show solution

(3) Between 0 and 60. If P, Q lie on each other's circumference, ∠AQP = 60°; as they move apart it decreases toward 0° (but A, Q, P can't be collinear). So 0° < ∠AQP < 60°.

HardCAT 2008

Two circles, both of radii 1 cm, intersect such that the circumference of each one passes through the centre of the circle of the other. What is the area (in sq cm) of the intersecting region?

(1) π/3 − √3/4
(2) 3π/3 + √3/2
(3) 4π/3 − √3/2
(4) 2π/3 − √3/2

Show solution

(4) 2π/3 − √3/2. The lens = 2 × (circular segment of 120°) = 2[(120/360)π·1² − ½·1²·sin120°] = 2[π/3 − √3/4] = 2π/3 − √3/2.

ModerateCAT 2017TITA

ABCD is a quadrilateral inscribed in a circle with centre O. If ∠COD = 120 degrees and ∠BAC = 30 degrees, then the value of ∠BCD (in degrees) is

Show solution

90. ∠CAD = ½·∠COD = 60°. ∠BAD = ∠BAC + ∠CAD = 30 + 60 = 90°. Cyclic quadrilateral ⇒ ∠BAD + ∠BCD = 180° ⇒ ∠BCD = 90°.

HardCAT 2017

Let ABC be a right-angled isosceles triangle with hypotenuse BC. Let BQC be a semi-circle, away from A, with diameter BC. Let BPC be an arc of a circle centered at A and lying between BC and BQC. If AB has length 6 cm then the area, in sq cm, of the region enclosed by BPC and BQC is

(1) 9π − 18
(2) 18
(3) 9π
(4) 9

Show solution

(2) 18. BC = 6√2, semicircle BQC area = ½·π·(3√2)² = 9π. Quadrant BPC (radius 6) = ¼·π·6² = 9π. △ABC = ½·6·6 = 18. Required = semicircle − quadrant + triangle = 9π − 9π + 18 = 18.

ModerateCAT 2018

In a circle, two parallel chords on the same side of a diameter have lengths 4 cm and 6 cm. If the distance between these chords is 1 cm, then the radius of the circle, in cm, is

(1) √13
(2) √14
(3) √11
(4) √12

Show solution

(1) √13. Let the 6-cm chord be x from centre: r² = x² + 3² = (x+1)² + 2². Solving x = 2 ⇒ r = √(4+9) = √13 cm.

HardCAT 2018

In a circle with center O and radius 1 cm, an arc AB makes an angle 60 degrees at O. Let R be the region bounded by the radii OA, OB and the arc AB. If C and D are two points on OA and OB, respectively, such that OC = OD and the area of triangle OCD is half that of R, then the length of OC, in cm, is

(1) (π/4√3)^½
(2) (π/6)^½
(3) (π/3√3)^½
(4) (π/4)^½

Show solution

(3) (π/(3√3))^½. R = (60/360)·π·1² = π/6. △OCD area = ½·OC²·sin60° = ½·R = π/12. So OC²·(√3/2) = π/6 ⇒ OC² = π/(3√3) ⇒ OC = √(π/(3√3)).

HardCAT 2019

AB is a diameter of a circle of radius 5 cm. Let P and Q be two points on the circle so that the length of PB is 6 cm, and the length of AP is twice that of AQ. Then the length, in cm, of QB is nearest to

(1) 9.3
(2) 7.8
(3) 9.1
(4) 8.5

Show solution

(3) 9.1. ∠APB = ∠AQB = 90° (semicircle). AP = √(10²−6²) = 8 ⇒ AQ = 4. QB = √(10²−4²) = √84 ≈ 9.1 cm.

HardCAT 2019

In a circle of radius 11 cm, CD is a diameter and AB is a chord of length 20.5 cm. If AB and CD intersect at a point E inside the circle and CE has length 7 cm, then the difference of the lengths of BE and AE, in cm, is

(1) 2.5
(2) 1.5
(3) 3.5
(4) 0.5

Show solution

(4) 0.5. CE·DE = AE·BE (intersecting chords). CE = 7, DE = 22 − 7 = 15, so 7·15 = AE·(20.5 − AE) ⇒ 2·AE² − 41·AE + 210 = 0 ⇒ AE = 10 or 10.5. Difference = 10.5 − 10 = 0.5.

HardCAT 2019

Two circles, each of radius 4 cm, touch externally. Each of these two circles is touched externally by a third circle. If these three circles have a common tangent, then the radius of the third circle, in cm, is

(1) 1/√2
(2) π/3
(3) √2
(4) 1

Show solution

(4) 1. For circle radius r tucked between two equal circles (radius 4) on a common tangent line: tangent-length relation 4² + (4−r)² = (4+r)² ⇒ 16 = 16r ⇒ r = 1 cm.

ModerateCAT 2020

A circle is inscribed in a rhombus with diagonals 12 cm and 16 cm. The ratio of the area of circle to the area of rhombus is:

(1) 3π/25
(2) 2π/15
(3) 6π/25
(4) 5π/18

Show solution

(3) 6π/25. Side = √(6²+8²) = 10. Inradius = (area)/(semiperimeter)·... = (½·12·16)/… giving r = (product of half-diagonals)/side = (6·8)/10 = 4.8. Circle area = π·4.8². Rhombus = ½·12·16 = 96. Ratio = π·23.04/96 = 6π/25.

ModerateCAT 2020

Let C be a circle of radius 5 metres having center at O. Let PQ be a chord of C that passes through points A and B where A is located 4 metres north of O and B is located 3 meters east of O. Then, the length of PQ, in meters, is nearest to:

(1) 7.2
(2) 6.6
(3) 8.8
(4) 7.8

Show solution

(3) 8.8. AB = √(4²+3²) = 5. Perpendicular distance OE from O to chord = (OA·OB)/AB = (4·3)/5 = 2.4. Half-chord = √(5² − 2.4²) = 4.4. PQ = 2·4.4 = 8.8 m.

ModerateCAT 2020TITA

Let C1 and C2 be concentric circles such that the diameter of C1 is 2 cm longer than that of C2. If a chord of C1 has length 6 cm and is a tangent to C2, then the diameter, in cm, of C1 is:

Show solution

10. Let radius of C₂ = r, C₁ = r + 1. Half-chord 3 is tangent to C₂, so r, 3, (r+1) form a right triangle: r² + 3² = (r+1)² ⇒ r = 4. C₁ radius = 5 ⇒ diameter = 10 cm.

HardCAT 2021 · Slot 1TITA

A circle of diameter 8 inches is inscribed in a triangle ABC where ∠ABC = 90°. If BC = 10 inches, then the area of the triangle in square inches is

Show solution

120. Inradius r = 4. BP = BQ = 4 (tangents); CQ = 10 − 4 = 6 = CR. Let AP = AR = x. In right △ABC: (x+4)² + 10² = (x+6)² ⇒ x = 20. Sides 24, 10, 26; s = 30. Area = r·s = 4·30 = 120 sq in.

HardCAT 2022 · Slot 3

In a triangle ABC, AB = AC = 8 cm. A circle drawn with BC as diameter passes through A. Another circle drawn with centre at A passes through B and C. Then the area, in sq cm, of the overlapping region between the two circles is:

(1) 16(π − 1)
(2) 32π
(3) 32(π − 1)
(4) 16π

Show solution

(3) 32(π − 1). Angle in semicircle ⇒ ∠BAC = 90°, so BC = 8√2, radius of C₁ = 4√2; A-centred circle radius = 8. Overlap = semicircle of C₁ + minor segment = ½·π·(4√2)² + [¼·π·8² − ½·8·8] = 16π + (16π − 32) = 32π − 32 = 32(π−1).

HardCAT 2023 · Slot 1

Let C be the circle x² + y² + 4x − 6y − 3 = 0 and L be the locus of the point of intersection of a pair of tangents to C with the angle between the two tangents equal to 60°. Then, the point at which L touches the line x = 6 is

(1) (6, 6)
(2) (6, 4)
(3) (6, 8)
(4) (6, 3)

Show solution

(4) (6, 3). Circle x² + y² + 4x − 6y − 3 = 0 has centre C(−2, 3) and radius √(4 + 9 + 3) = 4. For a point P from which the pair of tangents include 60°, sin30° = r/CP ⇒ CP = 2r = 8, so L is the concentric circle of radius 8: (x + 2)² + (y − 3)² = 64. Putting x = 6: (8)² + (y − 3)² = 64 ⇒ (y − 3)² = 0 ⇒ y = 3. Hence L touches x = 6 at (6, 3).

CAT 2024 & 2025, recent

HardCAT 2024 · Slot 2

Three circles of equal radii touch (but not cross) each other externally. Two other circles, X and Y, are drawn such that both touch (but not cross) each of the three previous circles. If the radius of X is more than that of Y, the ratio of the radii of X and Y is

(A) (7 + 4√3) : 1
(B) (4 + 2√3) : 1
(C) (4 + √3) : 1
(D) (2 + √3) : 1

Show solution

(A) (7 + 4√3) : 1. Let the three equal circles have radius 1; their centres form an equilateral triangle of side 2, whose circumradius (distance from the common centre to each of the three centres) is 2/√3. The big circle X encloses all three, so its radius = 2/√3 + 1; the small circle Y nestles in the middle, so its radius = 2/√3 − 1. Ratio = (2/√3 + 1)/(2/√3 − 1) = (2 + √3)/(2 − √3) = (2 + √3)² = 7 + 4√3, i.e. (7 + 4√3) : 1.

HardCAT 2025 · Slot 1

In a circle with center C and radius 6√2 cm, PQ and SR are two parallel chords separated by one of the diameters. If ∠PQC = 45°, and the ratio of the perpendicular distance of PQ and SR from C is 3 : 2, then the area, in sq. cm, of the quadrilateral PQRS is

(A) 4(3 + √14)
(B) 20(3√2 + √7)
(C) 20(3 + √14)
(D) 4(3√2 + √7)

Show solution

(C) 20(3 + √14). CP = CQ = 6√2 (radii), so △CPQ is isosceles; ∠PQC = 45° forces ∠PCQ = 90°, and the foot of the perpendicular from C bisects PQ. Distance of PQ from C = 6√2·sin45° = 6, and half-chord = 6, so PQ = 12. The two distances are in ratio 3 : 2, so SR is 4 from C; half of SR = √[(6√2)² − 4²] = √(72 − 16) = √56 = 2√14, so SR = 4√14. The chords lie on opposite sides of the diameter, so PQRS is a trapezium of height 6 + 4 = 10. Area = ½(PQ + SR)·height = ½(12 + 4√14)·10 = 20(3 + √14).

ModerateCAT 2025 · Slot 2 TITA

Two tangents drawn from a point P touch a circle with center O at points Q and R. Points A and B lie on PQ and PR, respectively, such that AB is also a tangent to the same circle. If ∠AOB = 50°, then ∠APB, in degrees, equals

Show solution

80. Let AB touch the circle at T. From the equal tangents, OA bisects ∠QOT and OB bisects ∠TOR, so ∠QOR = 2·∠AOB = 2·50° = 100°. In quadrilateral PQOR, ∠OQP = ∠ORP = 90° (radius ⊥ tangent), so ∠QPR = 360° − 90° − 90° − 100° = 80°. Hence ∠APB = ∠QPR = 80°.

HardCAT 2025 · Slot 3

ABCD is a trapezium in which AB is parallel to DC, AD is perpendicular to AB, and AB = 3DC. If a circle inscribed in the trapezium touching all the sides has a radius of 3 cm, then the area, in sq. cm, of the trapezium is

(A) 48
(B) 30√3
(C) 36√2
(D) 54

Show solution

(A) 48. AD ⊥ both parallel sides, so AD = height = 2r = 6. Let DC = c, AB = 3c. The slant side BC = √[AD² + (AB − DC)²] = √[36 + (2c)²]. A tangential quadrilateral has AB + DC = AD + BC, so 3c + c = 6 + √(36 + 4c²) ⇒ 4c − 6 = √(36 + 4c²). Squaring: 16c² − 48c + 36 = 36 + 4c² ⇒ 12c² = 48c ⇒ c = 4. Then DC = 4, AB = 12. Area = ½(AB + DC)·AD = ½(12 + 4)·6 = 48.

Quadrilaterals & Polygons · 35 CAT PYQs

Quadrilaterals & Polygons

Quadrilaterals & Polygons. Rectangles, parallelograms, rhombi, trapeziums; interior/exterior angles, diagonals and the regular hexagon.

HardCAT 1999

There is a circle of radius 1 cm. Each member of a sequence of regular polygons S₁(n), n = 4, 5, 6, …, where n is the number of sides of the polygon, is circumscribing the circle; and each member of the sequence of regular polygons S₂(n), n = 4, 5, 6, …, where n is the number of sides of the polygon, is inscribed in the circle. Let L₁(n) and L₂(n) denote the perimeters of the corresponding polygons of S₁(n) and S₂(n), then {L₁(13) + 2π}/L₂(17) is

(1) greater than π/4 and less than 1
(2) greater than 1 and less than 2
(3) greater than 2
(4) less than π/4

Show solution

(3) greater than 2. A circumscribing polygon's perimeter > the circle's circumference; an inscribed one's < circumference. With r = 1, circumference = 2π: so L₁(13) > 2π and L₂(17) < 2π. Hence {L₁(13) + 2π}/L₂(17) > (2π + 2π)/2π = 2, i.e. greater than 2.

HardCAT 2000

ABCDEFGH is a regular octagon. A and E are opposite vertices of the octagon. A frog starts jumping from vertex to vertex, beginning from A. From any vertex of the octagon except E, it may jump to either of the two adjacent vertices. When it reaches E, the frog stops and stays there. Let aₙ be the number of distinct paths of exactly n jumps ending in E. Then what is the value of a₍₂n−1₎?

(1) 0
(2) 4
(3) 2n − 1
(4) Cannot be determined

Show solution

(1) 0. To go from A to the opposite vertex E requires an even number of jumps (A and E are 4 steps apart either way around the octagon). An odd number of jumps (2n − 1) can never end at E, so a₍₂n−1₎ = 0.

ModerateCAT 2001

A square, whose side is 2 metres, has its corners cut away so as to form an octagon with all sides equal. Then the length of each side of the octagon, in metres is

(1) √2/(√2 + 1)
(2) 2/(√2 + 1)
(3) 2/(√2 − 1)
(4) √2/(√2 − 1)

Show solution

(2) 2/(√2 + 1). Let each corner cut have legs of length x, so the diagonal cut = x√2. The square's side splits as x + (octagon side) + x = 2, with octagon side = x√2. Then 2x + x√2 = 2 ⇒ x = 2/(2 + √2). Octagon side = x√2 = 2√2/(2 + √2) = 2/(√2 + 1) = 2(√2 − 1) m.

ModerateCAT 2002

In the following figure, the area of the isosceles right triangle ABE is 7 sq. cm. If EC = 3BE, then the area of rectangle ABCD is (in sq. cm.)

Rectangle ABCD with A top-left, D top-right, B bottom-left, C bottom-right; E on side BC, with right isosceles triangle ABE shaded

(1) 64
(2) 82
(3) 26
(4) 56

Show solution

(4) 56. △ABE is right-angled and isosceles, so AB = BE and area = ½·BE·AB = ½·BE² = 7 ⇒ BE·AB = 14. BC = BE + EC = BE + 3BE = 4BE. Area of rectangle = AB·BC = AB·4BE = 4·(AB·BE) = 4·14 = 56 sq cm.

ModerateCAT 2002

In order to cover less distance, a boy, rather than going along the longer and the shorter lengths of the rectangular path, goes by the diagonal. The boy finds that he saved a distance equal to half the longer side. The ratio of the breadth and length is

(1) 1/2
(2) 2/3
(3) 3/4
(4) 2/15

Show solution

(3) 3/4. Let breadth = b, length = l (l the longer side), diagonal = d. Saving = (l + b) − d = l/2, so d = l/2 + b. With d² = l² + b²: (l/2 + b)² = l² + b² ⇒ l²/4 + lb = l² ⇒ b = 3l/4. Ratio breadth : length = 3 : 4.

ModerateCAT 2002

In the diagram, ∠ABC = 90° = ∠DCH = ∠DOE = ∠EHK = ∠FKL = ∠GLM = ∠LMN, AB = BC = 2CH = 2CD = EH = FK = 2HK = 4KL = 2LM = MN. The ratio of the areas of the two quadrilaterals ABCD and DEFG is

Composite rectilinear figure with vertices A, D, E, F, G, N above base B, C, H, K, L, M, used for the angle FGO and area-ratio questions

(1) 1 : 2
(2) 2 : 1
(3) 12 : 7
(4) None of these

Show solution

(3) 12 : 7. Working from the given equal lengths, Area(trapezium ABCD) = 3·CD² and Area(trapezium DEFG) = (7/4)·CD². Ratio = 3 : 7/4 = 12 : 7.

ModerateCAT 2003

Each side of a given polygon is parallel to either the X or the Y axis. A corner of such a polygon is said to be convex if the internal angle is 90° or concave if the internal angle is 270°. If the number of convex corners in such a polygon is 25, the number of concave corners must be

(1) 20
(2) 0
(3) 21
(4) 22

Show solution

(3) 21. For a rectilinear (axis-parallel) polygon, (number of convex corners) − (number of concave corners) = 4. With convex = 25: concave = 25 − 4 = 21.

ModerateCAT 2003

In the figure below, ABCDEF is a regular hexagon and ∠AOF = 90°. FO is parallel to ED. What is the ratio of the area of the triangle AOF to that of the hexagon ABCDEF?

Regular hexagon ABCDEF with interior point O; FO parallel to ED and angle AOF = 90°

(1) 1/12
(2) 1/6
(3) 1/24
(4) 1/18

Show solution

(1) 1/12. The regular hexagon can be split into congruent small right triangles. △AOF is exactly one of these, which is 1/12 of the whole hexagon's area.

HardCAT 2004

A rectangular sheet of paper, when halved by folding it at the mid-point of its longer side, results in a rectangle, whose longer and shorter sides are in the same proportion as the longer and shorter sides of the original rectangle. If the shorter side of the original rectangle is 2, what is the area of the smaller rectangle?

(1) 4√2
(2) 2√2
(3) 2
(4) None of the above

Show solution

(2) 2√2. Let the original sides be a (longer) and b = 2 (shorter). Halving the longer side gives a rectangle of sides b and a/2. The proportion condition: a/b = b/(a/2) ⇒ a² = 2b² = 8 ⇒ a = 2√2. Smaller rectangle area = (a/2)·b = √2·2 = 2√2.

HardCAT 2008

Consider a square ABCD with midpoints E, F, G, H of AB, BC, CD and DA respectively. Let L denote the line passing through F and H. Consider points P and Q, on L and inside ABCD, such that the angles APD and BQC both equal 120°. What is the ratio of the area of ABQCDP to the remaining area inside ABCD?

(1) 4√2/3
(2) 2 + √3
(3) (10 − 3√3)/9
(4) 2√3 − 1

Show solution

(4) 2√3 − 1. Let the square have side x. With ∠APD = 120°, △APD inside the square has a fixed area, and similarly △BQC. Computing the area ABQCDP versus the remaining region gives the ratio 2√3 − 1.

ModerateCAT 2017

Let ABCDEF be a regular hexagon with each side of length 1 cm. The area (in sq cm) of a square with AC as one side is

(1) 3√2
(2) 3
(3) 4
(4) √3

Show solution

(2) 3. The shorter diagonal AC of a regular hexagon of side s equals √3·s = √3. Area of the square on AC = AC² = (√3)² = 3 sq cm.

EasyCAT 2018

Let ABCD be a rectangle inscribed in a circle of radius 13 cm. Which one of the following pairs can represent, in cm, the possible length and breadth of ABCD?

(1) 24, 10
(2) 25, 9
(3) 25, 10
(4) 24, 12

Show solution

(1) 24, 10. The diagonal equals the diameter = 26 cm, so L² + B² = 26² = 676. Only 24² + 10² = 576 + 100 = 676 satisfies this.

HardCAT 2018

Points E, F, G, H lie on the sides AB, BC, CD, and DA, respectively, of a square ABCD. If EFGH is also a square whose area is 62.5% of that of ABCD and CG is longer than EB, then the ratio of length of EB to that of CG is

(1) 3 : 8
(2) 2 : 5
(3) 4 : 9
(4) 1 : 3

Show solution

(4) 1 : 3. Take ABCD side = 10 (area 100), so EFGH area = 62.5. With EB = a and BF = b for the corner right triangle, the square EFGH side² = a² + b² = 62.5 and a + b = 10 (since EB + EA = 10 and by symmetry). Solving: a = 2.5, b = 7.5. As CG > EB, EB = 2.5 and CG = 7.5 ⇒ EB : CG = 1 : 3.

ModerateCAT 2018

In a parallelogram ABCD of area 72 sq cm, the sides CD and AD have lengths 9 cm and 16 cm, respectively. Let P be a point on CD such that AP is perpendicular to CD. Then the area, in sq cm, of triangle APD is

(1) 32√3
(2) 18√3
(3) 24√3
(4) 12√3

Show solution

(1) 32√3. Area = CD·AP = 72 ⇒ AP = 72/9 = 8. In right △APD: DP = √(AD² − AP²) = √(256 − 64) = √192 = 8√3. Area △APD = ½·AP·DP = ½·8·8√3 = 32√3.

ModerateCAT 2018

A parallelogram ABCD has area 48 sqcm. If the length of CD is 8 cm and that of AD is s cm, then which one of the following is necessarily true?

(1) s ≠ 6
(2) s ≥ 6
(3) 5 ≤ s ≤ 7
(4) s ≤ 6

Show solution

(2) s ≥ 6. Area = CD·AD·sin(∠ADC) = 48 ⇒ 8·s·sinθ = 48 ⇒ s·sinθ = 6. Since sinθ ≤ 1, s ≥ 6.

HardCAT 2020

In a trapezium ABCD, AB is parallel to DC, BC is perpendicular to DC and ∠BAD = 45°. If DC = 5 cm, BC = 4 cm, the area of the trapezium in sq. cm is

(1) 25
(2) 24
(3) 28
(4) 30

Show solution

(3) 28. Split the trapezium into rectangle DCEB and a right triangle. Since ∠BAD = 45° and the height BC = 4, the triangle is a 45-45-90 with both legs 4, so its area = ½·4·4 = 8. Rectangle DCEB area = DC·BC = 5·4 = 20. Area of trapezium = 20 + 8 = 28 sq cm.

ModerateCAT 2019

Corners are cut off from an equilateral triangle T to produce a regular hexagon H. Then, the ratio of the area of H to the area of T is

(1) 2 : 3
(2) 4 : 5
(3) 5 : 6
(4) 3 : 4

Show solution

(1) 2 : 3. Trisecting each side splits the triangle T into 9 equal small equilateral triangles; the regular hexagon H covers 6 of them and T is all 9. Ratio = 6 : 9 = 2 : 3.

HardCAT 2019TITA

Let A and B be two regular polygons having a and b sides, respectively. If b = 2a and each interior angle of B is 2 times each interior angle of A, then each interior angle, in degrees, of a regular polygon with a + b sides is ______.

Show solution

150. Interior angle of A = (a−2)·180/a; of B = (b−2)·180/b. With (b−2)·180/b = 2·(a−2)·180/a and b = 2a, solving gives a = 4, b = 8. For a + b = 12 sides: interior angle = (12 − 2)·180/12 = 150°.

ModerateCAT 2021 · Slot 1

Suppose the length of each side of a regular hexagon ABCDEF is 2 cm. If T is the mid point of CD, then the length of AT, in cm is

(1) √15
(2) √12
(3) √14
(4) √13

Show solution

(4) √13. The shorter diagonal AC of a regular hexagon of side 2 = √3·2 = 2√3, and AC ⊥ CD. T is the midpoint of CD, so TC = 1. By Pythagoras in right △ACT: AT² = AC² + TC² = (2√3)² + 1² = 12 + 1 = 13 ⇒ AT = √13.

ModerateCAT 2021 · Slot 1

If the area of a regular hexagon is equal to the area of an equilateral triangle of side 12 cm, then the length, in cm, of each side of the hexagon is

(1) 4√6
(2) 2√6
(3) 6√6
(4) √6

Show solution

(2) 2√6. Equilateral triangle area = (√3/4)·12² = 36√3. Regular hexagon area = (3√3/2)·x² = 36√3 ⇒ x² = 24 ⇒ x = 2√6.

HardCAT 2021 · Slot 2

If a rhombus has area 12 sq cm and side length 5 cm, then the length, in cm, of its longer diagonal is

(1) (√37 + √13)/2
(2) √37 + √13
(3) (√13 + √12)/2
(4) √13 + √12

Show solution

(2) √37 + √13. Let the half-diagonals be a and b. Area = ½·(2a)·(2b) = 2ab = 12, and a² + b² = 5² = 25. Then (a+b)² = 25 + 12 = 37 ⇒ a + b = √37, and (a−b)² = 25 − 12 = 13 ⇒ a − b = √13. Longer diagonal = 2a = (a+b) + (a−b) = √37 + √13.

ModerateCAT 2021 · Slot 2

The sides AB and CD of a trapezium ABCD are parallel, with AB being the smaller side. P is the midpoint of CD and ABPD is a parallelogram. If the difference between the areas of the parallelogram ABPD and the triangle BPC is 10 sq cm, then the area, in sq cm, of the trapezium ABCD is

(1) 40
(2) 30
(3) 25
(4) 20

Show solution

(2) 30. Since P is the midpoint of CD, DP = PC. Area(parallelogram ABPD) = DP·h; Area(△BPC) = ½·PC·h = ½·DP·h. Difference = DP·h − ½·DP·h = ½·DP·h = 10 ⇒ DP·h = 20. Trapezium ABCD = ABPD + △BPC = 20 + 10 = 30 sq cm.

HardCAT 2021 · Slot 3

The cost of fencing a rectangular plot is ₹200 per ft along one side, and ₹100 per ft along the three other sides. If the area of the rectangular plot is 60000 sft, then the lowest possible cost of fencing all four sides, in ₹, is

(1) 120000
(2) 100000
(3) 160000
(4) 90000

Show solution

(1) 120000. Let one pair of sides be l and the other b, with the ₹200/ft side being one of length b. Cost = 200b + 100b + 100l + 100l = 200l + 300b. For fixed l·b = 60000, the sum 200l + 300b is minimised when 200l = 300b, giving l = 300, b = 200. Cost = 200·300 + 300·200 = 60000 + 60000 = ₹120000.

HardCAT 2021 · Slot 3

Let ABCD be a parallelogram. The lengths of the side AD and the diagonal AC are 10 cm and 20 cm, respectively. If the angle ∠ADC is equal to 30°, then the area of the parallelogram, in sq cm, is

(1) 25(√3 + √15)
(2) 25(√5 + √15)/2
(3) 25(√3 + √15)/2
(4) 25(√5 + √15)

Show solution

(1) 25(√3 + √15). Drop AE ⊥ DC. △AED is 30-60-90 with hypotenuse AD = 10: AE = ½·AD = 5, DE = 5√3. In right △AEC: EC = √(AC² − AE²) = √(400 − 25) = √375 = 5√15. So DC = DE + EC = 5√3 + 5√15. Area = DC·AE = (5√3 + 5√15)·5 = 25(√3 + √15).

HardCAT 2021 · Slot 3TITA

A park is shaped like a rhombus and has area 96 sq m. If 40 m of fencing is needed to enclose the park, the cost, in ₹, of laying electric wires along its two diagonals, at the rate of ₹125 per m, is ______.

Show solution

3500. Side = 40/4 = 10 m. Area = ½·d₁·d₂ = 96 ⇒ d₁·d₂ = 192. Half-diagonals satisfy (d₁/2)² + (d₂/2)² = 10² ⇒ d₁² + d₂² = 400. (d₁ + d₂)² = 400 + 2·192 = 784 ⇒ d₁ + d₂ = 28. Cost = 28·125 = ₹3500.

ModerateCAT 2022 · Slot 1

All the vertices of a rectangle lie on a circle of radius R. If the perimeter of the rectangle is P, then the area of the rectangle is

(1) P²/16 − R²
(2) P²/8 − 2R²
(3) P²/2 − 2PR
(4) P²/8 − R²/2

Show solution

(2) P²/8 − 2R². Let the sides be x and y. Perimeter: 2(x + y) = P ⇒ x + y = P/2. Diagonal = diameter = 2R ⇒ x² + y² = 4R². Then (x + y)² = x² + y² + 2xy ⇒ P²/4 = 4R² + 2xy ⇒ xy = P²/8 − 2R², which is the area.

ModerateCAT 2022 · Slot 1TITA

A trapezium ABCD has side AD parallel to BC, ∠BAD = 90°, BC = 3 cm and AD = 8 cm. If the perimeter of this trapezium is 36 cm, then its area, in scm, is ______.

Show solution

66. AB is the height (⊥ to the parallel sides AD and BC). The horizontal difference of the parallel sides is AD − BC = 5. Let AB = h; the slant side CD = √(h² + 5²). Perimeter: AD + BC + AB + CD = 8 + 3 + h + √(h² + 25) = 36 ⇒ √(h² + 25) = 25 − h ⇒ h² + 25 = 625 − 50h + h² ⇒ h = 12. Area = ½·(AD + BC)·AB = ½·(8 + 3)·12 = 66 sq cm.

ModerateCAT 2022 · Slot 2

Let ABCDEF be a regular hexagon. What is the ratio of the area of the triangle ACE to that of the hexagon ABCDEF?

Regular hexagon ABCDEF with all diagonals drawn, highlighting triangle ACE joining alternate vertices

(1) 1/3
(2) 1/2
(3) 2/3
(4) 5/6

Show solution

(2) 1/2. Triangle ACE joins alternate vertices of the regular hexagon and is exactly half its area.

ModerateCAT 2022 · Slot 3

The lengths of all four sides of a quadrilateral are integer valued. If three of its sides are of length 1 cm, 2 cm and 4 cm, then the total number of possible lengths of the fourth side is

(1) 5
(2) 4
(3) 3
(4) 6

Show solution

(1) 5. For a valid quadrilateral, each side < sum of the other three. So x < 1 + 2 + 4 = 7, and 4 < 1 + 2 + x ⇒ x > 1. Integer x ∈ {2, 3, 4, 5, 6} ⇒ 5 possible values.

ModerateCAT 2022 · Slot 2TITA

Regular polygons A and B have number of sides in the ratio 1 : 2 and interior angles in the ratio 3 : 4. Then, the number of sides of B equals ______.

Show solution

10. Let the numbers of sides be n and 2n. Interior angles ratio: [(n−2)·180/n] : [(2n−2)·180/(2n)] = 3 : 4. Simplifying gives n = 5, so B has 2n = 10 sides.

HardCAT 2023 · Slot 2

In a rectangle ABCD, AB = 9 cm and BC = 6 cm. P and Q are two points on BC such that the areas of the figures ABP, APQ, and AQCD are in geometric progression. If the area of the figure AQCD is four times the area of triangle ABP, then BP : PQ : QC is

(1) 2 : 4 : 1
(2) 1 : 2 : 4
(3) 1 : 1 : 2
(4) 1 : 2 : 1

Show solution

(1) 2 : 4 : 1. Let the three areas be a, b, c in GP, so b² = ac, with c = 4a ⇒ b = 2a; areas in ratio 1 : 2 : 4 summing to 7a = total area 54 ⇒ a = 54/7. Areas of ABP and APQ are ½·BP·9 and ½·PQ·9, so BP = 12/7, PQ = 24/7, and QC = 6 − 12/7 − 24/7 = 6/7. Hence BP : PQ : QC = 12 : 24 : 6 = 2 : 4 : 1.

HardCAT 2023 · Slot 3

A rectangle with the largest possible area is drawn inside a semicircle of radius 2 cm. Then, the ratio of the lengths of the largest to the smallest side of this rectangle is

(1) 1 : 1
(2) 2 : 1
(3) √5 : 1
(4) √2 : 1

Show solution

(2) 2 : 1. Place the semicircle's diameter on the x-axis. The rectangle has half-width x and height y with x² + y² = 2² = 4; its area = 2xy. Maximising 2xy subject to x² + y² = 4 gives x = √2, y = √2. Full base = 2x = 2√2, height = √2. Ratio = 2√2 : √2 = 2 : 1.

ModerateCAT 2023 · Slot 3TITA

In a regular polygon, any interior angle exceeds the exterior angle by 120 degrees. Then, the number of diagonals of this polygon is ______.

Show solution

54. Interior = exterior + 120 ⇒ (x−2)·180/x = 360/x + 120 ⇒ 180x − 360 = 360 + 120x ⇒ 60x = 720 ⇒ x = 12. Number of diagonals = x(x−3)/2 = 12·9/2 = 54.

CAT 2024 & 2025, recent

ModerateCAT 2025 · Slot 1 TITA

If the length of a side of a rhombus is 36 cm and the area of the rhombus is 396 sq. cm, then the absolute value of the difference between the lengths, in cm, of the diagonals of the rhombus is

Show solution

60. Diagonals p, q: area = ½·p·q = 396 ⇒ pq = 792. Half-diagonals satisfy (p/2)² + (q/2)² = 36² ⇒ p² + q² = 5184. (p−q)² = p² + q² − 2pq = 5184 − 1584 = 3600 ⇒ |p − q| = 60.

HardCAT 2025 · Slot 2

Let ABCDEF be a regular hexagon and P and Q be the midpoints of AB and CD, respectively. Then, the ratio of the areas of trapezium PBCQ and hexagon ABCDEF is

(A) 6 : 19
(B) 5 : 24
(C) 6 : 25
(D) 7 : 24

Show solution

(B) 5 : 24. Take side a. Hexagon area = (3√3/2)·a². Placing the hexagon on coordinates with P the midpoint of AB and Q the midpoint of CD, the trapezium PBCQ has area (5√3/16)·a². Ratio = (5√3/16) : (3√3/2) = 5 : 24.

Mensuration, 2D · 8 CAT PYQs

Mensuration, 2D

Mensuration, 2D. Composite plane regions, tiling, scrap-area and perimeter problems combining squares, circles and arcs.

ModerateCAT 1998

Four identical coins are placed in a square. For each coin the ratio of area to circumference is same as the ratio of circumference to area. Then find the area of the square that is not covered by the coins.

Four identical coins (circles) packed in a square, two by two

(1) 16(π − 1)
(2) 16(8 − π)
(3) 16(4 − π)
(4) 16(4 − π/2)

Show solution

(3) 16(4 − π). πR²/2πR = 2πR/πR² ⇒ R = 2. Square side = 8, area = 64. Four coins cover 4·π·2² = 16π. Uncovered = 64 − 16π = 16(4 − π).

HardCAT 2003

Consider two different cloth-cutting processes. In the first one, n circular cloth pieces are cut from a square cloth piece of side a in the following steps: The original square of side a is divided into n smaller squares, not necessarily of the same size; then a circle of maximum possible area is cut from each of the smaller squares. In the second process, only one circle of maximum possible area is cut from the square of side a and the process ends there. The cloth pieces remaining after cutting the circles are scrapped in both the processes. The ratio of the total area of scrap cloth generated in the former to that in the latter is:

(1) 1 : 1
(2) 2 : 1
(3) n(4 − π)/(4n − π)
(4) (4n − π)/n(4 − π)

Show solution

(1) 1 : 1. For any square, scrap/area = 1 − π/4, a constant fraction independent of the square's size. So total scrap is the same fraction of the cloth in both processes ⇒ 1 : 1.

HardCAT 2005

A jogging park has two identical circular tracks touching each other, and a rectangular track enclosing the two circles. The edges of the rectangles are tangential to the circles. Two friends, A and B, start jogging simultaneously from the point where one of the circular tracks touches the smaller side of the rectangular track. A jog along the rectangular track, while B jogs along the two circular tracks in a figure of eight. Approximately, how much faster than A does B have to run, so that they take the same time to return to their starting point?

(1) 3.88%
(2) 4.22%
(3) 4.44%
(4) 4.72%

Show solution

(4) 4.72%. Rectangle perimeter = 12r; figure-of-eight = two circumferences = 4πr. For equal times, speed ratio = path ratio. (b−a)/a = (12r − 4πr)/4πr ≈ 0.0472 ⇒ ≈ 4.72%.

ModerateCAT 2005

Rectangular tiles each of size 70 cm by 30 cm must be laid horizontally on a rectangular floor of size 110 cm by 130 cm, such that the tiles do not overlap. A tile can be placed in any orientation so long as its edges are parallel to the edges of the floor. No tile should overshoot any edge of the floor. The maximum number of tiles that can be accommodated on the floor is:

(1) 4
(2) 5
(3) 6
(4) 7

Show solution

(3) 6. By trying mixed orientations of the tiles on the floor, the maximum that fit without overshooting any edge is 6.

ModerateCAT 2018

From a rectangle ABCD of area 768 sq cm, a semicircular part with diameter AB and area 72π sq cm is removed. The perimeter of the leftover portion, in cm, is:

(1) 80 + 16π
(2) 86 + 8π
(3) 88 + 12π
(4) 82 + 24π

Show solution

(3) 88 + 12π. Semicircle area = ½·π·(AB/2)² = 72π ⇒ AB = 24. BC = 768/24 = 32. Perimeter = AD + DC + BC + arc(AB) = 32 + 24 + 32 + π·12 = 88 + 12π.

HardCAT 2020

On a rectangular metal sheet of area 135 sq inch, a circle is painted such that the circle touches two opposite sides. If the area of the sheet left unpainted is two-thirds of the painted area then the perimeter of the rectangle in inches is:

(1) 5√π(3 + 9/π)
(2) 3√π(5 + 12/π)
(3) 4√π(3 + 9/π)
(4) 3√π(5/2 + 6/π)

Show solution

(2) 3√π(5 + 12/π). Let painted area = 3x; unpainted = 2x. Then 3x + 2x = 135 ⇒ x = 27, so circle area = 3×27 = 81 ⇒ πr² = 81 ⇒ r = 9/√π. Breadth = 2r = 18/√π, and L × B = 135 ⇒ L = √π·(15/2). Perimeter = 2[L + B] = 2[15√π/2 + 18/√π] = 3√π(5 + 12/π).

HardCAT 2020

The sum of the perimeters of an equilateral triangle and a rectangle is 90 cm. The area, T, of the triangle and the area, R, of the rectangle, both in sq cm, satisfy the relationship R = T². If the sides of the rectangle are in the ratio 1 : 3, then the length, in cm, of the longer side of the rectangle, is:

(1) 18
(2) 27
(3) 21
(4) 24

Show solution

(2) 27. Rectangle sides x, 3x: R = 3x²; triangle side a: T = (√3/4)a². R = T² ⇒ 3x² = (3/16)a⁴ ⇒ x = a²/4. With 8x + 3a = 90, solving gives a = 6, x = 9. Longer side = 3x = 27.

CAT 2024 & 2025, recent

HardCAT 2024 · Slot 1

In the XY-plane, the area, in sq. units, of the region defined by the inequalities y ≥ x + 4 and −4 ≤ x² + y² + 4(x − y) ≤ 0 is

(A) 2π
(B) 4π
(C) π
(D) 3π

Show solution

(A) 2π. The inequality is an annulus between circles of radii 2 and 2√2 centred at (−2, 2); the line y = x + 4 passes through the centre, halving the annulus: ½(8π − 4π) = 2π.

Mensuration, 3D · 14 CAT PYQs

Mensuration, 3D

Mensuration, 3D. Volumes & surface areas of cubes, cuboids, cylinders, cones, spheres, prisms and pyramids, plus recasting.

ModerateCAT 2003

A square tin sheet of side 12 inches is converted into a box with open top in the following steps: The sheet is placed horizontally. Then, equal sized squares, each of side x inches, are cut from the four corners of the sheet. Finally, the four resulting sides are bent vertically upwards in the shape of a box. If x is an integer, then what value of x maximizes the volume of the box?

(1) 3
(2) 4
(3) 1
(4) 2

Show solution

(4) 2. V = (12 − 2x)²·x. dV/dx = 0 gives x² − 8x + 12 = 0 ⇒ x = 2 or 6 (x = 6 gives V = 0). At x = 2, V = 8²·2 = 128, the maximum.

ModerateCAT 2003

Directions (Q. 2 to 4): Answer the questions on the basis of the information given below.
Consider a cylinder of height h cm and radius r = 2/π cms as shown in the figure (not drawn to scale). A string of a certain length, when wound on its cylindrical surface, starting at point A and ending at point B, gives a maximum of n turns (in other words, the string's length is the minimum length required to wind n turns.)
Cylinder with a string wound from A at the bottom to B at the top, marking turns 1, 2, 3, …, n
2. What is the vertical spacing in cms between two consecutive turns?

(1) h/n
(2) h/√n
(3) h/n²
(4) Cannot be determined with given information

Show solution

(1) h/n. The n turns are spread evenly over height h, so the spacing between successive turns is h/n.

ModerateCAT 2003

3. The same string, when wound on the exterior four walls of a cube of side n cms, starting at point C and ending at point D, can give exactly one turn (see figure, not drawn to scale). The length of the string, in cms, is:
Cube of side n with a string wound around its four exterior walls from C at a bottom corner to D at a top corner, completing one turn

(1) 2n
(2) √17 n
(3) n
(4) √13 n

Show solution

(2) √17·n. Unrolling the four faces gives a rectangle of width 4n and height n. The string is the diagonal: √((4n)² + n²) = √(17n²) = √17·n.

ModerateCAT 2003

4. In the setup of the previous two questions, how is h related to n?

(1) h = √2 n
(2) h = √17 n
(3) h = n
(4) h = √13 n

Show solution

(3) h = n. Since the cube has the same height as the cylinder and there is one turn matching n turns over the height, h = n.

HardCAT 2008

Consider a right circular cone of base radius 4 cm and height 10 cm. A cylinder is to be placed inside the cone with one of the flat surface resting on the base of the cone. Find the largest possible total surface area (in sq. cm) of the cylinder.

(1) 100π/3
(2) 80π/3
(3) 120π/7
(4) 130π/9

Show solution

(1) 100π/3. Similar triangles: h = 10 − 2.5r. TSA A = 2πr(r+h) = 20πr − 3πr². dA/dr = 0 ⇒ r = 10/3. A = 20π·(10/3) − 3π·(10/3)² = 100π/3.

ModerateCAT 2017

The base of a vertical pillar with uniform cross section is a trapezium whose parallel sides are of lengths 10 cm and 20 cm while the other two sides are of equal length. The perpendicular distance between the parallel sides of the trapezium is 12 cm. If the height of the pillar is 20 cm, then the total area, in sq cm, of all six surfaces of the pillar is

(1) 1300
(2) 1340
(3) 1480
(4) 1520

Show solution

(3) 1480. Slant sides = √(12²+5²) = 13. Base area = ½·12·(10+20) = 180. Perimeter = 10+20+13+13 = 56. Lateral = 56·20 = 1120. Total = 1120 + 180 + 180 = 1480.

ModerateCAT 2017

A solid metallic cube is melted to form five solid cubes whose volumes are in the ratio 1 : 1 : 8 : 27 : 27. The percentage by which the sum of the surface areas of these five cubes exceeds the surface area of the original cube is nearest to

(1) 10
(2) 50
(3) 60
(4) 20

Show solution

(2) 50. Volume ratio 1:1:8:27:27 ⇒ original volume 64 ⇒ side ratio 1:1:2:3:3 (orig 4). Area ratio 1:1:4:9:9 sums to 24 vs original 16. Excess = (24−16)/16 = 50%.

ModerateCAT 2017TITA

A ball of diameter 4 cm is kept on top of a hollow cylinder standing vertically. The height of the cylinder is 3 cm, while its volume is 9π cubic centimeters. Then the vertical distance, in cm, of the topmost point of the ball from the base of the cylinder is

Show solution

6. Volume = πr²·3 = 9π ⇒ r = √3. Ball radius R = 2; the centre of the ball sits 1 cm above the rim (since √(R²−r²) = √(4−3) = 1). Top point = 3 + 1 + 2 = 6 cm.

ModerateCAT 2018TITA

A right circular cone, of height 12 ft, stands on its base which has diameter 8 ft. The tip of the cone is cut off with a plane which is parallel to the base and 9 ft from the base. With π = 22/7, the volume, in cubic ft, of the remaining part of the cone is

Show solution

198. Full cone volume = ⅓·(22/7)·4²·12 = (22/7)·64. Small cone (height 3, radius 1) = (22/7)·1. Frustum = (22/7)·(64−1) = (22/7)·63 = 198 cubic ft.

HardCAT 2019

If the rectangular faces of a brick have their diagonals in the ratio 3 : 2√3 : √15, then the ratio of the length of the shortest edge of the brick to that of its longest edge is:

(1) √3 : 2
(2) 1 : √3
(3) 2 : √5
(4) √2 : √3

Show solution

(2) 1 : √3. With edges a, b, c: a²+b² = 9, b²+c² = 12, c²+a² = 15. Adding: a²+b²+c² = 18 ⇒ a = √6, b = √3, c = 3. Shortest : longest = b : c = √3 : 3 = 1 : √3.

ModerateCAT 2019

The base of a regular pyramid is a square and each of the other four sides is an equilateral triangle, length of each side being 20 cm. The vertical height of the pyramid, in cm, is:

(1) 12
(2) 10√2
(3) 8√3
(4) 5√5

Show solution

(2) 10√2. Each lateral edge = 20 (the faces are equilateral triangles of side 20). Half the base diagonal = (20√2)/2 = 10√2. The apex is above the centre, so h² + (10√2)² = 20² ⇒ h² = 400 − 200 = 200 ⇒ h = 10√2.

HardCAT 2019

A man makes complete use of 405 cc of iron, 783 cc of aluminium, and 351 cc of copper to make a number of solid right circular cylinders of each type of metal. These cylinders have the same volume and each of these has radius 3 cm. If the total number of cylinders is to be kept at a minimum, then the total surface area of all these cylinders, in sq cm, is:

(1) 1026(1 + π)
(2) 8464π
(3) 928π
(4) 1044(4 + π)

Show solution

(1) 1026(1 + π). Cylinder volume = HCF(405,783,351) = 27 ⇒ πr²h = 27 ⇒ h = 3/π. Counts: 15 + 29 + 13 = 57. TSA per cylinder = 2πr(r+h) = 2π·3·(3 + 3/π). Total = 57·that = 1026(π + 1).

ModerateCAT 2020

A solid right circular cone of height 27 cm is cut into two pieces along a plane parallel to its base at a height of 18 cm from the base. If the difference in volume of the two pieces is 225 cc, the volume, in cc, of the original cone is:

(1) 264
(2) 256
(3) 232
(4) 243

Show solution

(4) 243. Top small cone height 9 : full 27 = 1 : 3 ⇒ volume ratio 1 : 27. Let full = 27x; small = x, frustum = 26x. Difference 26x − x = 25x = 225 ⇒ x = 9. Original = 27·9 = 243 cc.

CAT 2024 & 2025, recent

ModerateCAT 2024 · Slot 1

The surface area of a closed rectangular box, which is inscribed in a sphere, is 846 sq cm, and the sum of the lengths of all its edges is 144 cm. The volume, in cubic cm, of the sphere is

(A) 1125π
(B) 750π
(C) 1125π√2
(D) 750π√2

Show solution

(C) 1125π√2. Edge sum 4(l+b+h) = 144 ⇒ l+b+h = 36; surface 2(lb+bh+hl) = 846. l²+b²+h² = 36² − 846 = 1296 − 846 = 450. The body diagonal = diameter, so (2R)² = 450 ⇒ R = 15/√2. Volume = (4/3)πR³ = (4/3)π·(15/√2)³ = 1125π√2.

Coordinate Geometry · 19 CAT PYQs

Coordinate Geometry

Coordinate Geometry. Distance, midpoint/section, area by coordinates, slopes and line/circle equations, often blended with pure geometry.

HardCAT 1999

Directions (Q. 1 and 2): Answer the questions based on the following information. A rectangle PRSU, is divided into two smaller rectangles PQTU, and QRST by the line TQ. PQ = 10 cm. QR = 5 cm and RS = 10 cm. Points A, B, F are within rectangle PQTU, and points C, D, E are within the rectangle QRST. The closest pair of points among the pairs (A, C), (A, D), (A, E), (F, C), (F, D), (F, E), (B, C), (B, D), (B, E) are 10√3 cm apart.

Which of the following statements is necessarily true?

(1) The closest pair of points among the six given points cannot be (F, C)
(2) Distance between A and B is greater than that between F and C.
(3) The closest pair of points among the six given points is (C, D), (D, E), or (C, E).
(4) None of the above

Show solution

(4) None of the above. We have not been given the distances between any two points, so none of the three statements can be guaranteed.

HardCAT 1999

(Same figure as Q1.) AB > AF > BF and CD > DE > CE, and BF = 6√5 cm. Which of the following is the closest pair of points among all the six given points?

(1) B, F
(2) C, D
(3) A, B
(4) None of these

Show solution

(4) None of these. Since CD > DE, option (2) cannot be the answer. Similarly, since AB > AF, option (3) cannot be the answer. We are not sure about the positions of points B and F. Hence, (1) cannot be the answer.

EasyCAT 1999

Directions (Q. 3 and 4): A robot is moved by feeding it with a sequence of instructions. GOTO (x, y), move to the point with coordinates (x, y) no matter where you are currently. WALKX(p), move parallel to the X-axis through a distance of p, in the positive direction if p is positive and in the negative direction if p is negative. WALKY(p), move parallel to the Y-axis through a distance of p, with the same sign convention.

The robot reaches point (6, 6) when a sequence of three instructions is executed, the first of which is a GOTO (x, y) instruction, the second is WALKX(2) and the third is WALKY(4). What are the value of x and y?

(1) 2, 4
(2) 0, 0
(3) 4, 2
(4) 2, 2

Show solution

(3) 4, 2. Work backwards from (6, 6). Before WALKY(4) the point was (6, 6−4) = (6, 2); before WALKX(2) it was (6−2, 2) = (4, 2). So GOTO(x, y) = GOTO(4, 2) ⇒ x = 4, y = 2.

EasyCAT 1999

(Same robot as Q3.) The robot is initially at (x, y), x > 0 and y < 0. The minimum number of instructions needed to be executed to bring it to the origin (0, 0) if you are prohibited from using the GOTO instruction is:

(1) 2
(2) 1
(3) x + y
(4) 0

Show solution

(1) 2. Two instructions are needed, one parallel to the X-axis and the other parallel to the Y-axis: WALKX(−x) then WALKY(−y).

ModerateCAT 2000

ABCD is a rhombus with the diagonals AC and BD intersecting at the origin on the xy-plane. The equation of the straight line AD is x + y = 1. What is the equation of BC?

(1) x + y = −1
(2) x − y = −1
(3) x + y = 1
(4) None of these

Show solution

(1) x + y = −1. AD ∥ BC (opposite sides of rhombus), so BC has the same slope: x + y = k. Diagonals bisect at the origin, so B and C are reflections of A and D ⇒ k = −1. BC: x + y = −1.

EasyCAT 2002

The area of the triangle with the vertices (a, a), (a + 1, a) and (a, a + 2) is:

(1) a³
(2) 1
(3) 0
(4) None of these

Show solution

(2) 1. The legs along the axes have lengths 1 and 2, with a right angle at (a, a). Area = ½·1·2 = 1 (independent of a).

ModerateCAT 2005

Four points A, B, C and D lie on a straight line in the XY-plane, such that AB = BC = CD, and the length of AB is 1 metre. An ant at A wants to reach a sugar particle at D. But there are insect repellents kept at points B and C. The ant would not go within one metre of any insect repellent. The minimum distance in metres the ant must traverse to reach the sugar particle is:

(1) 3√2
(2) 1 + π
(3) 4π/3
(4) 5

Show solution

(2) 1 + π. The ant detours around each repellent on a quarter-circle of radius 1 and crosses the gap on a straight 1 m. Total = 1 + 2·(π/2) = 1 + π.

ModerateCAT 2017

The shortest distance of the point (½, 1) from the curve y = |x − 1| + |x + 1| is:

(1) 1
(2) 0
(3) √2
(4) √(3/2)

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(1) 1. For |x| ≤ 1 the curve is the horizontal line y = 2; for |x| > 1 it slopes away. The nearest point to (½, 1) lies on y = 2 directly above, giving distance 2 − 1 = 1.

ModerateCAT 2017

The points (2, 5) and (6, 3) are two end points of a diagonal of a rectangle. If the other diagonal has the equation y = 3x + c, then c is:

(1) −5
(2) −6
(3) −7
(4) −8

Show solution

(4) −8. Diagonals bisect each other, so the other diagonal passes through the midpoint of the first = (4, 4). Substitute: 4 = 3·4 + c ⇒ c = −8.

ModerateCAT 2018

A triangle ABC has area 32 sq units and its side BC, of length 8 units, lies on the line x = 4. Then the shortest possible distance between A and the point (0, 0) is:

(1) 8 units
(2) 4 units
(3) 2√2 units
(4) 4√2 units

Show solution

(2) 4 units. Height from A to BC (line x = 4) = 2·Area/BC = 64/8 = 8, so A's x-coordinate is 4 ± 8 = 12 or −4. To minimise distance to origin, take A = (−4, 0): distance = 4 units.

HardCAT 2019TITA

With rectangular axes of co-ordinates, the number of paths from (1, 1) to (8, 10) via (4, 6), where each step from any point (x, y) is either to (x, y + 1) or to (x + 1, y), is:

Show solution

3920. (1,1)→(4,6): 3 right + 5 up ⇒ ⁸C₃ = 56 paths. (4,6)→(8,10): 4 right + 4 up ⇒ ⁸C₄ = 70 paths. Total = 56·70 = 3920.

ModerateCAT 2019TITA

Let T be the triangle formed by the straight line 3x + 5y − 45 = 0 and the co-ordinate axes. Let the circumcircle of T have radius of length L, measured in the same unit as the coordinate axes. Then, the integer closest to L is:

Show solution

9. Intercepts (15, 0), (0, 9) and origin form a right triangle. Circumradius = ½·hypotenuse = ½·√(15²+9²) = ½·√306 ≈ ½·17.49 = 8.74 ⇒ closest integer 9.

ModerateCAT 2020

The area, in sq. units, enclosed by the lines x = 2, y = |x − 2| + 4, the X-axis and the Y-axis is equal to:

(1) 6
(2) 8
(3) 12
(4) 10

Show solution

(4) 10. At x = 2, y = 4 (point A height); at x = 0, y = 6 (OC). It is a trapezium with parallel sides 4 and 6 and width 2. Area = ½·(4+6)·2 = 10.

ModerateCAT 2020

The vertices of a triangle are (0, 0), (4, 0) and (3, 9). The area of the circle passing through these three points is:

(1) 205π/9
(2) 123π/7
(3) 12π/5
(4) 14π/3

Show solution

(1) 205π/9. Sides a = PQ = 4, RO = √90, RQ = √82. Area of triangle = ½·4·9 = 18. R = (a·b·c)/(4·Area) ⇒ R² = (82·90)/(4·18)² ·… giving πR² = 205π/9.

ModerateCAT 2020

The points (2, 1) and (−3, −4) are opposite vertices of a parallelogram. If the other two vertices lie on the line x + 9y + c = 0, then c is:

(1) 15
(2) 12
(3) 13
(4) 14

Show solution

(4) 14. Diagonals bisect, so the centre = midpoint of (2,1) & (−3,−4) = (−½, −3/2) lies on the line. Substitute: −½ + 9·(−3/2) + c = 0 ⇒ c = 14.

ModerateCAT 2022 · Slot 1

Let ABCD be a parallelogram such that the coordinates of its three vertices A, B, C are (1, 1), (3, 4) and (−2, 8), respectively. Then, the coordinates of the vertex D are:

(1) (−4, 5)
(2) (−3, 4)
(3) (0, 11)
(4) (4, 5)

Show solution

(1) (−4, 5). Diagonals bisect: midpoint of AC = midpoint of BD. ((1−2)/2, (1+8)/2) = ((3+x)/2, (4+y)/2) ⇒ x = −4, y = 5. D = (−4, 5).

HardCAT 2023 · Slot 1

(1) (6, 6)
(2) (6, 4)
(3) (6, 8)
(4) (6, 3)

Show solution

(4) (6, 3). Centre (−2, 3), radius = √(4+9+3) = 4. For 60° between tangents, the external point is at distance 8 from centre (since sin30° = 4/CP ⇒ CP = 8). On x = 6: (6+2)² + (y−3)² = 64 ⇒ (y−3)² = 0 ⇒ y = 3. Point (6, 3).

HardCAT 2023 · Slot 2TITA

The area of the quadrilateral bounded by the Y-axis, the line x = 5, and the lines |x − y| − |x − 5| = 2, is:

Show solution

45. For 0 ≤ x ≤ 5, |x−5| = 5−x, so |x−y| = 7−x, giving lines y = 7 and 2x − y − 7 = 0. The trapezium has vertices D(0,7), C(5,7), B(5,3), A(0,−7). Area = ½·(AD + BC)·CD = ½·(14+4)·5 = 45.

CAT 2024 & 2025, recent

HardCAT 2025 · Slot 1

The (x, y) coordinates of vertices P, Q and R of a parallelogram PQRS are (−3, −2), (1, −5) and (9, 1), respectively. If the diagonal SQ intersects the x-axis at (a, 0), then the value of a is

(A) 13/4
(B) 29/9
(C) 10/3
(D) 27/7

Show solution

(B) 29/9. Diagonals bisect: midpoint of PR = (3, −0.5) = midpoint of QS → S = (5, 4). Line SQ slope 9/4; set y = 0 → x = 1 + 20/9 = 29/9.

Trigonometry · 1 CAT PYQs

Trigonometry

Trigonometry. Heights-and-distances and angle-of-elevation problems solved with sine/cosine and right-triangle ratios.

ModerateCAT 2003

A car is being driven, in a straight line and at a uniform speed, towards the base of a vertical tower. The top of the tower is observed from the car and, in the process, it takes 10 minutes for the angle of elevation to change from 45° to 60°. After how much more time will this car reach the base of the tower?

(1) 5(√3 + 1)
(2) 6(√3 + √2)
(3) 7(√3 − 1)
(4) 8(√3 − 2)

Show solution

(1) 5(√3 + 1). Let x be the distance of the second position of the car (at 60°) from the tower, and h the tower height. In △ABD, ∠ADB = 60°: h = x·tan 60° = x√3 = AB. Also, from the first position (at 45°), AB = BC·sin 45°, so BC = x√3. Then DC = BC − BD = x√3 − x = x(√3 − 1). The car covers DC in 10 minutes, so speed S = x(√3 − 1)/10. Remaining time = x/S = 10/(√3 − 1) = 5(√3 + 1) minutes.