◆ QA · Geometry

Triangles , formulas + CAT PYQs

Focused Geometry kit. The full chapter formula sheet (with explanations & basic examples) is tucked below; every CAT PYQ for Triangles is here.

38CAT PYQs
Geometrychapter
Formulas PYQs (38)↩ Full Geometry chapter

Geometry, formula sheet

Show the full Geometry formula sheet (explanations + basic examples)
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)
38 CAT questions

Triangles, 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.