◆ QA · Geometry

Quadrilaterals & Polygons , formulas + CAT PYQs

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

35CAT PYQs

Geometrychapter

Formulas PYQs (35)↩ 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)

Quadrilaterals & Polygons, 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.