# Geometry - Quadrilaterals & Polygons - Previous Year CAT/MBA Questions

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**XAT 2023 QADI | Geometry - Quadrilaterals & Polygons omet Question**

ABCD is a trapezoid where BC is parallel to AD and perpendicular to AB. Kindly note that BC < AD. P is a point on AD such that CPD is an equilateral triangle. Q is a point on BC such that AQ is parallel to PC. If the area of the triangle CPD is 4√3, find the area of the triangle ABQ.

- (a)
2√3

- (b)
4√3

- (c)
4

- (d)
4√3

- (e)
None of the above

Answer: Option A

**Explanation** :

Given that, CPD is an equilateral triangle

∠CPD = ∠PDC = ∠DCP = 60°

AQ || PC ⇒ ∠CPD = ∠QAP = 60°

BC || AD ⇒ ∠QAP = ∠AQB = 60°

⇒ AQCP is a parallelogram, hence AQ = PC = a

∆AQB is a 30°-60°-90° triangle.

⇒ BQ = 1/2 × AQ = a/2

Comapring ∆AQB and ∆CPD

Base AQ = 1/2 × PD while the heights of both triangles is same.

⇒ Area of ∆AQB = 1/2 × Area of ∆CPD = 2√3

Hence, option (a).

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**XAT 2020 QADI | Geometry - Quadrilaterals & Polygons omet Question**

A rectangular field is 40 meters long and 30 meters wide. Draw diagonals on this field and then draw circles of radius 1.25 meters, with centers only on the diagonals. Each circle must fall completely within the field. Any two circles can touch each other but should not overlap.

What is the maximum number of such circles that can be drawn in the field?

- (a)
39

- (b)
40

- (c)
37

- (d)
36

- (e)
38

Answer: Option C

**Explanation** :

Length of the diagonal = √(30^{2}+40^{2} ) = 50 m.

Each circle on the end of the diagonal will touch sides of the rectangular field

Using Pythagoras' theorem, the distance between the vertex of the rectangle and centre of the first circle drawn on the diagonal (OC) = 1.25√2

Distance between the vertex of the rectangle and circumference of the first circle drawn on the diagonal (OD) = 1.25√2 - 1.25 = 0.51 meters

Space that cannot be used to draw circle otherwise they will go outside rectangle on every diagonal = 0.51 × 2 = 1.02 meters

Space that can be used to draw circles = length of diagonal - unused space = 50 - 1.02 = 48.98 meters

On every diagonal, maximum number of such circles = usable length/diameter of each circle = 48.98/2.5 = 19.6

∴ Maximum 19 circles can be drawn on a diagonal.

Now, on every diagonal, one circle will be at the centre (intersection of diagonals) and 9 circles will be on each half of the diagonal

⇒ The circle in centre will be common for both diagonals.

∴ Total circles = 19 + 19 – 1 = 37.

Hence, option (c).

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**XAT 2020 QADI | Geometry - Quadrilaterals & Polygons omet Question**

Mohanlal, a prosperous farmer, has a square land of side 2 km. For the current season, he decides to have some fun. He marks two distinct points on one of the diagonals of the land. Using these points as centers, he constructs two circles. Each of these circles falls completely within the land, and touches at least two sides of the land. To his surprise, the radii of both the circles are exactly equal to 2/3 km. Mohanlal plants potatoes on the overlapping portion of these circles.

- (a)
5( 𝜋 + 4)/27

- (b)
2(2 𝜋 − 3 √3 )/27

- (c)
4(𝜋 − 3 √3 )/27

- (d)
𝟐(𝝅 − 2)/9

- (e)
(𝝅 − 2)/9

Answer: Option D

**Explanation** :

Length of the diagonal AC = 2√2

⇒ AX = ½ of diagonal = √2

Length of AC_{1} = 2/3 × √2 = 2√2/3

⇒ C_{1}X = AX – AC_{1} = √2 - 2√2/3 = √2/3

Using Pythagoras theorem:

XM = $\sqrt{{\left(\frac{2}{3}\right)}^{2}-{\left(\frac{\sqrt{2}}{3}\right)}^{2}}$ = $\frac{\sqrt{2}}{3}$

In ∆C_{1}XM,

C_{1}X = XM = √2/3 and ∠C_{1}XM = 90°

⇒ ∠XC_{1}M = 45°

And ∠LC_{1}M = 90°

∴ Area of ∆LC_{1}M = ½ × 2/3 × 2/3 = 2/9

Area of minor arc LM = Area of sector LC_{1}M - Area of ∆LC_{1}M

= $\frac{90}{360}\times \mathrm{\pi}\times {\left(\frac{2}{3}\right)}^{2}-\frac{2}{9}$ = $\frac{\mathrm{\pi}-2}{9}$

∴ Area of overlapping region = $2\left(\frac{\mathrm{\pi}-2}{9}\right)$

Hence, option (d).

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**IIFT 2020 QA | Geometry - Quadrilaterals & Polygons omet Question**

ABCD is a parallelogram whose diagonals are parallel to the lines 2y - x - 5 = 0 and y + 2x - 7 = 0 respectively.

Then ABCD is -

- (a)
Cyclic quadrilateral

- (b)
Rectangle

- (c)
Cube

- (d)
Rhombus

Answer: Option D

**Explanation** :

The slope of any line parallel to 2y - x - 5 = 0 would be 1/2

The slope of any line parallel to y + 2x - 7 = 0 would be -2

Product of slopes = -1 which shows that these 2 lines would be perpendicular i.e. we have a parallelogram

whose diagonals are perpendicular to each other. This means that the quadrilateral is atleast a rhombus

and it would be a square if we can get info that all sides are equal to each other.

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**XAT 2019 QADI | Geometry - Quadrilaterals & Polygons omet Question**

In the trapezium ABCD the sides AB and CD are parallel. The value of $\frac{\mathrm{sin}\angle BAC}{\mathrm{sin}\angle BAD}$ is

- (a)
AD/AC

- (b)
AB/CD

- (c)
BC/AD

- (d)
AC/AD

- (e)
AC/CD

Answer: Option A

**Explanation** :

Refer the diagram below:

In ∆ACL,

Sin∠BAC = Sin∠LAC = perpendicular/hypotenuse = CL/AC …(1)

Similarly, in ∆ADM,

Sin∠BAD = Sin∠MAD = DM/AD …(2)

(1) ÷ (2)

⇒ Sin∠BAC : Sin∠BAD = CL/AC : DM/AD

Now, CL = DM

Hence, Sin∠BAC : Sin∠BAD = 1/AC : 1/AD = AD : AC

Hence, option (a).

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**XAT 2019 QADI | Geometry - Quadrilaterals & Polygons omet Question**

In the picture below, EFGH, ABCD are squares, and ABE, BCF, CDG, DAH are equilateral triangles. What is the ratio of the area of the square EFGH to that of ABCD?

- (a)
√2 + 2

- (b)
3 + √2

- (c)
√2 + √3

- (d)
√3 + √2

- (e)
1 + √3

Answer: Option D

**Explanation** :

In □EFGH, EG is the diagonal.

Also, EI and GJ are the perpendicular bisectors of the equilateral triangles AEB and GCD.

Let us suppose AB = ’a’ units.

BC will also be ‘a’ units since AB and BC are sides of the same square.

In ∆ABE, EI = a√3/2 (perpendicular of an equilateral triangle is √3/2 times the side)

Similarly, GJ = a√3/2

Also, IJ = BC = a

∴ EG = EI + IJ + GJ = a√3/2 + a + a√3/2 = a(√3 + 1)

∴ Side of square EFGH = a(√3 + 1)/√2

⇒ Area of square EFGH = [a(√3 + 1)/√2]^{2} = (2 + √3)a^{2}

Also, area of square ABCD = a^{2}

Area of square ABCD = a^{2}

Ratio of area of EFGH to area of ABCD = (2 + √3)a^{2} : a^{2} = (2 + √3) : 1

Hence, option (d).

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**XAT 2018 QADI | Geometry - Quadrilaterals & Polygons omet Question**

If the diagonals of a rhombus of side 15 cm are in the ratio 3:4, find the area of the rhombus.

- (a)
54 sq. cm.

- (b)
108 sq. cm.

- (c)
144 sq. cm.

- (d)
200 sq. cm.

- (e)
None of the above

Answer: Option E

**Explanation** :

We know that the diagonals of a rhombus bisect each other at right angles.

So, let’s assume that the two diagonals have lengths 6x and 8x.

⇒ (3x)^{2} + (4x)^{2} = 152

⇒ 25x^{2} = 225

⇒ x^{2} = 9

⇒ x = 3

∴ The length of the diagonals are 18 and 24.

∴ Area of the rhombus = ½ × product of the diagonals = ½ × 18 × 24 = 216

Hence, option (e).

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**XAT 2017 QADI | Geometry - Quadrilaterals & Polygons omet Question**

AB, CD and EF are three parallel lines, in that order. Let d1 and d2 be the distances from CD to AB and EF respectively. d_{1} and d_{2} are integers, where d_{1} : d_{2} = 2 : 1. P is a point on AB, Q and S are points on CD and R is a point on EF. If the area of the quadrilateral PQRS is 30 square units, what is the value of QR when value of SR is the least?

- (a)
slightly less than 10 units

- (b)
10 units

- (c)
slightly greater than 10 units

- (d)
slightly less than 20 units

- (e)
slightly more than 20 units

Answer: Option E

**Explanation** :

Let d_{1} = x and d_{2} = 2x.

Now, SR would be shortest when SR = d1 = x

Area of PQRS = 30 square units.

∴ Area of PQRS = ½ × QS × 2x + ½ × QS × x

= (3/2) × QS × x = 30

⇒ QS × x = 20

Smallest value of x = 1 units.

⇒ QS = 20

∴ ∆QSR is right angled at S

⇒ QR^{2} = QS^{2} + SR^{2}

⇒ QR^{2} = 400 + 1

⇒ QR = slightly more than 20.

Hence, option (e).

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**XAT 2017 QADI | Geometry - Quadrilaterals & Polygons omet Question**

ABCD is a rectangle. P, Q and R are the midpoint of BC, CD and DA. The point S lies on the line QR in such a way that SR: QS = 1:3. The ratio of the area of triangle APS to area of rectangle ABCD is

- (a)
36/128

- (b)
39/128

- (c)
44/128

- (d)
48/128

- (e)
64/128

Answer: Option A

**Explanation** :

Let the length of the rectangle be ‘l’ and breadth be ‘b’.

The given figure is.

Let us draw a line MS parallel to DC

∆RMS ~ ∆RDQ

∴ RM : MD = RS : SQ = 1 : 3

Let the length of the rectangle (i.e., AB = CD) ABCD be 8l and width be 8b (i.e., AD = BC).

∴ AR = 4b and RM = b

Also, MS = l and SL = 7l.

Also, PL : LC = 1 : 3 [∵ RS : SQ = 1 : 3]

Area (∆ARS) = ½ × AR × SM = 2lb

Area (∆RDQ) = ½ × RD × DQ = 8lb

Area (∆ABP) = ½ × AB × BP = 16lb

Area (PSQC) = Area (∆PSL) + Area(∆SLQC)

= ½ × SL × PL + ½ × (SL + QC) × LC

= 3.5lb + 16.5lb

= 20lb

Area of ∆ASP = Area of rectangle ABCD – [Area (∆ARS) + Area (∆RDQ) + Area (∆ABP) + Area (PSQC)]

= 64lb – [2lb + 8lb + 16lb + 20lb]

= 64lb - 46lb = 18lb

∴ $\frac{Area(\u2206ASP)}{Area\left(ABCD\right)}$ = $\frac{18lb}{64lb}$ = $\frac{36}{128}$

Hence, option (a).

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**XAT 2016 QADI | Geometry - Quadrilaterals & Polygons omet Question**

∆ABC and ∆XYZ are equilateral triangles of 54 cm sides. All smaller triangles like ∆ANM, ∆OCP, ∆QPX etc. are also equilateral triangles. Find the area of the shape MNOPQRM.

- (a)
243√3 sq. cm.

- (b)
486√3 sq. cm.

- (c)
729√3 sq. cm.

- (d)
4374√3 sq. cm.

- (e)
None of the above

Answer: Option A

**Explanation** :

Since all triangles are equilateral triangles

This is possible when YM = MN = NZ = 1/3 of YZ = 54/3 = 18 cm.

∴ Side of the regular hexagon MNOPQRM is 18 cm.

∴ Area of the hexagon = 6 × √3/4 × (18)^{2} = 486√3.

Hence, option (b).

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**XAT 2016 QADI | Geometry - Quadrilaterals & Polygons omet Question**

ABCD is a quadrilateral such that AD = 9 cm, BC = 13 cm and ⎿DAB = ⎿BCD = 90°. P and Q are two points on AB and CD respectively, such that DQ : BP = 1 : 2 and DQ is an integer. How many values can DQ take, for which the maximum possible area of the quadrilateral PBQD is 150 sq.cm?

- (a)
14

- (b)
12

- (c)
10

- (d)
9

- (e)
8

Answer: Option D

**Explanation** :

Let DQ = x and PB = 2x.

From the diagram A(△BDP) = ½ × AD × PB = ½ × 9 × 2x.

Also A(△BDQ) = ½ × BC × DQ = ½ × 13 × x

A(ꐎBPDQ) = A(△BDQ) + A(△BDP) = ½ × 31 × x = 150

∴ x = 300 / 31

∴ x can have integer values from 1 to 9.

Hence , there are 9 values of DQ.

Hence, option (d).

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**IIFT 2016 QA | Geometry - Quadrilaterals & Polygons omet Question**

Let PQRSTU be a regular hexagon. The ratio of the area of the triangle PRT to that of the hexagon PQRSTU is

- (a)
0.3

- (b)
0.5

- (c)
1

- (d)
None of the above

Answer: Option B

**Explanation** :

A regular hexagon comprises six equilateral triangles.

∴ Area of hexagon = 6 × area of equilateral triangle = area of six equilateral triangles

P, R and T are alternate vertices of the hexagon. The triangle formed by joining them splits each equilateral triangle into half. Six half-equilateral triangles is equivalent to three equivalent triangles.

∴ Area of triangle PRT = area of three equilateral triangles.

∴ Required ratio = (area of three equilateral triangles) / (area of six equilateral triangles) = 0.5

Hence, option (b).

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**XAT 2015 QA | Geometry - Quadrilaterals & Polygons omet Question**

The parallel sides of a trapezoid ABCD are in the ratio of 4 : 5. ABCD is divided into an isosceles triangle ABP and a parallelogram PBCD (as shown below). ABCD has a perimeter equal to 1120 meters and PBCD has a perimeter equal to 1000 meters. Find Sin ∠ABC, given 2∠DAB = ∠BCD.

- (a)
4/5

- (b)
16/25

- (c)
5/6

- (d)
24/25

- (e)
A single solution is not possible

Answer: Option A

**Explanation** :

Let m∠DAB = *θ* ⇒ m∠BCD = 2*θ*

□PBCD is a parallelogram.

∴ m∠DPB = 2*θ*

m∠PBC = m∠PDC = (180 – 2*θ*)

∠DPB is an exterior angle of ∆PAB.

∴ By exterior angle theorem, m∠PBA = *θ*

as, m∠DAB = *θ*

Thus, in ∆PAB, PA = PB

∴ m∠ABC = *θ* + (180 – 2*θ*) = 180 – *θ*

According to the given conditions, 10*x* + *y* = 1120 and 10*x* = 1000

Solving the two equations, we get *x* = 100 and *y* = 120

sin (180 – *θ*) = sin *θ*

Applying sine rule to ∆PAB,

$\frac{100}{\mathrm{sin}\mathrm{\theta}}=\frac{120}{\mathrm{sin}(180-2\mathrm{\theta})}$

$\frac{}{}\phantom{\rule{0ex}{0ex}}\frac{100}{\mathrm{sin}\mathrm{\theta}}=\frac{120}{\mathrm{sin}2\mathrm{\theta}}.....\left(\mathrm{i}\right)$

sin 2θ = 2 × sin θ × cos θ

∴ (i) becomes

$\frac{100}{\mathrm{sin}\theta}=\frac{120}{2\times \mathrm{sin}\theta \times \mathrm{cos}\theta}$

$\frac{}{}\phantom{\rule{0ex}{0ex}}\therefore \mathrm{cos}\theta =\frac{3}{5}\Rightarrow \mathrm{sin}\theta =\frac{4}{5}$

$\frac{\mathrm{}}{}\phantom{\rule{0ex}{0ex}}\therefore \mathrm{sin}\angle ABC=\mathrm{sin}(180-\theta )=\mathrm{sin}\theta =\frac{4}{5}$

Hence, option (a).

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**IIFT 2009 QA | Geometry - Quadrilaterals & Polygons omet Question**

If there is threefold increase in all the sides of a cyclic quadrilateral, then the percentage increase in its area will be:

- (a)
81%

- (b)
9%

- (c)
900%

- (d)
None of the above

Answer: Option D

**Explanation** :

A square is a cyclic quadrilateral. So we can find the answer using a square.

The area of a square with side *x* is *x*^{2}

The area of a square with side 3*x* is 9*x*^{2}

∴ The percentage increase in area when all sides increase three fold = 800%.

Hence, option (d).

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