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

You can practice all previous year CAT questions from the topic Geometry - Quadrilaterals & Polygons. This will help you understand the type of questions asked in CAT. It would be best if you clear your concepts before you practice previous year CAT questions.

**CAT 2021 QA Slot 1 | Geometry - Quadrilaterals & Polygons**

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

- A.
√14

- B.
√12

- C.
√13

- D.
√15

Answer: Option C

**Explanation** :

Side of the regular hexagon = 2 cm.

Consider the figure below.

Consider the isosceles ∆ATF

TU is the altitude from T to AF.

We know, in a regular hexagon the distance between any two parallel sides = √3 × side.

∴ TU = √3 × 2 = 2√3 cm.

Since ATF is an isosceles triangle, U will be the mid-point of AF

∴ AU = 1 cm.

In ∆ATU

AT^{2} = TU^{2} + AU^{2}

⇒ AT^{2} = (2√3)^{2} + 1^{2}

⇒ AT^{2} = 13

⇒ AT = √13

Hence, option (c).

Workspace:

**CAT 2021 QA Slot 2 | Geometry - Quadrilaterals & Polygons**

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

- A.
20

- B.
30

- C.
40

- D.
25

Answer: Option B

**Explanation** :

Refer the diagram below.

ABPD is a parallelogram.

ABPD and ∆BPC have equal base and same height.

∴ Area (ABPD) = 2 × Area (∆BPC)

Also, Area (ABPD) - Area (∆BPC) = 10

⇒ 2 × Area (∆BPC) - Area (∆BPC) = 10

⇒ Area (∆BPC) = 10

∴ Area (ABPD) = 20

⇒ Area (ABCD) = Area (ABPD) + Area (∆BPC) = 20 + 10 = 30

Hence, option (b).

Workspace:

**CAT 2021 QA Slot 2 | Geometry - Quadrilaterals & Polygons**

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

- A.
√13 + √12

- B.
√37 + √13

- C.
(√13 + √12)/2

- D.
(√37 + √13)/2

Answer: Option B

**Explanation** :

Consider the diagram below.

In a rhombus diagonals perpendicularly bisect each other.

∴ Let OA = OC = x and OB = OD = y

In ∆AOB ⇒ x^{2} + y^{2} = 52^{2} …(1)

Also, area of the rhombus = ½ × 2x × 2y = 12

⇒ 2xy = 12 …(2)

(1) + (2)

⇒ x^{2} + y^{2} + 2xy = 25 + 12 = 37

⇒ (x + y)^{2} = 37

⇒ x + y = √37 …(3)

(1) - (2)

⇒ x^{2} + y^{2} - 2xy = 25 - 12 = 13

⇒ (x - y)^{2} = 13

⇒ x - y = √13 …(4)

Solving (3) and (4) we get,

2x = √37 + √13 and

2y = √37 - √13

∴ The longer diagonal = 2x = √37 + √13

Hence, option (b).

Workspace:

**CAT 2021 QA Slot 3 | Geometry - Quadrilaterals & Polygons**

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 INR, of laying electric wires along its two diagonals, at the rate of ₹125 per m, is

Answer: 3500

**Explanation** :

Side of the rhombus = ¼ × 40 = 10 cm.

Area of a rhombus = ½ d_{1}d_{2} = 96 [d_{1} and d_{2} are the two diagonals]

⇒ d_{1}d_{2} = 192

We know, (d_{1}/2)^{2} + (d_{2}/2)^{2} = 102

⇒ d_{1}^{2} + d_{2}^{2} = 400

⇒ (d_{1} + d_{2})^{2} - 2d_{1}d_{2} = 400

⇒ (d_{1} + d_{2})^{2} – 2 × 192 = 400

⇒ (d_{1} + d_{2})^{2} = 784

⇒ d_{1} + d_{2} = 28

Hence, the cost of laying electric wore along its two diagonals = 125 × (d_{1} + d_{2}) = 125 × 28 = Rs. 3,500

Hence, 3500.

Workspace:

**CAT 2021 QA Slot 3 | Geometry - Quadrilaterals & Polygons**

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 is sq. cm. is

- A.
25(√3 + √15)/2

- B.
25(√5 + √15)

- C.
25(√3 + √15)

- D.
25(√5 + √15)/2

Answer: Option C

**Explanation** :

Let AE be a perpendicular from A to DC.

∆ADE is a 30-60-90 triangle.

⇒ DE = √3/2 × 10 = 5√3

⇒ AE = 10/2 = 5

∆ACE is a 30-60-90 triangle.

⇒ CE^{2} = AC^{2} – AE^{2}

⇒ CE^{2} = 400 – 25 = 375

⇒ CE = 5√15

∴ Area of parallelogram = 2 × Area of triangle ADC

⇒ Area of parallelogram = 2 × (1/2 × CD × AE) = CD × AE

= (DE + EC) × AE

= (5√3 + 5√15) × 5

= 25(√3 + √15)

Hence, option (c).

Workspace:

**CAT 2020 QA Slot 1 | Geometry - Quadrilaterals & Polygons**

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

- A.
5π/18

- B.
2π/15

- C.
3π/25

- D.
6π/25

Answer: Option D

**Explanation** :

The following diagram can be drawn from the given information.

⇒ Area of the Rhombus = $\frac{1}{2}\times 12\times 16$ = 96

Radius of the circle is same as the height of ∆ABC

In a right triangle, altitude from right angle = $\frac{Productofshortersides}{hypotenuese}$ = $\frac{\left(6\times 8\right)}{10}$ = $\frac{24}{5}$

⇒ Area of circle = $\pi {\left(\frac{24}{5}\right)}^{2}$

∴ Area of Circle : Area of Rhombus = $\pi {\left(\frac{24}{5}\right)}^{2}:96$ = 6π : 25.

Hence, option (d).

Workspace:

**CAT 2020 QA Slot 2 | Geometry - Quadrilaterals & Polygons**

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

- A.
27

- B.
24

- C.
18

- D.
21

Answer: Option A

**Explanation** :

Let the side of equilateral triangle be a. Hence, perimeter = 3a.

Let the sides of the rectangle be x and 3x. Hence, perimeter = 8x

Given, 3a + 8x = 90 …(1)

Also given, R = T^{2}

⇒ x × 3x = ${\left[\frac{\sqrt{3}}{4}{a}^{2}\right]}^{2}$

⇒ 16x^{2} = a4

⇒ a^{2} = 4x

Substituting x in (1)

⇒ 3a + 2a^{2} = 90

⇒ 2a^{2} + 3a – 90 = 0

⇒ a = -15/2 or 6 (-ve value will be rejected)

∴ x = $\frac{{a}^{2}}{4}$ = 9

∴ Longer side of the rectangle = 3x = 27.

Hence, option (a).

Workspace:

**CAT 2020 QA Slot 3 | Geometry - Quadrilaterals & Polygons**

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

Answer: 28

**Explanation** :

The following diagram can be drawn using the information given.

Since, ∆ADE is an isosceles right triangle hence, DE = AE = 4 cm

⇒ AB = BE + AE = 5 + 4 = 9 cm

∴ Area of trapezium = $\frac{1}{2}\times (CD+AB)\times BC$ = $\frac{1}{2}\times 14\times 4$ = 28 cm^{2}.

Hence, 28.

Workspace:

**CAT 2020 QA Slot 3 | Geometry - Quadrilaterals & Polygons**

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

- A.
15

- B.
12

- C.
13

- D.
14

Answer: Option D

**Explanation** :

One of the diagonals of the parallelogram connects the points (2, 1) and (-3, -4).

Line x + 9y + c = 0, represents the other diagonal of the parallelogram.

We know, diagonals of a parallelogram bisect each other.

∴ x + 9y + c = 0 should pass through the midpoint of the line joining (2, 1) and (-3, -4).

The midpoint of the line joining (2, 1) and (-3, -4) = $\left(\frac{2-3}{2},\frac{1-4}{2}\right)$ = (-0.5, -1.5)

⇒ Point (-0.5 , -1.5) lies on x + 9y + c = 0.

⇒ -0.5 + 9 × (-1.5) + c = 0

⇒ -0.5 - 13.5 + c = 0

⇒ c = 14.

Hence, option (d).

Workspace:

**CAT 2019 QA Slot 2 | Geometry - Quadrilaterals & Polygons**

Let A and B be two regular polygons having a and b sides, respectively. If b = 2a and each interior angle of B is 3/2 times each interior angle of A, then each interior angle, in degrees, of a regular polygon with a + b sides is

Answer: 150

**Explanation** :

Sum of the interior angles of a n-sided Polygon = (n - 2) × 180

Measure of each interior angle = $\frac{(n-2)}{n}$× 180°

So, Interior angle of the polygon with side a = $\frac{(a-2)}{a}$ × 180°

Interior angle of the polygon with side b = $\frac{(2a-2)}{2a}$ × 180°

It is given that interior angle of side b is 3/2 times to that of the polygon with side A

⇒ $\frac{(2a-2)}{2a}$ × 180° = $\frac{3}{2}\left[\frac{(a-2)}{a}\times 180\xb0\right]$

⇒ (2a - 2) × 180 = (3a - 6) × 180

⇒ 2a - 2 = 3a - 6

⇒ a = 4. So, b = 8

Now, (a + b) sides = 8 + 4 = 12 sides

Interior angle =$\frac{12-2}{12}$× 180° = 150°

Hence, 150.

Workspace:

**CAT 2018 QA Slot 1 | Geometry - Quadrilaterals & Polygons**

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

- A.
2 : 5

- B.
4 : 9

- C.
3 : 8

- D.
1 : 3

Answer: Option D

**Explanation** :

y > x

Side of smaller square = $\sqrt{{x}^{2}+{y}^{2}}$

By the given condition,

x^{2} + y^{2} = 62.5% of (x + y)^{2}

x^{2} + y^{2} = 5/8 (x + y)^{2}

8x^{2} + 8y^{2} = 5x^{2} + 5y^{2} + 10xy

3x^{2} + 3y^{2} - 10xy = 0

3${\left(\frac{x}{y}\right)}^{2}$ -10$\left(\frac{x}{y}\right)$ + 3 = 0

$\left(\frac{x}{y}\right)$ = $\frac{10\pm \sqrt{{10}^{2}-4\times 3\times 3}}{6}$ = $\frac{10\pm 8}{6}$ = 9/3 of 1/3

As y > x, x/y < 1

∴ x : y = 1 : 3

Hence, option 4.

Workspace:

**CAT 2018 QA Slot 1 | Geometry - Quadrilaterals & Polygons**

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

- A.
12√3

- B.
24√3

- C.
18√3

- D.
32√3

Answer: Option D

**Explanation** :

Area of parallelogram = base × height

∴ A(□ABCD) = CD × AP

∴ 72 = 9 × AP

∴ AP = 8

Consider ∆APD.

PD = $\sqrt{{16}^{2}+{8}^{2}}$ = 8√3

A(APD) = 1/2 × 8√3 × 8 = 32√3 sq.cm.

Hence, option 4.

Workspace:

**CAT 2018 QA Slot 1 | Geometry - Quadrilaterals & Polygons**

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?

- A.
24, 12

- B.
24, 10

- C.
25, 10

- D.
25, 9

Answer: Option B

**Explanation** :

As □ABCD is a rectangle, AC = DB = 2 × 13 = 26 cm

AB^{2} + BC^{2} = 26^{2} = 676

(24^{2} + 10^{2}) = 676

Hence, option 2.

Workspace:

**CAT 2018 QA Slot 2 | Geometry - Quadrilaterals & Polygons**

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?

- A.
s ≠ 6

- B.
s ≥ 6

- C.
5 ≤ s ≤ 7

- D.
s ≤ 6

Answer: Option B

**Explanation** :

Area of parallelogram = Base × Height

∴ 48 = 8 × Height

∴ Height = 6 cm

If the parallelogram is a rectangle, AD is its height. In that case, s = 6.

Otherwise, using Pythagoras theorem,

AD > 6 or s > 6.

Therefore, s ≥ 6.

Hence, option 2.

Workspace:

**CAT 2018 QA Slot 2 | Geometry - Quadrilaterals & Polygons**

The area of a rectangle and the square of its perimeter are in the ratio 1 ∶ 25. Then the lengths of the shorter and longer sides of the rectangle are in the ratio ?

- A.
1 : 3

- B.
2 : 9

- C.
1 : 4

- D.
3 : 8

Answer: Option C

**Explanation** :

Suppose l = length of the rectangle and b = breadth of the rectangle.

Therefore, the area of the rectangle = lb and the perimeter of the rectangle = 2(l + b)

Therefore,

$\frac{\mathcal{l}b}{{\left[2\right(l+b\left)\right]}^{2}}=\frac{1}{25}i.e.\frac{\iota b}{{(l+b)}^{2}}=\frac{4}{25}$

Substituting the options, we can see that only option [3] i.e. 1:4 matches.

Hence, option 3.

Workspace:

**CAT 2017 QA Slot 2 | Geometry - Quadrilaterals & Polygons**

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

- A.
$3\sqrt{2}$

- B.
3

- C.
4

- D.
$\sqrt{3}$

Answer: Option B

**Explanation** :

ABCDEF is a hexagon with side 1 cm. Let us construct the diagram as shown below

Now if we look at triangle ABC

(∵ ∠ABC is an angle of regular hexagon ABCDEF)

Further, as AB = BC, △ ABC will become a 120° -30° -30° triangle and the ratio of it’s sides

will be $\sqrt{3}$ : 1 : 1

Now $\frac{AC}{AB}=\frac{\sqrt{3}}{1}$

∴ AC = $\sqrt{3}$ units

Area of square with side AC = $\sqrt{3}\times \sqrt{3}$ = 3 sq.units

Hence, option 2.

Workspace:

**CAT 2017 QA Slot 2 | Geometry - Quadrilaterals & Polygons**

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

Answer: 90

**Explanation** :

Now in ∆ COD, OC = OD (∵ they are radii of the circle)

∴ ∠OCD = ∠ODC

∴ ∠ODC = ∠OCD = $\frac{180-120}{2}$ = 30°

Further, since OB and OC are radius of the circle they will be equal

∴ ∠OBC will be equal to ∠OCB and

both will be equal to $\frac{180-60}{2}=\frac{120}{2}$ = 60°

Now ∠BOC = 180° - 120° = 60°

= 90∙

Answer: 90

Workspace:

**CAT 2008 QA | Geometry - Quadrilaterals & Polygons**

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?

- A.
$\frac{4\sqrt{2}}{3}$

- B.
$2+\sqrt{3}$

- C.
$\frac{10-3\sqrt{3}}{9}$

- D.
$1+\frac{1}{\sqrt{3}}$

- E.
$2\sqrt{3}-1$

Answer: Option E

**Explanation** :

Let the length of the sides of the square be 2s.

Consider ∆BQF,

BF = s

In 30° - 60° - 90° triangle,

QF = $\frac{s}{\sqrt{3}}$

∴ Area of ∆BQF = $\frac{1}{2}$ × s × $\frac{s}{\sqrt{3}}$

Area of ABQCDP = Area of square ABCD – 4 × Area of ∆BQF $=4{s}^{2}-4\left(\frac{1}{2}\times s\times \frac{s}{\sqrt{3}}\right)$

∴ Required ratio = $\frac{4{s}^{2}-4\left({\displaystyle \frac{1}{2}}\times s\times {\displaystyle \frac{s}{\sqrt{3}}}\right)}{4\left({\displaystyle \frac{1}{2}}\times s\times {\displaystyle \frac{s}{\sqrt{3}}}\right)}$ = $2\sqrt{3}-1$

Hence, option (e).

Workspace:

**CAT 2007 QA | Geometry - Quadrilaterals & Polygons**

**Each question is followed by two statements A and B. Answer each question using the following instructions.**

Mark (1) if the question can be answered by using statement A alone but not by using statement B alone.

Mark (2) if the question can be answered by using statement B alone but not by using statement A alone.

Mark (3) if the question can be answered by using both the statements together but not by using either of the statements alone.

Mark (4) if the question cannot be answered on the basis of the two statements.

Rahim plans to draw a square JKLM with a point O on the side JK but is not successful. Why is Rahim unable to draw the square?

A. The length of OM is twice that of OL.

B. The length of OM is 4 cm.

- A.
1

- B.
2

- C.
3

- D.
4

- E.
5

Answer: Option A

**Explanation** :

Let p be the side of square JKLM.

From Statement A,

OM = 2 × OL

OM is maximum when it is the diagonal of the square and has length $\sqrt{2}p$

When OM is maximum, $OM=\sqrt{2}\times OL$

∴ OM ≠ 2 × OL if O lies on JK.

∴ Rahim is unable to draw the square.

Statement B offers no additional or relevant information.

Hence, option (a).

Workspace:

**CAT 2006 QA | Geometry - Quadrilaterals & Polygons**

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

- A.
75

- B.
90

- C.
120

- D.
135

- E.
150

Answer: Option E

**Explanation** :

BP = PC = BC

m ∠BPC = m ∠PCB = m ∠PBC = 60°

Also, PC = CD = BP = AB

∆ ABP and ∆ PCD are isosceles triangles.

m ∠ABP = m ∠PCD = 90 – 60 = 30°

∴ m ∠APB = m ∠DPC = (180 – 30)/2 = 75°

∴ m ∠APD = 360 – (m ∠APB + m ∠DPC + m ∠BPC) = 360 – (75 + 75 + 60) = 150°

Hence, option 5.

Workspace:

**CAT 2003 QA - Leaked | Geometry - Quadrilaterals & Polygons**

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

- A.
20

- B.
0

- C.
21

- D.
22

Answer: Option C

**Explanation** :

Consider the polygon as shown in the figure. Here 4 corners A, D, G and J are the concave corners and remaining 8 corners are convex corners.

In general, for a polygon with sides parallel to either of the axes, if n is the number of concave corners and m is the number of convex corners, then we have,

m − n = 4

∵ m = 25

∴ n = 21

Hence, option 3.

Workspace:

**CAT 2003 QA - Leaked | Geometry - Quadrilaterals & Polygons**

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?

- A.
$\frac{1}{12}$

- B.
$\frac{1}{6}$

- C.
$\frac{1}{24}$

- D.
$\frac{1}{18}$

Answer: Option A

**Explanation** :

We can see that triangles AMF, AMB, BMC, CMD, DME and EMF are all equilateral triangles. Also, AE, XY and BD bisect MF, AB, ED and CM.

∴ All the small right triangles in the figure are congruent.

∴ A(∆AOF) = $\frac{1}{12}$ × the area of the hexagon ABCDEF

Hence, option 1.

Workspace:

**CAT 2003 QA - Retake | Geometry - Quadrilaterals & Polygons**

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?

- A.
3

- B.
4

- C.
1

- D.
2

Answer: Option D

**Explanation** :

When the tin sheet is cut across its corners as shown in the figure, the box formed will have a height of x inches and its base will be a square of side (12 – 2x) inches.

Let the volume of the box, *V* = (12 – 2*x*)^{2} × *x* = 4*x*^{3} − 48*x*^{2} + 144*x*

For V to be maximum, $\frac{dV}{dx}$ shoule be 0.

i.e. 12*x*^{2} − 96*x* + 144 = 0

∴ 12(x − 6)(x − 2) = 0

∴ x = 2 or x = 6

However, x cannot be 6 as the length of the side is (12 – 2x).

∴ x = 2

Hence, option 4.

Alternatively,

Since *V* = (12 – 2*x*)^{2} × *x* = [2(6 – *x*)]^{2} × *x* = 4*x*(6 – *x*)^{2},

Substituting values of *x* from 1 to 5, we get *V* maximum when *x* = 2 (i.e. *V* = 128)

Workspace:

**CAT 2003 QA - Retake | Geometry - Quadrilaterals & Polygons**

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

- A.
$\frac{1}{3}$

- B.
$\frac{1}{2}$

- C.
$\frac{2}{3}$

- D.
$\frac{5}{6}$

Answer: Option B

**Explanation** :

ABCDEF is a regular hexagon, join centre O with vertices A, C and E.

In quadrilateral AFEO, diagonal EA divides it into two equal area triangles.

i.e. Area (∆AFE) = Area (∆AOE)

Similarly, Area (∆ABC) = Area (∆AOC)

And, Area (∆CDE) = Area (∆COE)

$\therefore \frac{Area(\u2206ACE)}{Area(HexagonABCDEF)}=\frac{3}{6}=\frac{1}{2}$

Hence, option 2.

Workspace:

**CAT 2003 QA - Retake | Geometry - Quadrilaterals & Polygons**

The length of the circumference of a circle equals the perimeter of a triangle of equal sides, and also the perimeter of a square. The areas covered by the circle, triangle, and square are c, t, and s, respectively. Then,

- A.
s > t > c

- B.
c > t > s

- C.
c > s > t

- D.
s > c > t

Answer: Option C

**Explanation** :

Let radius of the circle be r, a side of the equilateral triangle be a, and a side of the square be x.

The circumference/perimeter of the circle, triangle and square are equal. Hence,

2πr = 3a = 4x = k

$\therefore r=\frac{k}{2\pi},a=\frac{k}{3},$ and $x=\frac{k}{4}$

The areas of the circle, triangle and square are c, t, s respectively. Hence,

$\begin{array}{rl}& c=\pi {r}^{2}=\frac{\pi {k}^{2}}{4{\pi}^{2}}=\frac{{k}^{2}}{4\pi},\\ & t=\frac{\sqrt{3}}{4}{a}^{2}=\frac{\sqrt{3}}{4}\times \frac{{k}^{2}}{9}=\frac{{k}^{2}}{12\sqrt{3}}\\ & s={x}^{2}=\frac{{k}^{2}}{16}\\ & \because \frac{1}{\pi}>\frac{1}{4}>\frac{1}{3\sqrt{3}}\\ & \therefore c>s>t\end{array}$

Hence, option 3.

Workspace:

## Feedback

**Help us build a Free and Comprehensive CAT/MBA Preparation portal by providing
us your valuable feedback about Apti4All and how it can be improved.**