# SSC > Area and Perimeter

Explore popular questions from Area and Perimeter for SSC. This collection covers Area and Perimeter previous year SSC questions hand picked by experienced teachers.

General Intelligence and Reasoning
General Awareness
Quantitative Aptitude
English Comprehension
Q 1.

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Incorrect-0.5

The sides of a triangle area in the ratio of {tex} \frac{1}{3}:\frac{1}{4}:\frac{1}{5} {/tex} and its perimeter is 94 Cm.Find the length of the smallest side of the triangle.

A

18 cm

B

22.5 cm

24 m

D

27 cm

##### Explanation

Given ratio = {tex} \frac{1}{3}:\frac{1}{4}:\frac{1}{5} {/tex}

Let lengths of the sides be 20x,15x and 12x.

Then, according to the question,

20x + 15x + 12x = 94 =>47x = 94

x = {tex} \frac{94}{47} {/tex}=2

Smallest side =12x = {tex} 12\times 2 {/tex} = 24 cm

Q 2.

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The ratio of length of each equal side and the third side of an isosceles triangle is 3:4. If the area of the triangle is {tex} 18\sqrt{5} {/tex} sq units, the third side is

A

{tex} 8\sqrt{2} {/tex}units

12 units

C

16 units

D

{tex} 5\sqrt{10} {/tex}units

##### Explanation

Let sides of isosceles triangle are 3x,3x and 4x.

Then, half-perimeter (s) = {tex} \frac{a+b+c}{2} {/tex}

= {tex} \frac{3x+3x+4x}{2} {/tex}=5x

Given, area of isosceles triangle

= {tex} 18\sqrt{5} {/tex}sq units

{tex} \sqrt{s(s-a)(s-b)(s-c)}=18\sqrt{5} {/tex}

{tex} \sqrt{5x(5x-3x)(5x-3x)(5x-4x)}=18\sqrt{5} {/tex}

{tex} \sqrt{5x\times 2x\times 2x\times x}=18\sqrt{5} {/tex}

=> {tex} 2\sqrt{5}x^{2}=18\sqrt{5} {/tex}

=> {tex} x^{2}=9 {/tex}

=> x=3

Third side of isosceles triangle

={tex} 4\times 3 {/tex}= 12units

Q 3.

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Three sides of a triangular field are of length 15 m, 20 m and 25 m long, respectively. Find the cost of sowing seeds in the field at the rate of RS. 5 per Sq m.

750

B

150

C

300

D

600

##### Explanation

Since,{tex} AC^{2}=AB^{2}+BC^{2} {/tex}

=> {tex} (25)^{2}=(15)^{2}+(20)^{2} {/tex}

=> 625=225+400

=> 625 = 625

So, the triangular field is right angled at B

Area of the field = {tex} \frac{1}{2}\times AB\times BC {/tex}

= {tex} \frac{1}{2}\times 15\times 20 {/tex}

={tex} 150m^{2} {/tex}

So, the cost of sowing seed is RS. 5 per sq m

Cost of sowing seed for

{tex} 150m^{2}=150\times 5 {/tex}

= RS. 750

Q 4.

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The length of a rectangular field is 100 m and its breadth is 40 m. What will be the area of the field?

A

{tex} (4\times 10^{2}) {/tex}sq m

B

{tex} (4\times 10) {/tex}sq m

C

{tex} (4\times 10^{4}) {/tex}sq m

{tex} (4\times 10^{3}) {/tex}sq m

##### Explanation

Requnred area = {tex} Length\times Breadth {/tex}

= {tex} 100\times 40 {/tex}=4000 sq m

= {tex} (4\times 10^{3}) {/tex} sq m

Q 5.

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The difference between the length and breadth of a rectangle is 23 m. If its perimeter is 206 m, then its area is

{tex} 2520 m^{2} {/tex}

B

{tex} 2480 m^{2} {/tex}

C

{tex} 2420 m^{2} {/tex}

D

None of the above

##### Explanation

Let the length of rectangle = L m

Breadth of rectangle = B m

using conditions from the question,

L - B = 23 ....(i)

2(L + B) = 206

L + B = 103 ....(ii)

On adding Eqs. (i) and (ii), we get

2L = 126

=> L = 63 m

=> B = 103-63=40m

Then, area of rectangle = {tex} L\times B {/tex}

= {tex} 63\times 40 {/tex}

= {tex} 2520 m^{2} {/tex}

Q 6.

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If the length of a rectangle decreases by 5 m and breadth increases by 3 m, then its area reduces 9 sq m. If length and breadth of this rectangle increased by 3 m and 2 m respectively, then its area increased by 67 sq m. What is the length of rectangle?

A

9 m

B

15.6 m

17 m

D

18.5 m

##### Explanation

Let length and breadth of a rectangle is x and y. Then, as per first condition,

(x-5)(y+3)=xy-9

=> xy-5y+3x-15=xy-9

=> 3x-5y=6 .....(i)

As per second condition,

(x+3)(y+2)=xy+67

=> xy+3y+2x+6=xy+67

=> 2x + 3y = 61 .....(ii)

On multiplying Eq. (i) by 3 and Eq. (ii) by 5 then adding, we get

9x - 15y = 18
10x + 15y = 305
----------------
=> 19x = 323

=> x = {tex} \frac{323}{19} {/tex}=17

Hence, the length of rectangle is 17 m

Q 7.

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The area of a rectangle whose length is 5 more than twice its width is 75 sq units. What is the perimeter of the rectangle?

40 units

B

30 units

C

24 units

D

20 units

##### Explanation

Let the width of the rectangle = x units

Length = (2 x + 5) units

According to the question,

Area = x(2x + 5)

=> 75 = {tex} 2x^{2}+5x {/tex}

=> {tex} 2x^{2} +5x-75=0{/tex}

=> {tex} 2x^{2} +15x-10x-75=0{/tex}

=> x(2x+15)-5(2x+15)=0

=> (2x+15)(x-5)=0

=> x = 5 and {tex} \frac{-15}{2} {/tex}

Width cannot be negative.

Width = 5 units

Length=2x+5 ={tex} 2\times 5+5 {/tex}=15 units

Perimeter of the rectangle

= 2(15 + 5) = 40 units

Q 8.

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Floor of a square room of side 10 m is to be completely covered with square tiles, each having length 50 cm. The smallest number of tiles needed is

A

200

B

300

400

D

500

##### Explanation

Area of square room = {tex} (10)^{2} {/tex} = 100 sq

= {tex} 100\times (100)^{2} {/tex} sq cm

= {tex} 100\times 100\times 100 {/tex} sq cm

Now, area of tile = {tex} (50)^{2}=50\times 50 {/tex} sq cm

Number of tiles needed

= {tex} \frac{Area \ of \ square \ room}{Area \ of \ tile} {/tex}

= {tex} \frac{100\times 100\times 100}{50\times 50} {/tex} =400

Hence, 400 tiles will be needed.

Q 9.

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The area of a trapezium is {tex} 384 cm^{2} {/tex}. If its parallel sides are in the ratio 3 : 5 and the perpendicular distance between them is 12 cm, the smaller of the parallel sides is

A

20 cm

24 cm

C

30 cm

D

36 cm

##### Explanation

Let the sides of trapezium be 5x and respectively.

According to the question,

{tex} \frac{1}{2}\times(5x+3x)\times 12 {/tex}=384

=> 8x={tex} \frac{384\times 2}{12} {/tex} => x={tex} \frac{64}{8} {/tex}=8 cm

Length of smaller of the parallel sides

= {tex} 8\times 3 {/tex} = 24 cm

Q 10.

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In a trapezium, the two non-parallel sides are equal in length, each being of 5 units. The parallel sides are at a distance of 3 units apart. If the smaller side of the parallel sides is of length 2 units, then the sum of the diagonals of the trapezium is

A

{tex} 10\sqrt{5} {/tex} units

{tex} 6\sqrt{5} {/tex} units

C

{tex} 5\sqrt{5} {/tex} units

D

{tex} 3\sqrt{5} {/tex} units

##### Explanation

In {tex} \triangle BCF {/tex}, by Pythagoras theorem,

{tex} (5)^{2}=(3)^{2}+(BF)^{2} {/tex}

BF=4 cm

AB=2+4+4=10 cm

Now, in {tex} \triangle ACF {/tex}, {tex} AC^{2}=CF^{2}+FA^{2} {/tex}

=> {tex} AC^{2}=3^{2}+6^{2} {/tex} => AC={tex} \sqrt{45} {/tex}cm

Similarly, BD = {tex} \sqrt{45} {/tex}cm

Sum of diagonals

=AC+BD= {tex} \sqrt{45}+\sqrt{45}=2\sqrt{45}=6\sqrt{5} {/tex}cm

Q 11.

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If area of a regular pentagon is {tex} 125\sqrt{3} {/tex} units sq cm, how long is its each side

10 cm

B

15 cm

C

16 cm

D

25 cm

##### Explanation

We know that,

Area of regular pentagon = {tex} 5a^{2}\frac{\sqrt{3}}{4} {/tex}

According to the question,

{tex} \frac{5a^{2}\sqrt{3}}{4}=125\sqrt{3} {/tex}=> {tex} a^{2}=\frac{125\sqrt{3}\times 4}{5\sqrt{3}} {/tex}=100

a = 10 cm

Q 12.

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The area of a sector of a circle of radius 36 cm is {tex} 72\pi cm^{2} {/tex}. The length of the Corresponding are of the sector 1s

A

{tex} \pi cm {/tex}

B

{tex} 2\pi cm {/tex}

C

{tex} 3\pi cm {/tex}

{tex} 4\pi cm {/tex}

##### Explanation

Given, area of sector = {tex} 72\pi cm^{2} {/tex}

or {tex} \frac{\pi r^{2}\theta}{360 degree}=72\pi {/tex}

{tex} \theta=\frac{72\times 360}{36\times 36} {/tex}=> {tex} \theta {/tex}=20 degree

Now,length of arc

={tex} \frac{\pi r\theta}{180 degree}=\frac{\pi\times 36\times 20 degree}{180 degree}=4\pi cm {/tex}

Q 13.

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The radius of a circle is so increased that its circumference increased by 5%. The area of the circle, then increases by

A

12.50%

10.25%

C

10.50%

D

11.25%

##### Explanation

Increase in Circumference of circle = 5%
Increase in radius is also 5%
Now, increase in area of circle
={tex} \left(2a+\frac{a^{2}}{100} \right)% {/tex}

={tex} \left(2\times 5+\frac{5\times 5}{100} \right)\% {/tex}=10.25%

Q 14.

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What is the total area of three equilateral triangles inscribed in a semi-circle of radius 2cm?

A

{tex} 12 cm^{2} {/tex}

B

{tex} \frac{3\sqrt{3}}{4}cm^{2} {/tex}

C

{tex} \frac{9\sqrt{3}}{4}cm^{2} {/tex}

{tex} 3 \sqrt{3}cm^{2} {/tex}

##### Explanation

since, {tex} \triangle's {/tex} AOB,BOC,COD are equilateral.

Sides = 2 cm

Now, total area = {tex} 3\times {/tex}{tex} \frac{\sqrt{3}}{4}(Side)^{2} {/tex}

= {tex} 3\times {/tex}{tex} \frac{\sqrt{3}}{4}\times (Side)^{2}=3\sqrt{3} cm^{2} {/tex}

Q 15.

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A circle of radius 10 cm has an equilateral triangle inscribed in it. The length of the perpendicular drawn from the centre to any side of the triangle is

A

{tex} 2.5\sqrt{3} {/tex}cm

B

{tex} 5\sqrt{3} {/tex}cm

C

{tex} 10\sqrt{3} {/tex}cm

None of these

##### Explanation

Height ={tex} \frac{10\times 3}{2} {/tex}=15 cm

So,length of perpendicular drawn from centre = 15 - 10 = 5 cm

Q 16.

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AB and CD are two diameters of a circle of radius r and they are mutually perpendicular. What is the ratio of the area of the circle to the area of the triangle ACD?

A

{tex} \frac{\pi}{2} {/tex}

{tex} \pi {/tex}

C

{tex} \frac{\pi}{4} {/tex}

D

{tex} 2\pi {/tex}

##### Explanation

Required ratio ={tex} \frac{Area \ of \ circle}{Area \ of \ \triangle ACD} {/tex}

={tex} \frac{\pi r^{2}}{\frac{1}{2}\times 2r\times r}=\pi {/tex}

Q 17.

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The area of a rectangle is 1.8 times the area of a square. The length of the rectangle is 5 times the breadth. The side of the square is 20 cm. What is the perimeter of the rectangle?

A

145 cm

144 cm

C

133 cm

D

135 cm

##### Explanation

Area of square ={tex} (Side)^{2}=20^{2} {/tex} = 400 sq cm

Area of rectangle ={tex} 1.8\times 400 {/tex} = 720 sq cm

Let length and breadth of rectangle be 5x and x, respectively.

Then, according to the question.

{tex} 5x\times x {/tex}=720

=> {tex} 5x^{2} {/tex}=720 => {tex} x^{2}=\frac{720}{5} {/tex}=144

={tex} \sqrt{144} {/tex}=12 cm

Perimeter of rectangle = 2 (5x + x) = 12x

={tex} 12\times 12 {/tex} = 144cm

Q 18.

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The side of a square is 5 cm which is 13 cm less than the diameter of a circle. What is the approximate area of the circle?

A

245 sq cm

B

235 sq cm

C

265 sq cm

255 sq cm

##### Explanation

Diameter of the circle = 13 + 5 = 18cm

Area of the circle = {tex} \pi r^{2}=\frac{22}{7}\times 9^{2} {/tex}

= {tex} \frac{22\times 81}{7}=\frac{1782}{7} {/tex}=254.57 sq cm.

=255 sq cm.

Q 19.

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If area of a square is 64 sq cm, then find the area of the circle formed by the same perimeter.

A

{tex} \frac{215}{\pi} {/tex}sq cm

B

{tex} \frac{216}{\pi} {/tex}sq cm

{tex} \frac{256}{\pi} {/tex}sq cm

D

{tex} \frac{318}{\pi} {/tex}sq cm

##### Explanation

Area of square = 64 sq cm

{tex} (Side)^{2} {/tex}= 64

Side = {tex} \sqrt{64} {/tex} = 8 cm

According to the question,

=> {tex} 2\pi r=4\times 8 {/tex} => r={tex} \frac{4\times 8}{2\pi}=\frac{16}{\pi} {/tex}

Area of the circle

={tex} \pi \times 16/ \pi\times 16/ \pi=\frac{256}{\pi} {/tex}sq cm.

Q 20.

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The perimeter of a square is twice the perimeter of a rectangle. If the perimeter of the square is 72 cm and the length of the rectangle is 12 cm. what is the difference between the breadth of the rectangle and the Side of the square?

A

9 cm

12 cm

C

18 cm

D

3 cm

##### Explanation

Perimeter of square = 72 cm

Perimeter of rectangle

={tex} \frac{72}{2} {/tex}=36 cm

=> {tex} 2\times (l+b) {/tex}=36

[l and b are dimension of rectangle]

{tex} 2\times (12+b) {/tex}=36

=> 12 + b = 18

b=18-12=6cm

And side of square = {tex} \frac{72}{4} {/tex}=18

Difference between breadth of rectangle and side of the square

= 18 - 6 = 12 cm.

Q 21.

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The area of a square is twice the area of a circle. The area of the circle is 392 sq cm. Find the length of the side of the square.

28 cm

B

26 cm

C

24 cm

D

22 cm

##### Explanation

According to the question,

Area of Square ={tex} 2\times 392 {/tex}

=> {tex} a^{2} {/tex}=784

a={tex} \sqrt{784} {/tex}=28 cm.

Q 22.

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The area of a rectangle is equal to the area of a circle with circumference equal to 39.6 m. What is the length of the rectangle if its breadth is 4.5 m?

A

33.52 m

B

21.63 m

C

31.77 m

27.72 m

##### Explanation

Let the radius of the circle be r.

Perimeter of given circle = 39.6 m

{tex} 2\pi r {/tex}=39.6 => ={tex} \frac{7\times 39.6}{22\times 2} {/tex}=6.3m

Area of circle = {tex} \frac{22}{7}\times 6.3\times 6.3 {/tex}

= {tex} \frac{873.18}{7}=124.74 m^{2} {/tex}

Now, area of a rectangle = Area of a circle

{tex} l\times b {/tex}=124.74 => {tex} l\times 4.5 {/tex}=124.74

l={tex} \frac{124.74}{4.5} {/tex}=27.72 m.

Q 23.

Correct2

Incorrect-0.5

If the area of a circle is equal to the area of square with side {tex} 2\sqrt{\pi} {/tex} units, what is the diameter of the circle ?

A

1 unit

B

2 units

4 units

D

8 units

##### Explanation

Area of the circle = Area of the square

={tex} (Side)^{2} {/tex}

{tex} \pi r^{2}=(2\sqrt{\pi})^{2} {/tex}

=> {tex} \pi r^{2}=4\pi {/tex}

=> {tex} r^{2}=\frac{4\pi}{\pi} {/tex}=4

=> r={tex} \sqrt{4} {/tex}=2 units

Diameter of circle (d) = {tex} 2\times r=2\times 2 {/tex}

= 4 units

Q 24.

Correct2

Incorrect-0.5

Consider the following statements.
The area of square is greater than the area of the triangle.
The area of circle is less than the area of triangle.
Which of the statement is/are correct?

Only I

B

Only II

C

Both l and II

D

Neither I nor II

##### Explanation

Let the radius of circle is 'r' and a side of a square is 'a', then given condition

{tex} 2\pi r {/tex}=4a => a={tex} \frac{\pi r}{2} {/tex}

Area of square

={tex} \left(\frac{\pi r}{2}\right)^{2}=\frac{\pi^{2}r^{2}}{4}=\frac{9.86 r^{2}}{4}=2.46 r^{2} {/tex}

and area of circle = {tex} \pi r^{2}=3.14 r^{2} {/tex}

and let the side of equilateral triangle is x.

Then, given condition,

3x={tex} 2\pi r {/tex}=> x={tex} \frac{2\pi r}{3} {/tex}

Area of equilateral triangle = {tex} \frac{\sqrt{3}}{4}x^{2} {/tex}

={tex} \frac{\sqrt{3}}{4}{/tex}{tex} \times {/tex}{tex} \frac{4\pi^{2}r^{2}}{9} {/tex}

={tex} \frac{\pi^{2}}{3\sqrt{3}}r^{2}=1.89 r^{2} {/tex}

Hence, Area of circle > Area of square > Area of equilateral triangle.

Q 25.

Correct2

Incorrect-0.5

Which one of the following is a Pythagorean triple in which one side differs from the hypotenuse by two units?

A

{tex} (2n+1,4n,2n^{2}+2n) {/tex}

B

{tex} (2n,4n,n^{2}+1) {/tex}

C

{tex} (2n^{2},2n,2n+1) {/tex}

{tex} (2n,n^{2}-1,n^{2}+1) {/tex}

##### Explanation

By hit and trial method.

Put,n = 2 in option (d)

= {tex} (2\times 2),(2)^{2}-1,(2)^{2}+1] {/tex}=(4,3,5)

Which satisfy pythagoras theorem and one side differes from hypnotenuse by 2 units.