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**NCERT solutions for Maths Perimeter and Area ****Download as PDF**

## NCERT Solutions for Class 7 Maths Perimeter and Area

**Class –VII Mathematics (Ex. 11.3)**

**Question 1.**Find the circumference of the circles with the following radius: {tex}\left( {{\text{Take }}\pi {\text{ = }}\frac{{22}}{7}} \right){/tex}

(a) 14 cm

(b) 28 mm

(c) 21 cm

**Answer:**

(a) Circumference of the circle = {tex}2\pi r{/tex} = {tex}2 \times \frac{{22}}{7} \times 14{/tex} = 88 cm

(b) Circumference of the circle = {tex}2\pi r{/tex} = {tex}2 \times \frac{{22}}{7} \times 28{/tex} = 176 mm

(c) Circumference of the circle = {tex}2\pi r{/tex} = {tex}2 \times \frac{{22}}{7} \times 21{/tex} = 132 cm

NCERT Solutions for Class 7 Maths Exercise 11.3

**Question 2.**Find the area of the following circles, given that: {tex}\left( {{\text{Take }}\pi {\text{ = }}\frac{{22}}{7}} \right){/tex}

(a) radius = 14 mm

(b) diameter = 49 m

(c) radius 5 cm

**Answer:**

(a) Area of circle = {tex}\pi {r^2}{/tex} = {tex}\frac{{22}}{7} \times 14 \times 14{/tex} = 22 x 2 x 14 = 616 {tex}m{m^2}{/tex}

(b) Diameter = 49 m

{tex}\therefore {/tex} radius = {tex}\frac{{49}}{2}{/tex} = 24.5 m

{tex}\therefore {/tex} Area of circle = {tex}\pi {r^2}{/tex} = {tex}\frac{{22}}{7} \times 24.5 \times 24.5{/tex} = 22 x 3.5 x 24.5 = 1886.5 {tex}{m^2}{/tex}

Area of circle = {tex}\pi {r^2}{/tex} = {tex}\frac{{22}}{7} \times 5 \times 5{/tex} = = {tex}\frac{{550}}{7}{/tex} {tex}c{m^2}{/tex}

NCERT Solutions for Class 7 Maths Exercise 11.3

**Question 3.**If the circumference of a circular sheet is 154 m, find its radius. Also find the area of the sheet. {tex}\left( {{\text{Take }}\pi {\text{ = }}\frac{{22}}{7}} \right){/tex}

**Answer:**

Circumference of the circular sheet = 154 m

{tex} \Rightarrow {/tex} {tex}2\pi r{/tex} = 154 m {tex} \Rightarrow {/tex} {tex}r = \frac{{154}}{{2\pi }}{/tex}

{tex} \Rightarrow {/tex} {tex}r = \frac{{154 \times 7}}{{2 \times 22}}{/tex} = 24.5 m

Now Area of circular sheet = {tex}\pi {r^2}{/tex} = {tex}\frac{{22}}{7} \times 24.5 \times 24.5{/tex}

= = 22 x 3.5 x 24.5 = 1886.5 {tex}{m^2}{/tex}

Thus, the radius and area of circular sheet are 24.5 m and 1886.5 {tex}{m^2}{/tex}respectively.

NCERT Solutions for Class 7 Maths Exercise 11.3

**Question 4.**A gardener wants to fence a circular garden of diameter 21 m. Find the length of the rope he needs to purchase, if he makes 2 rounds of fence. Also, find the costs of the rope, if it cost Rs. 4 per meter. {tex}\left( {{\text{Take }}\pi {\text{ = }}\frac{{22}}{7}} \right){/tex}

**Answer:**

Diameter of the circular garden = 21 m

{tex}\therefore {/tex} Radius of the circular garden = {tex}\frac{{21}}{2}{/tex} m

Now Circumference of circular garden = {tex}2\pi r{/tex} = {tex}2 \times \frac{{22}}{7} \times \frac{{21}}{2}{/tex}

= 22 x 3 = 66 m

The gardener makes 2 rounds of fence so the total length of the rope of fencing

= 2 x {tex}2\pi r{/tex} = 2 x 66 = 132 m

Since the cost of 1 meter rope = Rs. 4

Therefore, cost of 132 meter rope = 4 c 132 = Rs. 528

NCERT Solutions for Class 7 Maths Exercise 11.3

**Question 5.**From a circular sheet of radius 4 cm, a circle of radius 3 cm is removed. Find the area of the remaining sheet. {tex}\left( {{\text{Take }}\pi {\text{ = 3}}{\text{.14}}} \right){/tex}

**Answer:**

Radius of circular sheet (R) = 4 cm and radius of removed circle {tex}\left( r \right){/tex} = 3 cm

Area of remaining sheet = Area of circular sheet – Area of removed circle

= {tex}\pi {{\text{R}}^2} – \pi {r^2}{/tex} = {tex}\pi \left( {{{\text{R}}^2} – {r^2}} \right){/tex}

= {tex}\pi \left( {{4^2} – {3^2}} \right){/tex} = {tex}\pi \left( {16 – 9} \right){/tex}

= 3.14 x 7 = 21.98{tex}\;c{m^2}{/tex}

Thus, the area of remaining sheet is 21.98{tex}\;c{m^2}{/tex}.

NCERT Solutions for Class 7 Maths Exercise 11.3

**Question 6.**Saima wants to put a lace on the edge of a circular table cover of diameter 1.5 m. Find the length of the lace required and also find its cost if one meter of the lace costs Rs. 15. {tex}\left( {{\text{Take }}\pi {\text{ = 3}}{\text{.14}}} \right){/tex}

**Answer:**

Diameter of the circular table cover = 1.5 m

{tex}\therefore {/tex} Radius of the circular table cover = {tex}\frac{{1.5}}{2}{/tex} m

Circumference of circular table cover = {tex}2\pi r{/tex} = {tex}2 \times 3.14 \times \frac{{1.5}}{2}{/tex} = 4.71 m

Therefore the length of required lace is 4.71 m.

Now the cost of 1 m lace = Rs. 15

Then the cost of 4.71 m lace = 15 x 4.71 = Rs. 70.65

Hence, the cost of 4.71 m lace is Rs. 70.65.

NCERT Solutions for Class 7 Maths Exercise 11.3

**Question 7.**Find the perimeter of the adjoining figure, which is a semicircle including its diameter.

**Answer:**

Diameter = 10 cm

{tex}\therefore {/tex} Radius = {tex}\frac{{10}}{2}{/tex} = 5 cm

According to question,

Perimeter of figure = Circumference of semi-circle + diameter

= {tex}\pi r{/tex} + D

= {tex}\frac{{22}}{7} \times 5 + 10{/tex} = {tex}\frac{{110}}{7} + 10{/tex}

= {tex}\frac{{110 + 70}}{7} = \frac{{180}}{7}{/tex} = 25.71 cm

Thus, the perimeter of the given figure is 25.71 cm.

NCERT Solutions for Class 7 Maths Exercise 11.3

**Question 8.**Find the cost of polishing a circular table-top of diameter 1.6 m, if the rate of polishing is Rs. 15/{tex}{m^2}{/tex}. {tex}\left( {{\text{Take }}\pi {\text{ = 3}}{\text{.14}}} \right){/tex}

**Answer:**

Diameter of the circular table top = 1.6 m

{tex}\therefore {/tex} Radius of the circular table top = {tex}\frac{{1.6}}{2} = {/tex} 0.8 m

Area of circular table top = {tex}\pi {r^2}{/tex}

= 3.14 x 0.8 x 0.8 = 2.0096 {tex}{m^2}{/tex}

Now cost of 1 m2 polishing = Rs. 15

Then cost of 2.0096{tex}{m^2}{/tex} polishing = 15 x 2.0096 = Rs. 30.14 (approx.)

Thus, the cost of polishing a circular table top is Rs. 30.14 (approx.)

NCERT Solutions for Class 7 Maths Exercise 11.3

**Question 9.**Shazli took a wire of length 44 cm and bent it into the shape of a circle. Find the radius of that circle. Also find its area. If the same wire is bent into the shape of a square, what will be the length of each of its sides? Which figure encloses more area, the circle or the square? {tex}\left( {{\text{Take }}\pi {\text{ = }}\frac{{22}}{7}} \right){/tex}

**Answer:**

Total length of the wire = 44 cm

{tex}\therefore {/tex} the circumference of the circle = {tex}2\pi r{/tex} = 44 cm

{tex} \Rightarrow {/tex} {tex}2 \times \frac{{22}}{7} \times r = 44{/tex} {tex} \Rightarrow {/tex} {tex}r = \frac{{44 \times 7}}{{2 \times 22}}{/tex} = 7 cm

Now Area of the circle = {tex}\pi {r^2}{/tex} = {tex}\frac{{22}}{7} \times 7 \times 7{/tex} = 154 cm2

Now the wire is converted into square.

Then perimeter of square = 44 cm

{tex} \Rightarrow {/tex} 4 x side = 44 {tex} \Rightarrow {/tex} side = {tex}\frac{{44}}{4}{/tex} = 11 cm

Now area of square = side x side = 11 x 11 = 121 {tex}c{m^2}{/tex}

Therefore, on comparing, the area of circle is greater than that of square, so the circle enclosed more area.

**Question 10.**From a circular card sheet of radius 14 cm, two circles of radius 3.5 cm and a rectangle of length 3 cm and breadth 1 cm are removed (as shown in the adjoining figure). Find the area of the remaining sheet. {tex}\left( {{\text{Take }}\pi {\text{ = }}\frac{{22}}{7}} \right){/tex}

**Answer:**

Radius of circular sheet (R) = 14 cm and Radius of smaller circle {tex}\left( r \right){/tex} = 3.5 cm

Length of rectangle {tex}\left( l \right){/tex} = 3 cm and breadth of rectangle {tex}\left( b \right){/tex} = 1 cm

According to question,

Area of remaining sheet=Area of circular sheet– (Area of two smaller circle + Area of rectangle)

= {tex}\pi {{\text{R}}^2} – \left[ {2\left( {\pi {r^2}} \right) + \left( {l \times b} \right)} \right]{/tex}

= {tex}\frac{{22}}{7} \times 14 \times 14 – \left[ {\left( {2 \times \frac{{22}}{7} \times 3.5 \times 3.5} \right) – \left( {3 \times 1} \right)} \right]{/tex}

= 22 x 14 x 2 – [44 x 0.5 x 3.5 + 3]

= 616 – 80

= 536 {tex}c{m^2}{/tex}

Therefore the area of remaining sheet is 536 cm2.

NCERT Solutions for Class 7 Maths Exercise 11.3

**Question 11.**A circle of radius 2 cm is cut out from a square piece of an aluminium sheet of side 6 cm. What is the area of the left over aluminium sheet? {tex}\left( {{\text{Take }}\pi {\text{ = 3}}{\text{.14}}} \right){/tex}

**Answer:**

Radius of circle = 2 cm and side of aluminium square sheet = 6 cm

According to question,

Area of aluminium sheet left = Total area of aluminium sheet – Area of circle

= side x side – {tex}\pi {r^2}{/tex}

= 6 x 6 – {tex}\frac{{22}}{7}{/tex} x 2 x 2

= 36 – 12.56

= 23.44 {tex}c{m^2}{/tex}

Therefore, the area of aluminium sheet left is 23.44 {tex}c{m^2}{/tex}.

NCERT Solutions for Class 7 Maths Exercise 11.3

**Question 12.**The circumference of a circle is 31.4 cm. Find the radius and the area of the circle. {tex}\left( {{\text{Take }}\pi {\text{ = 3}}{\text{.14}}} \right){/tex}

**Answer:**

The circumference of the circle = 31.4 cm

{tex} \Rightarrow {/tex} {tex}2\pi r{/tex} = 31.4 {tex} \Rightarrow {/tex} 2 x 3.14 x {tex}r{/tex} = 31.4

{tex} \Rightarrow {/tex} {tex}r = \frac{{31.4}}{{2 \times 3.14}}{/tex} = 5 cm

Then area of the circle = {tex}\pi {r^2}{/tex} = 3.14 x 5 x 5

= 78.5 {tex}c{m^2}{/tex}

Therefore, the radius and the area of the circle are 5 cm and 78.5{tex}c{m^2}{/tex} respectively.

NCERT Solutions for Class 7 Maths Exercise 11.3

**Question 13.**A circular flower bed is surrounded by a path 4 m wide. The diameter of the flower bed is 66 m. What is the area of this path? {tex}\left( {{\text{Take }}\pi {\text{ = 3}}{\text{.14}}} \right){/tex}

**Answer:**

Diameter of the circular flower bed = 66 m

{tex}\therefore {/tex} Radius of circular flower bed {tex}\left( r \right) = \frac{{66}}{2}{/tex} = 33 m

{tex}\therefore {/tex} Radius of circular flower bed with 4 m wide path (R) = 33 + 4 = 37 m

According to the question,

Area of path = Area of bigger circle – Area of smaller circle

= {tex}\pi {{\text{R}}^2} – \pi {r^2}{/tex} = {tex}\pi \left( {{{\text{R}}^2} – {r^2}} \right){/tex}

= {tex}\pi \left[ {{{\left( {37} \right)}^2} – {{\left( {33} \right)}^2}} \right]{/tex}

= 3.14 [ (37 + 33) (37 – 33)] {tex}\left[ {\because {a^2} – {b^2} = \left( {a + b} \right)\left( {a – b} \right)} \right]{/tex}

= 3.14 x 70 x 4

= 879.20 {tex}{m^2}{/tex}

Therefore, the area of the path is 879.20 {tex}{m^2}{/tex}.

NCERT Solutions for Class 7 Maths Exercise 11.3

**Question 14.**A circular flower garden has an area of 314 m2. A sprinkler at the centre of the garden can cover an area that has a radius of 12 m. Will the sprinkler water the entire garden? {tex}\left( {{\text{Take }}\pi {\text{ = 3}}{\text{.14}}} \right){/tex}

**Answer:**

Circular area by the sprinkler = {tex}\pi {r^2}{/tex} = 3.14 x 12 x 12

= 3.14 x 144 = 452.16 {tex}{m^2}{/tex}

Area of the circular flower garden = 314 {tex}{m^2}{/tex}

Since Area of circular flower garden is smaller than area by sprinkler.

Therefore the sprinkler will water the entire garden.

NCERT Solutions for Class 7 Maths Exercise 11.3

**Question 15.**Find the circumference of the inner and the outer circles, shown in the adjoining figure. {tex}\left( {{\text{Take }}\pi {\text{ = 3}}{\text{.14}}} \right){/tex}

**Answer:**

Radius of outer circle {tex}\left( r \right){/tex} = 19 m

{tex}\therefore {/tex} Circumference of outer circle = {tex}2\pi r{/tex} = 2 x 3.14 x 19

= 119.32 m

Now radius of inner circle {tex}\left( {r’} \right){/tex} = 19 – 10 = 9 m

{tex}\therefore {/tex} Circumference of inner circle = {tex}2\pi r'{/tex} = 2 x 3.14 x 9

= 56.52 m

Therefore the circumferences of inner and outer circles are 56.52 m and 119.32 m respectively.

NCERT Solutions for Class 7 Maths Exercise 11.3

**Question 16.**How many times a wheel of radius 28 cm must rotate to go 352 m? {tex}\left( {{\text{Take }}\pi {\text{ = }}\frac{{22}}{7}} \right){/tex}

**Answer:**

Let wheel must be rotate {tex}n{/tex} times of its circumference.

Radius of wheel = 28 cm and Total distance = 352 m = 35200 cm

{tex}\therefore {/tex} Distance covered by wheel = {tex}n{/tex} x circumference of wheel

{tex} \Rightarrow {/tex} 35200 = {tex}n \times 2\pi r{/tex}

{tex} \Rightarrow {/tex} 35200 = {tex}n \times 2 \times \frac{{22}}{7} \times 28{/tex}

{tex} \Rightarrow {/tex} {tex}n = \frac{{35200 \times 7}}{{2 \times 22 \times 28}}{/tex}

{tex} \Rightarrow {/tex} {tex}n{/tex} = 200 revolutions

Thus wheel must rotate 200 times to go 352 m.

NCERT Solutions for Class 7 Maths Exercise 11.3

**Question 17.**The minute hand of a circular clock is 15 cm long. How far does the tip of the minute hand move in 1 hour? {tex}\left( {{\text{Take }}\pi {\text{ = 3}}{\text{.14}}} \right){/tex}

**Answer:**

In 1 hour, minute hand completes one round means makes a circle.

Radius of the circle {tex}\left( r \right){/tex} = 15 cm

Circumference of circular clock = {tex}2\pi r{/tex}

= 2 x 3.14 x 15 = 94.2 cm

Therefore, the tip of the minute hand moves 94.2 cm in 1 hour.

## NCERT Solutions for Class 7 Maths Exercise 11.3

NCERT Solutions Class 7 Maths PDF (Download) Free from myCBSEguide app and myCBSEguide website. Ncert solution class 7 Maths includes text book solutions from Class 7 Maths Book . NCERT Solutions for CBSE Class 7 Maths have total 15 chapters. 7 Maths NCERT Solutions in PDF for free Download on our website. Ncert Maths class 7 solutions PDF and Maths ncert class 7 PDF solutions with latest modifications and as per the latest CBSE syllabus are only available in myCBSEguide.

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