NCERT Solutions for Class 7 Maths Chapter 9 Exercise 9.2 - Perimeter and Area

1. Find the circumference of the circle with the following radius: (Take \[\pi  = \dfrac{{22}}{7}\])

(a) \[14cm\]

Ans: Radius is \[14cm\] .

As, we have given the radius of circle therefore, to calculate the circumference of circle we will use the formula \[2\pi r\] and substitute the value of \[\pi  = \dfrac{{22}}{7}\] as per given in the question.

Circumference of circle \[ = 2\pi r\]

Circumference of circle \[ = 2 \times \dfrac{{22}}{7} \times r\]

Circumference of circle \[ = 2 \times \dfrac{{22}}{7} \times 14cm\]

Circumference of circle \[ = 88cm\]

(b) \[28cm\]

Ans: Radius is \[28cm\] .

As, we have given the radius of circle therefore, to calculate the circumference of circle we will use the formula \[2\pi r\] and substitute the value of \[\pi  = \dfrac{{22}}{7}\] as per given in the question.

Circumference of circle \[ = 2\pi r\]

Circumference of circle \[ = 2 \times \dfrac{{22}}{7} \times r\]

Circumference of circle \[ = 2 \times \dfrac{{22}}{7} \times 28cm\]

Circumference of circle \[ = 176cm\]

(c) \[21cm\]

Ans: Radius is \[21cm\] .

As, we have given the radius of circle therefore, to calculate the circumference of circle we will use the formula \[2\pi r\] and substitute the value of \[\pi  = \dfrac{{22}}{7}\] as per given in the question.

Circumference of circle \[ = 2\pi r\]

Circumference of circle \[ = 2 \times \dfrac{{22}}{7} \times r\]

Circumference of circle \[ = 2 \times \dfrac{{22}}{7} \times 21cm\]

Circumference of circle \[ = 132cm\]

2. Find the area of the following circles, given that (Take \[\pi  = \dfrac{{22}}{7}\])

(a) Radius \[ = 14mm\]

Ans: Radius is \[14mm\].

As, we have given the radius of circle therefore, to calculate the area of circle we will use the formula \[\pi {r^2}\] and substitute the value of \[\pi  = \dfrac{{22}}{7}\] as per given in the question.

Now,

Area of circle \[ = \pi {r^2}\]

Area of circle \[ = \dfrac{{22}}{7} \times {\left( {14} \right)^2}\]

Area of circle \[ = 616m{m^2}\]

(b) Diameter \[ = 49m\]

Ans: As we have given the diameter of the circle so, first we will calculate the radius of the circle. Radius of the circle is half of its diameter. To calculate the area of a circle we will use the formula \[\pi {r^2}\] and substitute the value of \[\pi  = \dfrac{{22}}{7}\] as per given in the question.

\[Radius = \dfrac{{Diameter}}{2}\]

So, \[Radius = \dfrac{{49}}{2}m\]

\[Radius = 24.5m\]

Area of circle \[ = \pi {r^2}\]

Area of circle \[ = \dfrac{{22}}{7} \times {\left( {24.5m} \right)^2}\]

Area of circle \[ = 1886.5{m^2}\]

(c) Radius \[ = 5cm\]

Ans: As, we have given the radius of circle therefore, to calculate the area of circle we will use the formula \[\pi {r^2}\] and substitute the value of \[\pi  = \dfrac{{22}}{7}\] as per given in the question.

Radius is \[5cm\].

Area of circle \[ = \pi {r^2}\]

Area of circle \[ = \dfrac{{22}}{7} \times {\left( {5cm} \right)^2}\]

Area of circle \[ = 78.57c{m^2}\]

3. If the circumference of a circular sheet is \[154m\], find its radius. Also find the area of the sheet. (Take \[\pi  = \dfrac{{22}}{7}\])

Ans: Radius of the circular sheet \[ = ?\]

Circumference of the circular sheet \[ = 154m\]

As we have given the circumference of the circular sheet and we have to calculate the radius of the circular sheet.

Therefore, to calculate the radius of a circular sheet we will use the formula of circumference of the circle.

Circumference of circle \[ = 2\pi r\]

\[2\pi r = 154m\]

Substituting the value of \[\pi  = \dfrac{{22}}{7}\] .

\[2 \times \dfrac{{22}}{7} \times r = 154m\]

\[r = \dfrac{{154m \times 4}}{{2 \times 22}}\]

\[r = 24.5m\]

Therefore, the radius of the circular sheet is \[24.5m\].

Now, calculating the area of a circular sheet.

As, we have calculated radius of circular sheet therefore, to calculate the area of circle we will use the formula \[\pi {r^2}\] and substitute the value of \[\pi  = \dfrac{{22}}{7}\] as per given in the question.

Area circular sheet \[ = \pi {r^2}\]

Area circular sheet\[ = \dfrac{{22}}{7} \times {\left( {24.5m} \right)^2}\]

Area circular sheet \[ = 1886.5{m^2}\]

Therefore, the area of the circular sheet is \[1886.5{m^2}\] .

4. A gardener wants to fence a circular garden of diameter 21m. Find the length of the rope he needs to purchase, if he makes 2 rounds of fence. Also find the cost of the rope, if it costs ₹ 4 per meter. (Take \[\pi  = \dfrac{{22}}{7}\])

Ans: We have given the radius of the circular garden so, first we will calculate the radius of circular radius.

Therefore, radius of circular radius is given as,

Diameter of the garden is \[21m\] .

Radius of the garden will be \[\dfrac{{21}}{2}m\] i.e. \[10.5m\] .

Now, we will calculate the circumference of the circular garden in order to know the length of rope required for fencing.

Therefore, the circumference of the garden \[ = 2\pi r\]

Circumference of garden \[ = 2 \times \dfrac{{22}}{7} \times 10.5m\]

Circumference of garden \[ = 66m\]

Therefore, the length of rope required for fencing of garden \[ = 2 \times 66m\]

Length of rope required \[ = 132m\]

Now, we have to calculate the cost of \[132m\] rope purchased for fencing.

Therefore,

Cost of 1m rope \[ = {\text{ Rs}}{\text{. }}4\]

Cost of 132m rope \[ = {\text{Rs }}132 \times 4\]

Cost of 132m rope \[ = {\text{ Rs 528}}\]

So, the cost of \[132m\] is Rs 528.

5. From a circular sheet of radius 4 cm, a circle of radius 3 cm is removed. Find the area of the remaining sheet. (Take \[\pi  = 3.14\])

Ans: We have given a circular sheet of radius 4 cm and out of which a circular piece of radius 3 cm is removed therefore, we have to calculate the area of the remaining sheet.

Therefore, calculating the area of a circular sheet of radius 4 cm.

Radius of circular  sheet \[ = 4cm\]

Area of circular sheet \[ = \pi {r^2}\]

Substituting the value of radius and \[\pi  = 3.14\] .

Area of circular sheet \[ = 3.14 \times {\left( {4cm} \right)^2}\]

Area of circular sheet \[ = 50.24c{m^2}\]

Now, calculating the area of a circle of radius 3 cm removed.

Radius of circle removed \[ = 3cm\]

Area of circle removed \[ = \pi {r^2}\]

Substituting the value of radius and \[\pi  = 3.14\] .

Area of circle removed \[ = 3.14 \times {\left( {3cm} \right)^2}\]

Area of circle removed \[ = 28.26c{m^2}\]

Therefore, the area of the remaining part of the circular sheet is calculated by subtracting the removed area from the area of the circular sheet.

So,

The remaining area of sheet \[ = 50.24c{m^2} - 28.26c{m^2}\]

Therefore, the remaining area of the sheet is \[21.98c{m^2}\] .

6. Saima wants to put a lace on the edge of a circular table cover of 1.5 m diameter. Find the length of the lace required and also find its cost if one meter of the lace costs ₹ 15. (Take \[\pi  = 3.14\])

Ans: As we have given the diameter of the circular table first we will calculate the radius of the circular table.

Therefore,

Diameter of circular table \[ = 1.5m\]

Radius of the circular table is \[\dfrac{{1.5}}{2}m\] i.e. \[0.75m\] .

Now, Saima will put a lace on the edge of a circular table for that we have to calculate the circumference of the circular table.

Therefore,

Circumference of circular table \[ = 2\pi r\]

Substituting the value of radius and \[\pi  = 3.14\] .

Circumference of circular table \[ = 2 \times 3.14 \times 0.75m\]

Circumference of circular table \[ = 4.71m\]

Therefore, \[4.71m\] of lace is required for a circular table.

Now, we have to calculate the cost of lace used.

Therefore,

Cost of 1m lace is Rs 15.

Cost of \[4.71m\] \[ = 4.71 \times 15\]

Therefore, the cost of  \[4.71m\] is Rs \[70.65\] .

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

Ans: As we have given the diameter of the circular table first we will calculate the radius of the circular table.

Therefore,

Diameter of the semicircle is 10m.

Radius of the semicircle is \[\dfrac{{10}}{2}cm\] i.e. \[5cm\] .

Now, we will calculate the circumference of the semicircle.

Therefore,

Circumference of semicircle \[ = \pi r\]

Substituting the value of radius and \[\pi  = \dfrac{{22}}{7}\] .

Circumference of semicircle \[ = \dfrac{{22}}{7} \times 5cm\]

Circumference of semicircle \[ = 15.71cm\]

This is the circumference of the circular part but the perimeter of the semi-circle diameter is also included.

Therefore,

Perimeter of semicircle \[ = \] Circumference of semicircle \[ + \] Diameter of semicircle

Perimeter of semicircle \[ = 15.71cm + 10cm\]

Perimeter of semicircle \[ = 25.71cm\]

8. Find the cost of polishing a circular table-top of diameter 1.6 m, if the rate of polishing is ₹15/m2. (Take \[\pi  = 3.14\])

Ans: As we have given the diameter of the circular table first we will calculate the radius of the circular table.

Therefore,

Diameter of the circular table is \[1.6m\] .

Radius of the circular table is \[\dfrac{{1.6}}{2}m\] i.e. \[0.8m\] .

Now, we have to polish the circular table-top for that, we have to calculate the area of the circular table.

Therefore,

Area of circular table \[ = \pi {r^2}\]

Substituting the value of radius and \[\pi  = 3.14\] .

Area of circular table \[ = 3.14 \times {\left( {0.8m} \right)^2}\]

Area of circular table \[ = 2.0096{m^2}\]

Now, we have to calculate the cost of polishing the circular table-top.

Therefore,

Cost of polishing \[1{m^2}\] is Rs 15.

Cost of polishing \[2.0096{m^2}\]\[ = 2.0096 \times 15\]

So, the cost of polishing the circular table is Rs \[30.144\] .

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? (Take \[\pi  = \dfrac{{22}}{7}\])

Ans: Shazli took a wire of length 44 cm. First he converts the wire into circular shape and then converts it into a square.

When \[44cm\] wire is converted into a circle then, the circumference of circular shaped wire will be 44 cm. So, we have to calculate the radius of the circular wire in order to calculate the area of the circular area. We will use the formula of the circumference of the circle and substitute \[\pi  = \dfrac{{22}}{7}\] .

Therefore,

Circumference of circle \[ = 44cm\] .

\[44cm = 2\pi r\]

Substituting the value of \[\pi  = \dfrac{{22}}{7}\] .

\[44cm = 2 \times \dfrac{{22}}{7} \times r\]

\[r = \dfrac{{44cm \times 7}}{{22 \times 2}}\]

\[r = 7cm\]

Therefore, the radius of the circle is \[7cm\] .

Now, calculating the area of circular wire.

Therefore,

Area of circular wire \[ = \pi {r^2}\]

Substituting the value of radius and \[\pi  = \dfrac{{22}}{7}\] .

Area of circular wire \[ = \dfrac{{22}}{7} \times {\left( {7cm} \right)^2}\]

Area of circular wire \[ = 154c{m^2}\]

Therefore, the area of the circle is \[154c{m^2}\] .

Now, the same wire is converted into a square therefore, the perimeter of the square will be 44 cm. In order to calculate the area of the square we have to calculate the side of the square then, we’ll calculate the area of the square.

Since the perimeter of the square \[ = 4a\] .

\[4a = 44cm\]

\[a = \dfrac{{44cm}}{4}\]

\[a = 11cm\]

So, the side of the square is \[11cm\] .

Now, calculating the area of the square.

Area of square \[ = {a^2}\]

Area of square \[ = {\left( {11cm} \right)^2}\]

Area of square \[ = 121c{m^2}\]

Now, on comparing the area of circle and square, the area of circle is greater than square.

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 1cm are removed. (as shown in the adjoining figure). Find the area of the remaining sheet. (Take \[\pi  = \dfrac{{22}}{7}\])

Ans: Here, we have given the circular card sheet of radius 14 cm in which two circles of radius \[3.5cm\] and one rectangle of length 3 cm and breadth 1 cm. We have to calculate the area of remaining sheet.

Therefore, first calculating the area of a circular card sheet of radius 14 cm.

Area of circular card is given as,

Radius of the circular card is \[14cm\].

Area of circular card \[ = \pi {r^2}\]

Substituting the value of radius and \[\pi  = \dfrac{{22}}{7}\] .

Area of circular card \[ = \dfrac{{22}}{7} \times {\left( {14cm} \right)^2}\]

Area of circular card \[ = 616c{m^2}\]

So, the area of the circular card is \[616c{m^2}\].

Now, calculating the area removes two circles and a rectangle.

Area of two circle removed is given as,

Radius of the small circle removed is \[3.5cm\] .

Area of two circle removed \[ = 2 \times \pi {r^2}\]

Substituting the value of radius and \[\pi  = \dfrac{{22}}{7}\] .

Area of two circle removed \[ = 2 \times \dfrac{{22}}{7} \times {\left( {3.5cm} \right)^2}\]

Area of two circles removed \[ = 77c{m^2}\] .

Therefore, area of two removed circle is \[77c{m^2}\]

Now, area of rectangle removed is given as,

Length is \[3cm\] .

Breath is \[1cm\] .

Area of rectangle \[ = length \times breadth\]

Area of rectangle \[ = 3cm \times 1cm\] .

So, the area of the rectangle \[ = 3c{m^2}\] .

Therefore, the total area removed is \[77c{m^2} + 3c{m^2}\] .

Total area removed from the circular card is \[80c{m^2}\] .

Now, calculating the remaining area of the remaining circular card .

Remaining area of circular card \[{\text{ = Area of circular card }} - {\text{ Area of removed part}}\]

Remaining area of circular card \[ = 616c{m^2} - 80c{m^2}\] .

Therefore, the remaining area of the circular card is \[536c{m^2}\] .

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 leftover aluminium sheet? (Take \[\pi  = 3.14\])

Ans: Here we have given a square aluminium sheet of side 6 cm from which a circle of radius of 2 cm is cut out from square aluminium sheet. We have to calculate the area of the remaining aluminium sheet.

Therefore,

Area of square aluminium sheet.

Side of the square piece is 6 cm.

Area of square aluminium sheet \[ = {a^2}\]

Area of square aluminium sheet \[ = {\left( {6cm} \right)^2}\]

Area of square aluminium sheet \[ = 36c{m^2}\]

Therefore, area of square aluminium sheet is \[36c{m^2}\]

Now calculating the area of a circle of radius 2 cm removed from square aluminium sheet.

Area of circle cut out from square aluminium sheet is given as,

Radius of the circle is 2 cm.

Area of two circle \[ = \pi {r^2}\]

Substituting the value of radius and \[\pi  = \dfrac{{22}}{7}\] .

Area of two circle \[ = \dfrac{{22}}{7} \times {\left( {2cm} \right)^2}\]

Area of two circles \[ = 12.56c{m^2}\] .

Now, the remaining area of square aluminium sheet is calculated by subtracting removed area from area of square aluminium sheet.

So, the area of the remaining part of the aluminium sheet is \[36c{m^2} - 12.56c{m^2}\].

Therefore, the remaining area of the aluminium sheet is \[23.44c{m^2}\] .

12. The circumference of a circle is 31.4 cm. Find the radius and the area of the circle? (Take \[\pi  = 3.14\])

Ans: Radius of the circular sheet \[ = ?\]

As we have given the circumference of the circular sheet and we have to calculate the radius of the circular sheet.

Therefore, to calculate the radius of a circular sheet we will use the formula of circumference of the circle.

Circumference of the circle is \[31.4cm\] .

Circumference of circle \[ = 2\pi r\]

\[31.4cm = 2\pi r\]

Substituting the value of \[\pi  = 3.14\] .

\[31.4cm = 2 \times 3.14 \times r\]

\[r = \dfrac{{31.4}}{{3.14 \times 2}}\]

\[r = 5cm\]

Therefore, the radius of the circle is 5 cm.

Now, calculating the area of a circular sheet.

As we have calculated the radius of the circular sheet therefore, to calculate the area of the circle we will use the formula \[\pi {r^2}\] and substitute the value of \[\pi  = 3.14\] as per given in the question.

Area of two circle \[ = \pi {r^2}\]

Substituting the value of radius and \[\pi  = 3.14\] .

Area of two circle \[ = 3.14 \times {\left( {5cm} \right)^2}\]

Area of two circles \[ = 78.5c{m^2}\] .

Therefore, the area of the circle is \[78.5c{m^2}\] .

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? (Take \[\pi  = 3.14\])

Ans: We have to calculate the area of the path around the flower bed. In order to calculate the area of the path first, we have to calculate area of the flower bed of the diameter 66m then area of flower bed of diameter 66m including the path of 4m wide.

The diameter of the flower bed is 66m.

Calculating the area of the flower bed.

First we have to calculate the radius of the flower bed.

Therefore,

Radius of the flower bed will be \[\dfrac{{66}}{2}m\] i.e. 33m.

Area of flower bed \[ = \pi {r^2}\]

Substituting the value of radius and \[\pi  = 3.14\] .

Area of flower bed \[ = 3.14 \times {\left( {33m} \right)^2}\]

Area of flower bed \[ = 3419.46{m^2}\]

Therefore, the area  of the flower bed is \[3419.46{m^2}\] .

Now, calculate the area of the flower bed of diameter 66m including a path of 4m wide.

So, the new radius of flower bed including path \[ = \left( {33 + 4} \right)m\]

Area of flower bed with path \[ = \pi {r^2}\]

Substituting the value of radius and \[\pi  = 3.14\] .

Area of flower bed with path \[ = 3.14 \times {\left( {37} \right)^2}\]

Area of flower bed with path \[ = 4298.66{m^2}\]

Therefore, the area of path \[ = \] Area of flower bed with path \[ - \] Area of flower bed

Area of path \[ = \left( {4298.66 - 3419.46} \right){m^2}\]

Area of path \[ = 879.2{m^2}\]

Therefore, the area of path is \[879.2{m^2}\]

14. A circular flower garden has an area of \[314{m^2}\]. 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? (Take \[\pi  = 3.14\])

Ans: Here, a circular flower garden of area \[314{m^2}\] is given in which there is a sprinkler at the centre of the garden that can cover a distance of radius 12m. We have to decide whether the sprinkler will water the entire garden or not.

For that we have to calculate the radius of the circular flower garden using the area given.

Therefore,

Area of the flower garden is \[ = 314{m^2}\] .

Area of flower garden \[ = \pi {r^2}\]

\[314{m^2} = \pi {r^2}\]

Substituting the value of radius and \[\pi  = 3.14\] .

\[\dfrac{{314{m^2}}}{{3.14}} = {r^2}\]

\[100{m^2} = {r^2}\]

Taking square root on both side

\[\sqrt {100{m^2}}  = r\]

\[r = 10m\]

Therefore, the radius of the flower garden is 10m.

And the radius  of sprinkler water is 12m.

Therefore, the sprinkler will water the entire garden.

15. Find the circumference of the inner and the outer circles, shown in the adjoining figure? (Take \[\pi  = 3.14\])

Ans: Here, we have given an inner circle and outer circle. Radius of the outer circle is 19m and the radius of the inner circle is calculated by subtracting 10m from 19m. Then, the circumference of the inner and outer circle is to be calculated.

Circumference of outer circle,

Radius of the outer circle is 19m.

Circumference of outer circle \[ = 2\pi r\]

Substituting the value of radius and \[\pi  = 3.14\] .

Circumference of outer circle \[ = 2 \times 3.14 \times 19m\]

Circumference of outer circle \[ = 119.32m\]

Therefore, circumference of outer circle is \[119.32m\] ,

Circumference of inner circle.

Radius of inner circle \[ = \left( {19 - 10} \right)m\]

Radius of inner circle \[ = 9m\]

Circumference of inner circle \[ = 2\pi r\]

Substituting the value of radius and \[\pi  = 3.14\] .

Circumference of inner circle \[ = 2 \times 3.14 \times 9m\]

Circumference of inner circle \[ = 56.52m\]

Therefore, the circumference of the inner circle is \[56.52m\] .

16. How many times a wheel of radius 28 cm must rotate to go 352 m? (Take \[\pi  = \dfrac{{22}}{7}\])

Ans: Here, in this question a wheel of radius of 28 cm is given which covers a distance of 352 m. We have to calculate the number of revolutions taken by the wheel to cover a distance of 352m.

As, distance covered by the wheel in one revolution will be equal to its circumference. Therefore, calculating the circumference of wheel,

Radius of wheel \[ = 28cm\]

Circumference of wheel \[ = 2\pi r\]

Substituting the value of radius and \[\pi  = \dfrac{{22}}{7}\] .

Circumference of wheel \[ = 2 \times \dfrac{{22}}{7} \times 28cm\]

Circumference of wheel \[ = 176cm\]

Therefore, distance covered in 1 revolution \[ = 176cm\]

Distance covered by wheel \[ = 352m\]

Distance covered by wheel \[ = 35200cm\]

Number of rotation done by wheel \[ = \dfrac{{{\text{Total distance covered by wheel}}}}{{{\text{Distance covered in revolution}}}}\]

So, number of rotation \[ = \dfrac{{35200cm}}{{176cm}}\]

Therefore, total number of revolution done by wheel \[ = 200\]

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? (Take \[\pi  = 3.14\])

Ans: Minute hand of circular clock takes 1 hour to complete one round of circular clock. Radius of a circular clock is equivalent to it’s minute hand therefore the radius is 15m.

Now, the distance covered by the minute hand is equivalent to the circumference of the circular clock therefore, we will calculate the circumference of the circular clock.

Length of the minute hand of the circular clock \[ = 15cm\]

Therefore,

Distance travelled by the tip of minute hand in 1 hour \[ = \] circumference of the clock

circumference of the clock \[ = 2\pi r\]

Substituting the value of radius and \[\pi  = 3.14\] .

circumference of the clock \[ = 2 \times 3.14 \times 15cm\]

circumference of the clock \[ = 94.2cm\]

Therefore, the distance travelled by the tip of a minute hand in 1 hour is \[94.2cm\] .

Conclusion

Exercise 9.3 of Class 7 Maths Chapter 9 focuses on practical applications of perimeter and area concepts. This exercise is crucial for understanding how to calculate the perimeter and area of composite figures and involves problems related to real-life scenarios. It is important to accurately use formulas and ensure consistency in units of measurement. Practicing these problems will enhance your problem-solving skills and help you master the fundamentals of geometry​

Class 7 Maths Chapter 9: Exercises Breakdown

CBSE Class 7 Maths Chapter 9 Other Study Materials

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