S.chand publication New Learning Composite mathematics solution of class 7 Chapter 14 Perimeter and Area Exercise 14F

 Exercise 14F

Question 1

(a) {3.14×(10)²} cm²

={3.14×100} cm²

=314 cm²

(b) {3.14×(5)²} cm²

={3.14×25}cm²

=78.50cm²

(c) {3.14×(20)²} cm²

={3.14×400} cm²

=1256 cm²

(d) {3.14×(25)²} cm²

={3.14×625} cm²

=1962.50 cm²

Question 2

(a) $\left\{\frac{1}{4} \pi (28)^2\right\}$ cm²

$=\left(\frac{1}{4} \times \frac{22}{7} \times 28 \times 28\right)$ cm²

=616 cm²

(b) $\left\{\frac{1}{4} \pi (28)^2\right\}$ cm²

$=\left\{\frac{1}{4}\times \frac{22}{7} \times 313600 \right}$ cm²

=246400 cm²

(c) $\left\{\frac{1}{4} \pi (28)^2\right\}$ cm²

$=\left\{\frac{1}{4}\times \frac{22}{7} \times (8.4)^2\right\}$ cm²

=55.44 cm²

(d) $\left\{\frac{1}{4} \pi (28)^2\right\}$ cm²

$=\left\{\frac{1}{4} \times \frac{22}{7} \times 4.41\right\}$ cm²

=3.465 cm²

Question 3

(a)

πr²=113.04

$r^2=\frac{113.04}{3.14}$

r²=36

r=√36

∴r=6

(b)

πr²=3.14

3.14×r²=3.14

$r^2=\frac{3.14}{3.14}$

r²=1

r=1

(c) 

πr²=28.26

3.14×r²=28.26

$r^2=\frac{28.26}{9.14}$

r²=9

r=3

Question 4

The area of the region$=\left\{\frac{22}{7} \times (77)^2\right\}$ km²

$=\left(\frac{22}{7} \times 77 \times 77\right)$ km²

=18634  km²

Question 5

Area of the shaded portion$=\left\{\frac{1}{4} \times \frac{22}{7} \times (7)^2\right\}$ cm²

$=\left\{\frac{1}{4} \times \frac{22}{7} \times 7 \times 7\right\}$ cm²

=38.5 cm²

Question 6

(a)

Area of the smaller circle$=\left\{\frac{22}{7} \times (7)^2\right\}$ cm²

$=\frac{22}{7} \times 7\times 7$ cm²

=154 cm²

Area of the larger circle$=\left\{\frac{22}{7} \times (14)^2\right\}$ cm²

$=\left(\frac{22}{7} \times 14 \times 14\right)$ cm²

=616 cm²

∴Area of the shaded portion=(616-154)=462 cm²

(b)

Area of the triangle$=\left(\frac{1}{2}\times 16 \times 12\right)$ cm²

=96 cm²

Area of the shaded portion=(314-96)=218 cm²

(c)

Area of one circle$=\left\{\frac{1}{4} \times 3.14 \times 8^2\right\}$ m²

$=\frac{1}{4} \times 3.14 \times 8 \times 8$ m²

=50.24 m²

Area of two circle=(2×50.24)=100.48 m²

∴Area of the rectangle=(8²+8²)m²

=(64+64)m²

=128 m²

∴Area of shaded portion=(128-100.48) m²

=27.52 m²

Area of the larger semi-circle$=\left\{\dfrac{\frac{1}{4}\times \frac{22}{7} \times (14)^2}{2}\right\}$ cm²

$=\dfrac{\frac{1}{4} \times \frac{22}{7} \times 14 \times 14}{2}$ cm²

$=\left\{\frac{154}{2}\right}$ cm²

=77 cm²

Area of the smaller semi-circle$=\dfrac{\frac{1}{4}\times \frac{22}{7} \times (7)^2}{2}$ cm²

$=\dfrac{\frac{1}{4} \times \frac{22}{7} \times 7 \times 7}{2}$ cm²

$=\frac{38.5}{2}$ cm²

=19.25  cm²

∴Area of the shaded portion=(77-19.25) cm²

=57.75  cm²

Question 7

Figure to be added

The combined area of two circle is 308 cm²

The area of one circle$=\frac{308}{2}$ cm²

Let, radius of circle r=154 cm²

ATQ,

$\frac{22}{7} \times r^2=154$

$r^2=154\times \frac{7}{22}$

r²=49

∴r=7

∴Diameter of the circle=(7×2)=14 cm

∴Perimeter of the circle={14+(14+14)} cm

=(14+28) cm

=84 cm

Question 8

Figure to be added

Area of the circular card sheet$=\left\{\frac{22}{7} \times (14)^2\right\}$ cm²

$=\frac{22}{7} \times 14 \times 14$ cm²

=616 cm²

Area of the two small circles$=2\times \frac{22}{7} \times 3.5 \times 3.5$ cm²

=2×38.5 cm²

=77 cm²

Area of the rectangle=(3×1) cm

=3 cm

∴Area of the remaining portion=616-(77+3) cm²

=(616-80) cm²

=536  cm²

Question 9

Let, the radius of the circle be r

ATQ,

$2\times \frac{22}{7} \times r=88$

$r=\frac{88}{2} \times \frac{7}{22}$

∴r=14

∴Area of the circle$=\left(\frac{22}{7}\times 14 \times 14\right)$ cm²

=616 cm²

If the same piece of wire is bent into the shape of a square 

Let the one side of the square be a

4a=88

$a=\frac{88}{4}$

∴a=22

∴Area of the square=(22)² cm²

=484 cm²

Ans : Circle have more area

Question 10

∴Area of the field that the cow can graze$=\left(\frac{1}{4} \times \frac{22}{7} \times (14)^2 \right)$ m²

$=\left(\frac{1}{4} \times \frac{22}{7} \times 14 \times 14\right)$ m²

=154 m²