Find the perimeters of the following shapes taking the arcs to be quarter or half or three-quarters Class 9

Find the perimeters of the following shapes taking the arcs to be quarter or half or three-quarters Class 9

Question 1.

Find the perimeters of the following shapes (taking the arcs to be quarter or half or three-quarters of a circle, as appropriate) (Fig. i to ix): Class 9

Solution:

i.

Diameter of Semicircles = 60 m

∴ radius = 30 m

Length of rectangle = 80 m

∵ Circumference of 2 Semicircle

= circumference of full circle

= 2πr

$=2 \times \frac{22}{7} \times 30=\frac{1320}{7}$

Perimeter of Straight parts = 80 + 80 = 160

∴ Total Perimeter of shape = Circumference of 2 Semicircles + Perimeter of straight path

$\begin{aligned} & =160+\frac{1320}{7} \\ & =\frac{1120+1320}{7} \\ & =\frac{2440}{7}=348.57 \mathrm{~m}\end{aligned}$

∴ Total Perimeter of shape = 348.57 m

ii.

Diameter of outer Semicircle = 12 cm.

Radius of outer Semicircle = 6 cm,

Diameter of inner Semicircle = 8 cm.

∴ Radius of inner Semicircle = 4 cm

∵ Perimeter of shape = Circumference of outer semicircle + Circumference of inner semicircle + width of shape ... (i)

Circumference of Outer Semicircle

$=\pi r=\frac{22}{7} \times 6=\frac{132}{7}$

Circumference of Inner Semicircle

$=\pi r=\frac{22}{7} \times 4=\frac{88}{7}$

Width of shape = 6 - 4 = 2 cm

Perimeter of shape

$\begin{aligned} & =\frac{132}{7}+\frac{88}{7}+2 \\ & =\frac{220}{7}+2 \\ & =\frac{220+14}{7} \\ & =\frac{234}{7}\end{aligned}$

= 33.42 cm

∴ Perimeter of shape = 33.42 cm

iii.

Diameter of Semicircle = 10 cm

∴ radius = 5 cm

∵ Length of one semicircle

$=\pi r=\frac{22}{7} \times 5=\frac{110}{7} \mathrm{~cm}$.

Perimeter of Shape

= 4 × Length of one Semicircle

= 4 × $\frac{110}{7}=\frac{440}{7}$

= 62.86 cm.

∴ Perimeter of Shape = 62.86 cm

iv.

Diameter of semicircle = 12 cm

∴ radius = 6 cm

∵ Length of one semicircle

$=\pi r=\frac{22}{7} \times 6=\frac{132}{7} \mathrm{~cm}$.

Perimeter of shape = 3 × Length of semicircle

$\begin{aligned} & =3 \times \frac{132}{7} \\ & =\frac{396}{7}=56.57 \mathrm{~cm}\end{aligned}$

∴ Perimeter of shape = 56.57 cm

v.

Diameter of semicircle = 14 cm

∴ radius = 7 cm

Radius of quarter circle = 14 cm

Number of semicircles = 4

Number of quarter circles = 4

∵ Length of one semicircle = πr

= $\frac{22}{7}$ × 7 = 22 cm

∴ Total length of 4 semicircles

= 4 × 22 = 88 cm ...(i)

∵ Length of one 4 quarter circles

$\begin{aligned} & =\frac{1}{4} \times 2 \pi r \\ & =\frac{1}{4} \times 2 \times \frac{22}{7} \times 14 \\ & =\frac{1}{2} \times \frac{22}{7} \times 14 \\ & =\frac{1}{2} \times 44=22 \mathrm{~cm}\end{aligned}$

∴ Total length of 4 quarter circles

= 4 × 22 = 88 cm ....(ii)

Now, Perimeter of shape = Length of 4 Semicircles + Length of 4 quarter circles

= 88 cm + 88 cm

= 176 cm ...[From (i) and (ii)]

∴ The perimeter of the shape is 176 cm

vi.

Diameter of bigger semicircle = 28 cm

∴ radius = 14 cm

Diameter of smaller semicircle = $\frac{28}{4}$ = 7 cm

∴ radius = 3.5 cm

Number of small semicircles = 4

Length of bigger semicircle = πr

= $\frac{22}{4}$ × 14

= 44 cm ....(i)

Length of smaller semicircle = πr

∴ = $\frac{22}{4}$ × 3.5

= 11 cm

Total length of 4 small semicircles = 4 × 11

= 44 cm ...(ii)

∴ Perimeter of shape = Length of bigger semicircle + length of 4 smaller semicircles

= 44 + 44

= 88 cm ... [From (i) and (ii)]

∴ The perimeter of the shape is 88 cm

vii.

Three semicircles are constructed on the sides of a right angled triangle Base = 8 cm, Height = 6 cm

∴ Hypotenuse

$\begin{aligned} & =\sqrt{8^2+6^2}=\sqrt{64+36} \\ & =\sqrt{100}=10 \mathrm{~cm}\end{aligned}$

Each side of triangle is diameter of a semicircle

∵ Length of one semicircle = πr

∴ Length of semicircle with diameter 8 cm

$=\frac{22}{7} \times 4=\frac{88}{7} \mathrm{~cm}$.

∴ Length of semicircle with diameter 6 cm

$=\frac{22}{7} \times 3=\frac{66}{7} \mathrm{~cm}$.

∴ Length of semicircle with diameter 10 cm.

$=\frac{22}{7} \times 5=\frac{110}{7} \mathrm{~cm}$.

∴ Perimeter of shape

$=\frac{88}{7}+\frac{66}{7}+\frac{110}{7}=\frac{264}{7}=37.71 \mathrm{~cm}$.

∴ The perimeter of the shape is 37.71 cm.

viii.

Diameter of bigger semicircle = 4 + 4 + 4

= 12 cm.

∴ radius = 6 cm

∴ Diameter of smaller semicircle = 4 cm.

∴ radius = 2 cm

Number of small semicircles = 3

Length of bigger semicircle = πr = $\frac{22}{7}$ × 6

= $\frac{132}{7}$ cm.

Length of one smaller semicircle

$=\pi r=\frac{22}{7} \times 2=\frac{44}{7} \mathrm{~cm}$.

Length of 3 smaller semicircles

$=3 \times \frac{44}{7}=\frac{132}{7} \mathrm{~cm}$.

∴ Perimeter of shape = $\frac{132}{7}+\frac{132}{7}=\frac{264}{7}$

= 37.71 cm.

∴ The perimeter of the shape is 37.71 cm.

ix.

∵ Diameter of bigger semicircle

= 10 + 10 = 20 cm.

∴ radius = 10 cm.

∴ Diameter of smaller semicircle = 10 cm

∴ radius = 5 cm

Number of smaller semicircles = 2

∴ Length of bigger semicircle = πr

$=\frac{22}{7} \times 10=\frac{220}{7}$

∵ Length of one smaller semicircle = πr

$\begin{aligned} & =\frac{22}{7} \times 5 \\ & =\frac{110}{7} \mathrm{~cm} .\end{aligned}$

Total length of 2 semicircles

$\begin{aligned} & =2 \times \frac{110}{7} \\ & =\frac{220}{7} \mathrm{~cm} .\end{aligned}$

∴ Perimeter of shape = $\frac{220}{7}+\frac{220}{7}$

= $\frac{440}{7}$ = 62.86 cm

∴ The perimeter of the shape is 62.86 cm