A brooch is made with silver wire in the form of a circle with diameter 35 mm. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors as shown in figure.

Given:

A brooch is made with silver wire in the form of a circle with diameter 35 mm. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors.

To do:

We have to find

(i) the total length of the silver wire required.

(ii) the area of each sector of the brooch.

Solution:

(i) Diameter of the brooch $(d) = 35\ mm$

Total length of the silver wire required $=$ Circumference of brooch $+ 5 \times$ diameter

$= \pi d + 5d$

$= (\pi + 5) \times 35$

$= (\frac{22}{7} + 5) \times 35$

$= (\frac{22 + 5(7)}{7} \times 35$

$= 57 \times 5$

$= 285\ mm$

The total length of the silver wire required is $285\ mm$.

(ii) Diameter of the brooch $(d) = 35\ mm$

This implies,

Radius of the brooch $(r) = \frac{35}{2}\ mm$

The wire divides the brooch into 10 equal sectors.

This implies,

Angle of each sector $(\theta) = \frac{360^o}{10}$

$= 36^o$

Therefore,

Area of each sector of the brooch $= \frac{36^o}{360^o} \times \pi r^2$

$= \frac{1}{10} \times \frac{22}{7} \times (\frac{35}{2})^2$

$= 96.25\ mm^2$

The area of each sector of the brooch is $96.25\ mm^2$. 

Tutorialspoint

Updated on: 10-Oct-2022

3K+ Views

Kickstart Your Career

Get certified by completing the course

Get Started