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 $96.25\ mm^2$.
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