Question

9. A brooch is made with silver wire in the
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 Fig. 5.12. Find:
(1) the total length of the silver wire required.
(ii) the area of each sector of the brooch.

Answer 1

In the above Question , the following information is given -

A brooch is made with silver wire in the circle with diameter 35 mm.

The wire is also used in making 5 diameters which divide the circle into 10 equal sectors .

To find -

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

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

Solution -

Here ,

The diameter of the brooch is 35 millimetres .

Now ,

Diameter = 2 × Radius

=> 2R = 35 mm

=> radius = 17.5 mm .

Now ,

Perimeter of the brooch

=> 2 π r

=> π d

=> 110 mm

Now ,

For making 1 diameter , length of wire used = 35 mm .

For making 5 diameters -

Length of wire used

=> 35 × 5 mm

=> 175 mm

Now ,

Total length of silver wire used -

=> 110 mm + 175 mm

=> 285 mm .

Now ,

Area of the given brooch

=> π r²

=> π × ( 35 ² ) / 4

=> 306.25 π mm².

Now ,

As there are 10 sectors made -

Area Of Each Sector

=> ( 1 / 10 ) th of area of circle

=> ( 1 / 10 ) th of 306.25 π cm².

=> 3.0625 π mm² .

__________

Answer -

[ 1 ] Total length of silver wire used is 285 mm .

[ 1 ] Area of each sector is 3.0625 π mm² .

____________

Answer 2

QuestioN :-

A brooch is made with silver wire in the  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 Fig. 5.12. Find:

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

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

SolutioN :-

Diameter of the silver wire, d = 35 mm

Radius, r = 35/2 mm

(1) Total length of the silver wire required = Circumference of the wire.

⇒Circumference = 2πr

⇒Circumference = 2 × (22/7) × (35/2)

⇒Circumference = 2 × 11 × 5

⇒Circumference = 110 mm

Then, according to the question -

Length of the 5 diameters = 5 × 35

⇒Length of the 5 diameters = 175 mm

Length of silver wire required = 175 + 110

⇒Length of silver wire required = 285 mm

$\rule{190}3$

(2) Area of each sector of the brooch :-

$\sf{A=\dfrac{\theta}{360^\circ}\times \pi r^2 }$

As it is divided into 10 equal sectors, so

θ = 360°/10 = 36°

$\sf{\hookrightarrow A=\dfrac{36^\circ}{360^\circ}\times \pi (\dfrac{35}{2})^2 }$

$\sf{\hookrightarrow A=\dfrac{1}{10} \times \dfrac{22}{7}\times \dfrac{1225}{4} }\\\\\sf{\hookrightarrow A= \dfrac{1}{5}\times 11 \times \dfrac{175}{4} }\\\\\sf{\hookrightarrow A= \dfrac{1925}{20} }\\\\\boxed{\boxed{\sf{\hookrightarrow A= 96.25\ mm^2}\\}}$