# Ex.12.2 Q9 Areas Related to Circles Solution - NCERT Maths Class 10

## Question

A brooch is made with silver wire in the form of a circle with diameter \(35\,\rm{mm.}\) The wire is also used in making \(5\) diameters which divide the circle into \(10\) equal sectors as shown in the figure. Find:

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

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

## Text Solution

**What is known?**

A brooch is made with silver wire in the form of a circle with diameter \(35 \,\rm{mm.}\) The wire is also used in making \(5\) diameters which divides the circle into \(10\) equal sectors.

**What is unknown?**

(i) The total length of silver were required

(ii) Area of each sector of the brooch

**Reasoning:**

(i) Since the silver wire is used in making the \(5\) diameters and perimeter of the circular brooch.

\(\therefore \) Total length of silver were required

\(=\) Circumference of circle \(+ \,5\, \times\,\)diameter

\(= \pi {{d}} + 5{{d}}\) Where \({d}\) is diameter of the brooch

\(={d}\,(\pi + 5)\)

(ii) To find area of each sector of the brooch, we first find the angle made by each sector at the centre of the circle.

Since the wire divides into \(10\) equal sectors

\(\therefore\) Angle of sector \(\begin{align} \theta =\frac{360^\circ }{10^\circ } \end{align}\)

Radius of the brooch \(\begin{align}= {\frac{d}{2}} \end{align}\)

\(\therefore\) Area of each sector of brooch \(\begin{align}= \frac{\theta }{{{{360}^\circ }}} \times \pi {r^2} \end{align}\)

**Steps:**

(i) Diameter of the brooch \({(d) = 35 \,\rm{mm}}\)

Total length of silver wire required \(= \) Circumference of brooch \(+\, 5\,\) diameter

\[\begin{align}&= \pi d + 5d\\& = (\pi + 5)35\,\\ &= \left( {\frac{{22 + 35}}{7}} \right) \times 35\\ &= 57 \times 5\\ &= 285\,\,\rm{{mm}}\end{align}\]

ii) Radius of brooch \(\begin{align}\left( r \right) = \frac{{35}}{2} = 17.5\,\rm{{mm}}\end{align}\)

Since the wire divides the brooch into \(10 \) equal sectors

\(\therefore \) Angle of sector \(\begin{align} \theta = \frac{{{{360}^\circ }}}{{{{10}^\circ }}}\,\, = {36^\circ } \end{align}\)

\(\therefore\) Area of each sector of the brooch

\[\begin{align}&= \frac{{{{36}^\circ }}}{{{{360}^\circ }}} \times \pi {r^2}\\& = \frac{1}{{10}} \times \frac{{22}}{7} \times 17.5 \times 17.5\\& = 2.2 \times 2.5 \times 17.5\\ &= 96.25\,\rm{{m}}{{{m}}^2}\end{align}\]