# Ex.12.3 Q8 Areas Related to Circles Solution - NCERT Maths Class 10

## Question

Figure depicts a racing track whose left and right ends are semicircular. The distance between the two inner parallel line segments is \(\text{60 m}\) and they are each \(\text{106 m}\) long. If the track is \(\text{10 m}\) wide, find:

(i) The distance around the track along its inner edge

(ii) The area of the track.

## Text Solution

**What is known?**

(i) A racing track whose right and left ends are semicircular.

(ii) The distance between the \(2\) inner parallel line segments is \(\text{60 m}\) and they are \(\text{106 m}\) long

(iii) Width of track \(\text{= 10 cm}\)

**What is unknown?**

(i) The distance around the track along its inner edge.

(ii) Area of the track.

**Reasoning:**

(i) Draw a figure along with dimensions to visuals the track properly.

(ii) Visually the distance around the track along its inner edge.

\({ GH + {\rm{arc}}\, HIJ + JK + {\rm{arc}}\, KLG}\)

Where \(GH = JK = \text{106 m}\)

And arc \(HIJ = {\text{arc}}\, KLG =\) circumference of semicircle with diameter \(\text{60 m.}\)

So, it can be easily found by substituting the required values.

(iii) To find area of track

Visually it’s clear that

Area of the tack = Area of rectangle \(ABHG \;+\) Area of rectangle \(KJDE \;+\) (Area of semicircle\( BCD\; –\) Area of semicircle \(HIJ\)) + (Area of semicircle \(EFA\; – \) Area of semicircle \(KLG\))

Radii of semicircles \(HIJ\) and \(KLG \)\(\begin{align} {\text{ = }}\frac{{60}}{{\text{2}}}{\text{ = 30 m}}\end{align}\)

Radii of semicircles \(BCD\) and \(EFA =\text{ 30 m + 10 m = 40 m}\)

And \(JK = GH =\text{ 106 m}\)

And \(DJ = HB = \text{10 m}\)

Area of the track

\[\begin{align} = {GH} \times {HB} + {JK} \times {DJ} + \left( {\frac{1}{2}\pi {{(40)}^2} - \frac{1}{2}{{(30)}^2}} \right) + \left( {\frac{1}{2}\pi {{(40)}^2} - \frac{1}{2}{{(30)}^2}} \right)\end{align}\]

**Steps:**

Diameter of semicircle \(HIJ =\) Diameter of semicircle \(KLG = \rm{60 m}\)

\(\therefore\;\) their radius \(\begin{align}\left( {{r_1}} \right) = \frac{{60}}{2} = 30\,{\text{m}}\end{align}\)

(i) The distance around the track along its inner edge.

\[\begin{align}& =GH+ arc HIJ + JK + arc KLG\\&= 106 + \frac{{2\pi {r_1}}}{2} + 106 + \frac{{2\pi {r_1}}}{2}\\ &= {106 + \pi \times 30 + 106 + \pi \times 30}\\&={ 212 + \frac{{1320}}{7}}\\ &= \frac{{1484 + 1320}}{7}\\ &= \frac{{2804}}{7}\,{\text{m}}\end{align}\]

(ii) Radius of semicircle \(BCD =\) Radius of semicircle \(EFA\)

\[\begin{align}\left( {{r_2}} \right) &= 30\,\rm{m }+ 10m\\ &= 40\,\rm{m }\end{align}\]

Area of the tack \(=\) Area of rectangle \(ABHG \,+\) Area of rectangular \(KJDE \;+\) (Area of semicircle \(BCD\;-\) Area of semicircle \(HIJ\)) \(+ \)(Area of semicircle \(EFA\; -\) Area of semicircle \(KLG\))

\[\begin{align} &= \left( {106 \times 10} \right) + \left( {106 \times 10} \right) + \left[ {\frac{1}{2}\pi {{\left( {40} \right)}^2} - \frac{1}{2}\pi {{\left( {30} \right)}^2}} \right] + \left[ {\frac{1}{2}\pi {{\left( {40} \right)}^2} - \frac{1}{2}\pi {{\left( {30} \right)}^2}} \right]\\ &= 1060 + 1060 + \left[ {\frac{1}{2}\pi \left( {1600 - 900} \right)} \right] + \left[ {\frac{1}{2}\pi \left( {1600 - 900} \right)} \right]\\ &= 1060 + 1060 + \frac{\pi }{2} \times 700 + \frac{\pi }{2} \times 700\\ &= 2120 + 700\pi \\ &= 2120 + 700 \times \frac{{22}}{7}\\ &= 2120 + 2200\\ &= 4320{{\text{m}}^{\text{2}}}\end{align}\]