# Areas Related to Circles - NCERT Class 10 Maths

Areas Related to Circles

Exercise 12.1

## Question 1

The radii of two circles are \(19\, \rm{cm}\) and \(9\,\rm{cm}\) respectively. Find the radius of the circle which has circumference equal to the sum of the circumferences of the two circles.

### Solution

**Video Solution**

**What is known?**

Radii of two circles.

**What is unknown?**

Radius of \(3^\rm{rd}\) circle.

**Reasoning:**

Using the formula of circumference of circle \({C = 2}\pi r\) we find the radius of the circle.

**Steps:**

Radius \(({r_1})\)of \(1^ \rm{st}\) circle \(= 19 \,\rm{cm}\)

Radius \(({r_2})\) or \(2^\rm{nd} \) circle \(=9\,\rm{cm}\)

Let the radius of \(3^ \rm{rd}\) circle be \(r.\)

Circumference of \(1^\rm{st}\) circle

\(= 2 \pi{{\text{r}}_{\text{1}}}{\text{ = 2 }}\pi (19) = 38 \pi\)

Circumference of\(2^\rm{ nd }\) circle

\(= 2 \pi{{\text{r}}_{\text{2}}}{\text{ = 2 }}\pi (9) = 18 \pi\)

Circumference of \(3^\rm{rd}\)circle \(=2\pi r\)

Given that,

Circumference of \(3^\rm{rd}\) \(\rm{}circle\) \(=\) Circumference of \(1^\rm{st }\) \(\rm{}circle\) \(+\)Circumference of \(2^\text{nd}\)\(\rm{}circle.\)

\[\begin{align} 2 \pi r &=38 \pi+18 \pi \\ &=56 \pi \\ r &=\frac{56 \pi}{2 \pi} \\ &=28 \end{align}\]

Therefore, the radius of the circle which has circumference equal to the sum of the circumference of the given two circles is \(28\, \rm{cm.}\)

## Question 2

The radii of two circles are \(8\,\rm{cm}\) and \(6\,\rm{cm}\) respectively. Find the radius of the circle having area equal to the sum of the areas of the two circles.

### Solution

**Video Solution**

**What is known?**

Radii of two circles.

**What is unknown?**

Radius of \(3^\rm{rd}\) circle.

**Reasoning:**

Using the formula of area of circle \(A = \pi {r^2}\) we find the radius of the circle.

**Steps:**

Radius of \((r_1)\) \(1^\rm{st}\) circle \(= 8\,\rm{cm}\)

Radius of \((r_2)\)\(2^\rm{nd}\) circle \(= 6\,\rm{cm}\)

Let the radius of \(3^\rm{rd}\) circle \(=r.\)

Area of

\(1^\rm{st} \)circle \(=\pi \rm{r}_{1}^{2}= \pi (8)^2= 64\pi\)

Area of

\(2^\rm{nd}\) circle \(=\pi \rm{r}_{2}^{2}= \pi (6)^2= 36\pi\)

Given that,

Area of \(3^\rm{rd}\) circle \(=\) Area of \(1^\rm{st}\) circle \(+\) Area of \(2^\rm{nd}\) circle

\[\begin{align}{\pi{{{r}}^{{2}}}}& = {\pi {{r}}_{{1}}^{{2}}\,{{ + }}\,\pi {{r}}_{{2}}^{{2}}}\\{\pi {{{r}}^{{2}}}} &= {64\pi \,{{ + }}\,36\pi }\\{\pi{{{r}}^{{2}}}} &= {{100\pi }}\\{\,\,{{{r}}^{{2}}}} &={{{100}}}\\{\,\,{{r}}} &= \,\pm \,10\end{align}\]

However, the radius cannot be negative. Therefore, the radius of the circle having area equal to the sum of the areas of the two circles is \(10\,\rm{cm.}\)

## Question 3

Given figure depicts an archery target marked with its five scoring areas from the centre outwards as Gold, Red, Blue, Black and White. The diameter of the region representing Gold score is \(21 \,\rm{cm}\) and each of the other bands is \(10.5 \,\rm{cm}\) wide. Find the area of each of the five scoring regions.

[Use \(\pi= \,\frac{22}{7}\)]

### Solution

**Video Solution**

**What is known?**

Diameter of the gold region and width of the other regions.

**What is unknown?**

Area of each scoring region.

**Reasoning:**

Area of the region between \(2\) concentric circles is given by \(\begin{align}\pi {\text{r}}_2^2 - \pi {\text{r}}_1^2\end{align}\).

**Steps:**

Radius\(({r_1})\) of gold region (i.e., \(1^\rm{st}\)^{ }circle)\( = \frac{{21}}{2} = 10.5\,{\text{cm}}\)

Given that each circle is \(10.5\,\rm{cm}\) wider than the previous circle.

Therefore,

Radius \((r_2)\) of \(2^\rm{nd}\) circle

\[\begin{align}&= 10.5 + 10.5\\&=21 \,\rm{cm}\end{align}\]

Radius \(({r_3})\) of \(3^\rm{rd}\)circle

\[\begin{align}&= 21 + 10.5\\ &= 31.5\,{\text{cm}}\end{align}\]

Radius \((r_4)\) of \(4^\rm{th}\) circle

\[\begin{align}&= 31.5 + 10.5\\ &= 42\,{\text{cm}}\end{align}\]

Radius \((r_5)\) of \(5^\rm{th}\) circle

\[\begin{align}&= 42 + 10.5\\ &= 52.5\,\,{\text{cm}} \end{align}\]

Area of gold region\(=\) Area of \(1^\rm{st}\) circle \(= \pi {r}_1^2 = \pi {(10.5)^2} = 346.5\;\rm{cm^2}\)

Area of red region \(=\)Area of \(2^\text{nd }\)circle \(-\) Area of \(1^\rm{ st }\)circle

\[\begin{align}& = \pi {\text{r}}_2^2 - \pi {\text{r}}_1^2\\& = \pi {{(21)}^2} - {{(10.5)}^2}\\& = 441\pi - 110.25\pi = 330.75\pi \\& = 1039.5\,{\text{c}}{{\text{m}}^2}\end{align}\]

Area of blue region \(=\) Area of \(3^\text{rd}\) circle \( -\) Area of \(2^\rm{nd}\)circle

\[\begin{align}&= \pi _{13}^2 - \pi {\text{r}}_1^2\\& = \pi {{(31.5)}^2} - \pi {{(21)}^2}\\&= 992.25\pi - 441\pi = 551.25\pi \\&= 1732.5\,{\text{c}}{{\text{m}}^2}\end{align}\]

Area of black region\(=\) Area of \(4^\rm{th}\) circle \(-\)Area of \(3^\rm{rd}\)circle

\[\begin{align}& = \pi r_4^2 - \pi r_3^2\\& = \pi {{(42)}^2} - \pi {{(31.5)}^2}\\&= 1764\pi - 992.25\pi \\&= 771.75\pi \\ &= 2425.5\,{\text{c}}{{\text{m}}^2}\end{align}\]

Area of white region \(=\) Area of \(5^\rm{th}\) circle \(-\) Area of \(4^\rm{th}\)circle

\[\begin{align}&= \pi {\text{r}}_5^2 - \pi \pi _4^2\\&= \pi {{(52.5)}^2} - \pi {{(42)}^2}\\&= 2756.25\pi - 1764\pi \\&= 992.25\pi \\ &= 3118.5\,{\text{c}}{{\text{m}}^2}\end{align}\]

Therefore,areas of gold, red, blue, black, and white regions are \(346.5\, \rm{cm^2},\)\(1039.5 \,\rm{cm^2},\) \(1732.5 \,\rm{cm^2},\) \(2425.5\,\rm{cm^2},\)and\(3118.5 \,\rm{cm^2}\) respectively.

## Question 4

The wheels of a car are of diameter \(80\, \rm{cm}\) each. How many complete revolutions does each wheel make in \(10\) minutes when the car is traveling at a speed of \(66\, \rm{km}\) per hour?

### Solution

**Video Solution**

**What is known?**

Diameter of the wheel of the car and the speed of the car.

**What is unknown?**

Revolutions made by each wheel.

**Reasoning:**

Distance travelled by the wheel in one revolution is nothing but the circumference of the wheel itself.

**Steps:**

Diameter of the wheel of the car \(= 80\,\rm{cm}\)

Radius \((r)\) of the wheel of the car \(= 40\,\rm{cm}\)

Distance travelled in \(1\) revolution \(=\) Circumference of wheel

Circumference of wheel

\[\begin{align}&= 2\pi \,{ r}\\& = 2\pi \left( {{\text{40}}} \right)\\&= 80\pi\, \rm{cm}\end{align}\]

Speed of car\(= 66\, \text{km/hour}\)

\[\begin{align}&= \frac{{66 \times {\text{ }}100000}}{{60}}\,{\text{cm/}}\,{\text{min}}\\&= 110000 {\text{ cm/min}}\end{align}\]

Distance travelled by the car in \(10\) minutes

\[\begin{align}&= {\text{ }}110000{\text{ }} \times {\text{ }}10{\text{ }}\\&= {\text{ }}1100000{\text{ cm}}\end{align}\]

Let the number of revolutions of the wheel of the car be \(n\)

**\(\rm{n} \times\)**Distance travelled in\(1\) revolution \(=\)Distance travelled in \(10\) minutes

\[\begin{align}\\\rm{n} \times 80\pi &= 1100000\\{\text{n}} &= \frac{{1100000 \times 7}}{{80 \times 22}}\\ &= \frac{{35000}}{8}\\&= 4375\end{align}\]

Therefore, each wheel of the car will make \(4375\) revolutions.

## Question 5

Tick the correct answer in the following and justify your choice: If the perimeter and the area of a circle are numerically equal, then the radius of the circle is

(A) \(2\) units

(B) \(\pi \) units

(C) \(4\) units

(D) \(7\) units

### Solution

**Video Solution**

**What is known?**

Perimeter and area of the circle are numerically equal.

**What is unknown?**

Radius of circle.

**Reasoning:**

Given that Perimeter and area of the circle are numerically equal. We get \(2\pi \rm{r}= \pi \rm{r^2}.\)

Using this relation we find the radius.

**Steps:**

Let the radius of the circle \(= r.\)

Circumference of circle \( = 2\pi r\)

Area of circle \(= \pi \rm{r^2}\)

Given that, the circumference of the circle and the area of the circle are equal.

This implies \[\begin{align}2\pi \rm{r} &= \pi \rm{r^2}\\2 &= \rm{r}\end{align}\]

Therefore, the radius of the circle is \(2\) units.

Hence, the correct answer is \(\rm{A.}\)

In the chapter 12, Areas Related to Circles, the discussion begins with a review of the concepts of the perimeter (circumference) and area of a circle. Then these concepts are applied to determine the area of the sector, the length of an arc of a sector and area of the segment. The last section deals with the calculation of areas of combinations of plane figures. There are two exercises in this chapter - the first one deals with the problems to determine the area of the sector, segment and the length of the arc; and the second exercise deals with the problems to determine the area of a combination of plane figures.

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