NCERT Solutions for Class 10 Maths Chapter 11 Areas Related to Circles

Q1. Find the area of a sector of a circle with a radius of 6 cm if the angle of the sector is 60°.

Answer:

We know that the area of a sector having radius r and angle $\theta$ is given by:

Area $=\frac{\theta }{{360}^{\circ }}\times \pi r^2$

Thus, the area of the given sector is:

$$\begin{gathered} =\frac{{60}^{\circ }}{{360}^{\circ }}\times \pi \times 6^2 \\ =\frac{132}{7} \end{gathered}$$ cm²

Q2. Find the area of a quadrant of a circle whose circumference is 22 cm.

Answer:

We know the circumference of the circle = $2\pi r$

Thus, $2\pi r=22$
$\Rightarrow r=\frac{11}{\pi }$ cm

Also, we know that the area of a sector is given by:

Area $=\frac{\theta }{{360}^{\circ }}\times \pi r^2$

It is given that we need to find the area of a quadrant.
So, $\theta ={90}^{\circ }$

Hence, the area becomes:

$$\begin{gathered} =\frac{{90}^{\circ }}{{360}^{\circ }}\times \pi {\left(\frac{11}{\pi }\right)}^2 \\ =\frac{77}{8} \end{gathered}$$ cm²

Q3. The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.

Answer:

The minute hand rotates 360 degrees in one hour.

We need to find rotation in 5 min.

So, $\frac{{360}^{\circ }}{60}\times 5={30}^{\circ }$

The area of the sector
$=\frac{\theta }{{360}^{\circ }}\times \pi r^2$
$=\frac{{30}^{\circ }}{{360}^{\circ }}\times \pi \times {14}^2$
$=\frac{154}{3}$ cm²

Hence, the area swept by the minute hand in 5 minutes is $\frac{154}{3}$ cm².

Q4. A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding:

(i) Minor segment

(ii) Major sector. (Use π = 3.14)

Answer :

(i)The angle in the minor sector is 90°.

Thus, the area of the sector
$=\frac{\theta }{{360}^{\circ }}\times \pi r^2$
$=\frac{{90}^{\circ }}{{360}^{\circ }}\times \pi \times {10}^2$
$=\frac{1100}{14}=78.5$ cm²

Here, the base and height of the triangle are equal to the radius of the circle.
Now the area of a triangle is:
$=\frac{1}{2}\times$ Base $\times$ Height $=\frac{1}{2}\times 10\times 10=50$ cm²

Thus, the area of the minor segment
= Area of the sector - The area of a triangle
$=78.5-50=28.5$ cm²

(ii)The area of the major sector can be found directly by using the formula:

Area $=\frac{\theta }{{360}^{\circ }}\times \pi r^2$

In this case, the angle $=360°-90°=270°$

Thus, the area is:

$$\begin{gathered} =\frac{{270}^{\circ }}{{360}^{\circ }}\times \pi \times 10\times 10 \\ =\frac{3300}{14}=235.7 \end{gathered}$$ cm²

Q5. In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find:

(i) the length of the arc

(ii) area of the sector formed by the arc

(iii) the area of the segment formed by the corresponding chord.

Answer:

(i)The length of the arc is given by:-

Length of the arc $=\frac{\theta }{{360}^{\circ }}\times 2\pi r$

$$\begin{gathered} =\frac{{60}^{\circ }}{{360}^{\circ }}\times 2\times \pi \times 21 \\ =22 \end{gathered}$$ cm

Hence, the length of the arc is 22 cm.

(ii) Area of the sector

$=\frac{\theta }{{360}^{\circ }}\times \pi r^2$

$$\begin{gathered} =\frac{{60}^{\circ }}{{360}^{\circ }}\times \pi \times {21}^2 \\ =231 \end{gathered}$$ cm²

Thus, the area of the sector is 231 cm².

(iii) For the area of the segment, we need to subtract the area of the triangle attached to the area of the arc.

Thus, consider the triangle:-

It is given that the angle of the arc is 60°, or we can say that all angles are 60°. (Since the two sides are equal).
Hence, it is an equilateral triangle.

The area of the triangle is:

$$\begin{gathered} =\frac{\sqrt{3}}{4}\times {\text{Side}}^2 \\ =\frac{\sqrt{3}}{4}\times {21}^2 \\ =\frac{441\sqrt{3}}{4} \end{gathered}$$ cm²

Hence, the area of the segment is:

$=(231-\frac{441\sqrt{3}}{4})$ cm²

Q6. A chord of a circle of radius 15 cm subtends an angle of 60° at the centre. Find the areas of the corresponding minor and major segments of the circle.
(Use $\pi =3.14$ and $\sqrt{3}=1.73$ )

Answer:

The area of the sector is :

$$\begin{gathered} =\frac{\theta }{{360}^{\circ }}\times \pi r^2 \\ =\frac{{60}^{\circ }}{{360}^{\circ }}\times \pi \times {15}^2 \\ =117.75 \end{gathered}$$ cm²

Now consider the triangle; the angle of the sector is 60°.

This implies that it is an equilateral triangle. (As two sides are equal, they will have the same angle. This is possible only when all angles are equal i.e., 60°).

Thus, the area of the triangle is:-

$=\frac{\sqrt{3}}{4}\times {15}^2$

$=\frac{1.73\times 225}{4}$
$=97.31$ cm²

Hence area of the minor segment $=117.75-97.31=20.44$ cm²

The area of the major segment is:

$=\pi r^2-20.44$
$=\pi \times {15}^2-20.44$
$=(3.14\times 225)-20.44$
$=706.5-20.44$
$=686.06$ cm²

Q7. A chord of a circle of radius 12 cm subtends an angle of 120° at the centre. Find the area of the corresponding segment of the circle.
(Use $\pi =3.14$ and $\sqrt{3}=1.73$ )

Answer:

For the area of the segment, we need the area of the sector and the area of the associated triangle.

So, the area of the sector is :

$=\frac{{120}^{\circ }}{{360}^{\circ }}\times \pi \times {12}^2$

$=150.72$ cm²

Now, consider the triangle:-

Draw a perpendicular from the centre of the circle on the base of the triangle (let it be h).
r be the radius and b be the base of the triangle.

Using geometry, we can write,

$\frac{h}{r}=\cos {60}^{\circ }$

$\Rightarrow h=6$ cm

Similarly, $\frac{\frac{b}{2}}{r}=\sin {60}^{\circ }$

$\Rightarrow b=12\sqrt{3}$ cm

Thus, the area of the triangle is :

$=\frac{1}{2}\times 12\sqrt{3}\times 6$

$=62.28$ cm²

Hence, the area of the segment is: $=150.72-62.28=88.44$ cm²

Q8. A horse is tied to a peg at one corner of a square-shaped grass field of side 15 m using a 5 m long rope (see Fig). Find
(i) the area of that part of the field in which the horse can graze.

(ii) the increase in the grazing area if the rope were 10 m long instead of 5 m. (Use $\pi =3.14$)

Answer:

(i)The part grazed by the horse is given by = Area of sector

$=\frac{\theta }{{360}^{\circ }}\times \pi r^2$

$=\frac{{90}^{\circ }}{{360}^{\circ }}\times \pi \times 5^2$

$=\frac{25\pi }{4}$ m²

(ii)When the length of the rope is 10 m, the area grazed will be:-

$=\frac{\theta }{{360}^{\circ }}\times \pi r^2$

$=\frac{{90}^{\circ }}{{360}^{\circ }}\times \pi \times {10}^2$

$=25\pi$ m²

Hence, the change in the grazing area is given by :

$=25\pi -\frac{25\pi }{4}=58.88$ m²

Q9. A brooch is made with silver wire in the form of a circle with a diameter of 35 mm. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors, as shown in Fig.
Find:

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

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

Answer:

(i)The total wire required will be for 5 diameters and the circumference of the brooch.

The circumference of the brooch:-

$$\begin{gathered} =2\pi r \\ =2\times \pi \times \frac{35}{2} \\ =110 \end{gathered}$$ mm

Total required wire $=110+(5\times 35)=285$ mm

(ii) The total number of lines present in the brooch is 10 (line starting from the centre).

Thus, the angle of each sector is 36°.

The area of the sector is given by:

$=\frac{\theta }{{360}^{\circ }}\times \pi r^2$
$=\frac{{36}^{\circ }}{{360}^{\circ }}\times \pi \times {\left(\frac{35}{2}\right)}^2$
$=\frac{385}{4}$
$=96.25$ mm²

Q10. An umbrella has 8 ribs that are equally spaced (see Fig). Assuming the umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.

Answer:

It is given that the umbrella has 8 ribs so the angle of each sector is 45°.

Thus, the area of the sector is given by:

$=\frac{\theta }{{360}^{\circ }}\times \pi r^2$
$=\frac{{45}^{\circ }}{{360}^{\circ }}\times \pi \times {45}^2$
$=\frac{22275}{28}$ cm²

Hence, the area between two consecutive ribs is $\frac{22275}{28}$ cm².

Q11. A car has two wipers that do not overlap. Each wiper has a blade of length 25 cm sweeping through an angle of 115°. Find the total area cleaned at each sweep of the blades.

Answer:

The area cleaned by one wiper is:

$=\frac{\theta }{{360}^{\circ }}\times \pi r^2$

$=\frac{{115}^{\circ }}{{360}^{\circ }}\times \pi \times {25}^2$

$=\frac{158125}{252}$ cm²

Hence, the required area (area cleaned by both blades) is given by:-

$=2\times \frac{158125}{252}=\frac{158125}{126}$ cm²

Q12. To warn ships of underwater rocks, a lighthouse spreads a red-coloured light over a sector of angle 80° to a distance of 16.5 km. Find the area of the sea over which the ships are warned. (Use π = 3.14)

Answer:

The area of the sector is given by:

$=\frac{\theta }{{360}^{\circ }}\times \pi r^2$

In this case, the angle is ${80}^{\circ }$.

Thus, the area is:-

$=\frac{{80}^{\circ }}{{360}^{\circ }}\times \pi \times {16.5}^2$

$=189.97$ km²

Q13. A round table cover has six equal designs as shown in Fig. If the radius of the cover is 28 cm, find the cost of making the designs at the rate of Rs. 0.35 per cm². (Use $\sqrt{3}=1.7$ )

Answer:

The angle of each of the six sectors is 60° at the centre. $(\because \frac{{360}^{\circ }}{6}={60}^{\circ })$

The area of the sector is given by:-

$=\frac{\theta }{{360}^{\circ }}\times \pi r^2$

$=\frac{{60}^{\circ }}{{360}^{\circ }}\times \pi \times {28}^2$

$=410.66$ cm²

And the area of the equilateral triangle associated with the segment:-

$=\frac{\sqrt{3}}{4}\times$ Side²

$=\frac{\sqrt{3}}{4}\times {28}^2=333.2$ cm²

Hence, the area of segment $=410.66-333.2=77.46$ cm²

Thus the total area of design $=6\times 77.46=464.76$ cm²

So, the total cost for the design $=0.35\times 464.76=$ Rs. 162.66

Q14. Tick the correct answer in the following :

The area of a sector of angle p (in degrees) of a circle with radius R is:

(A) $\frac{p}{{180}^{\circ }}\times 2\pi R$

(B) $\frac{p}{{180}^{\circ }}\times \pi R^2$

(C) $\frac{p}{{360}^{\circ }}\times 2\pi R$

(D) $\frac{p}{{720}^{\circ }}\times 2\pi R^2$

Answer:

Area of a sector

$=\frac{\theta }{{360}^{\circ }}\times \pi r^2$, where $\theta =$ Central angle, $r$ = radius
Here, $\theta =p$ and $r=R$

$=\frac{p}{{360}^{\circ }}\times \pi R^2$
$=\frac{p}{{720}^{\circ }}\times 2\pi R^2$

Hence, option (D) is correct.

NCERT Solutions for Class 10 Maths Chapter Wise