Areas Related to Circles | Exercise 11.1

Question 8

A horse is tied to a peg at one corner of a square shaped grass field of side $15\text{ m}$ by means of a $5\text{ m}$ long rope (see Fig. 11.8). 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 $10\text{ m}$ long instead of $5\text{ m}$ . (Use $\pi = 3.14$ )

Solution

Understand the Question

The grass field is square, so each corner angle is $\theta = 90^\circ$ .
Tied to a corner peg, the horse can only graze within a sector of a circle (a quadrant) of radius equal to the rope length $r$ .
The area of a sector is given by: $\text{Area} = \dfrac{\theta}{360^\circ} \times \pi r^2 = \dfrac{90^\circ}{360^\circ} \times \pi r^2 = \dfrac{1}{4} \pi r^2$
For part (i), calculate the area with $r = 5\text{ m}$ .
For part (ii), calculate the new area with $r = 10\text{ m}$ and subtract the initial area to find the increase.

(i) the area of that part of the field in which the horse can graze.

Step 1 · Calculate Initial Grazing Area

The grazing area forms a sector of a circle with radius $r = 5\text{ m}$ and central angle $\theta = 90^\circ$ . Diagram 1

\begin{aligned} \text{Area} &= \dfrac{\text{Central Angle}}{360^\circ} \times \pi r^2 \\[0.6em] &= \dfrac{90^\circ}{360^\circ} \times 3.14 \times (5\text{ m})^2 \\[0.6em] &= \dfrac{1}{4} \times 3.14 \times 25 \\[0.6em] &= 0.25 \times 78.5 \\[0.6em] &= 19.625\text{ m}^2 \end{aligned}

Answer

(i) $19.625\text{ m}^2$

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

Step 1 · Calculate New Grazing Area

When the rope length is increased to $r = 10\text{ m}$ :

\begin{aligned} \text{New Area} &= \dfrac{90^\circ}{360^\circ} \times 3.14 \times (10\text{ m})^2 \\[0.6em] &= \dfrac{1}{4} \times 3.14 \times 100 \\[0.6em] &= 0.25 \times 314 \\[0.6em] &= 78.50\text{ m}^2 \end{aligned}

Step 2 · Find the Increase in Grazing Area

Subtract the initial grazing area from the new grazing area:

\begin{aligned} \text{Increase in Area} &= \text{New Area} - \text{Initial Area} \\[0.6em] &= 78.50\text{ m}^2 - 19.625\text{ m}^2 \\[0.6em] &= 58.875\text{ m}^2 \end{aligned}

Answer

(ii) $58.875\text{ m}^2$

Common Mistakes

Full Circle vs. Quadrant: Calculating the area of a full circle ( $\pi r^2$ ) instead of a quadrant ( $\dfrac{1}{4}\pi r^2$ ) formed at the $90^\circ$ corner of the square.
Using the Field's Side Length as Radius: Using the field side length ( $15\text{ m}$ ) instead of the rope length ( $5\text{ m}$ or $10\text{ m}$ ) as the radius $r$ .
Value of $\pi$ : Using $\pi = \dfrac{22}{7}$ instead of the specified $\pi = 3.14$ , which results in a slight calculation discrepancy.
Reporting Total Area Instead of Increase: Forgetting to subtract the initial area from the new area in part (ii).

More questions in Exercise 11.1

Unless stated otherwise, use $\pi = \dfrac{22}{7}$ .

Find the area of a sector of a circle with radius $6 \text{ cm}$ if angle of the sector is $60^\circ$ .

Find the area of a quadrant of a circle whose circumference is $22\text{ cm}$ .

The length of the minute hand of a clock is $14\text{ cm}$ . Find the area swept by the minute hand in $5$ minutes.

A chord of a circle of radius $10\text{ cm}$ subtends a right angle at the centre. Find the area of the corresponding :

(i) minor segment

(ii) major sector. (Use $\pi = 3.14$ )

In a circle of radius $21 \text{ cm}$ , an arc subtends an angle of $60^\circ$ at the centre. Find:

(i) the length of the arc

(ii) area of the sector formed by the arc

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

A chord of a circle of radius $15\text{ cm}$ subtends an angle of $60^\circ$ at the centre. Find the areas of the corresponding minor and major segments of the circle.

(Use $\pi = 3.14$ and $\sqrt{3} = 1.73$ )

A chord of a circle of radius $12\text{ cm}$ subtends an angle of $120^\circ$ at the centre. Find the area of the corresponding segment of the circle.

(Use $\pi = 3.14$ and $\sqrt{3} = 1.73$ )

A horse is tied to a peg at one corner of a square shaped grass field of side $15\text{ m}$ by means of a $5\text{ m}$ long rope (see Fig. 11.8). 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 $10\text{ m}$ long instead of $5\text{ m}$ . (Use $\pi = 3.14$ )

A brooch is made with silver wire in the form of a circle with diameter $35\text{ mm}$ . The wire is also used in making 5 diameters which divide the circle into 10 equal sectors as shown in Fig. 11.9. Find :

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

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

Q10

An umbrella has 8 ribs which are equally spaced (see Fig. 11.10). Assuming umbrella to be a flat circle of radius $45\text{ cm}$ , find the area between the two consecutive ribs of the umbrella.

Q11

A car has two wipers which do not overlap. Each wiper has a blade of length $25\text{ cm}$ sweeping through an angle of $115^\circ$ . Find the total area cleaned at each sweep of the blades.

Q12

To warn ships for underwater rocks, a lighthouse spreads a red coloured light over a sector of angle $80^\circ$ to a distance of $16.5\text{ km}$ . Find the area of the sea over which the ships are warned. (Use $\pi = 3.14$ )

Q13

A round table cover has six equal designs as shown in Fig. 11.11. If the radius of the cover is $28\text{ cm}$ , find the cost of making the designs at the rate of ₹ $0.35$ per $\text{cm}^2$ . (Use $\sqrt{3} = 1.7$ )

Q14

Tick the correct answer in the following :

Area of a sector of angle $p$ (in degrees) of a circle with radius $R$ is

(A) $\dfrac{p}{180} \times 2\pi R$ (B) $\dfrac{p}{180} \times \pi R^2$ (C) $\dfrac{p}{360} \times 2\pi R$ (D) $\dfrac{p}{720} \times 2\pi R^2$