CBSE Class 10 Mathematics Areas Related to Circles VBQs Set 04

Very short Questions

Question. Show that if the circumferences of two circles are equal, then their areas are also equal.
Answer: Let radii of two circles be ‘r’ and ‘R’.
We are given that
\( 2\pi r = 2\pi R \)
\( \implies \) \( r = R \)
and hence \( \pi r^2 = \pi R^2 \)
\( \implies \) The areas are equal.

Question. The diameter of a cycle wheel is 21 cm. How many revolutions will it make in moving 66 m?
Answer: Since Circumference of the wheel = \( 2\pi R \)
\( = 2 \times \frac{22}{7} \times 10.5 = 66 \text{ cm} \)
Hence, to cover a distance of 66 m, the wheel will make \( = \frac{66 \times 100}{66} = 100 \text{ revolutions} \).

Question. Is it true to say that area of a segment of a circle is less than the area of its corresponding sector? Why? 
Answer: No.
It is true only in case of minor segment. In case of major segment, area of segment is always greater than the area of its corresponding sector.

Question. Is it true to say that the area of a square inscribed in a circle of diameter p cm is p\(^2\) cm\(^2\)? Why? 
Answer: No.
When the square is inscribed in the circle, the diameter of the circle is equal to the diagonal of the square but not the side of the square.
Let side of square be a.
∴ Length of diagonal = \( \sqrt{a^2 + a^2} = \sqrt{2}a \)
\( \implies \) diameter of circle, \( p = \sqrt{2}a \)
\( \implies \) \( p^2 = 2a^2 \)
∴ Area of square \( a^2 = \frac{p^2}{2} \)

Question. Find the area of a sector of a circle of diameter 56 cm and central angle 45º. 
Answer: Given: diameter of the circle (d) = 56 cm
Then, the radius of circle (r) = 28 cm
Central angle, \( \theta = 45^{\circ} \)
The area of the sector = \( \frac{\theta}{360^{\circ}} \times \pi r^2 \)
\( = \frac{45}{360} \times \frac{22}{7} \times 28 \times 28 \)
\( = 308 \text{ cm}^2 \)
Hence, the area of the sector is 308 cm\(^2\).

Question. Find the area of a sector of a circle of radius 28 cm and central angle 45°.
Answer: Given: Radius of circle, r = 28 cm and central angle, \( \theta = 45^{\circ} \)
We know that,
If \( \theta \) is measured in degrees then
∴ Area of sector = \( \frac{\theta}{360^{\circ}} \times \pi r^2 \)
\( = \frac{45}{360} \times \frac{22}{7} \times 28 \times 28 \)
\( = 308 \text{ cm}^2 \)
Hence, the required area of the sector of the circle is 308 cm\(^2\).

Short Answer(SA-I) Type Questions

Question. The perimeter of a sector of a circle of radius 5.2 cm is 16.4 cm. Find the area of the sector. 
Answer: Let \( \theta \) be the central angle of the sector. Then,
length of arc of circle = \( \frac{\theta}{360^{\circ}} \times 2\pi r \)
Perimeter = \( \frac{\theta}{360^{\circ}} \times 2\pi r + 2r \)
\( 16.4 = \frac{\theta}{360^{\circ}} \times 2\pi (5.2) + 2 \times 5.2 \)
[\( \because r = 5.2 \text{ cm} \)]
\( \implies \) \( 5.2 \pi \theta = 1080 \text{ (on simplification)} \)
Now, area of the sector = \( \frac{\theta}{360^{\circ}} \times \pi (5.2)^2 \)
\( = \frac{1080 \times 5.2}{360} \text{ sq cm.} \)
\( = 15.6 \text{ sq. cm.} \)

Question. The diameter of two circles with centre A and B are 16 cm and 30 cm respectively. If area of another circle with centre C is equal to the sum of areas of these two circles, then find the circumference of the circle with centre C.
Answer: Area of circle = \( \pi r^2 \)
Let the radius of circle with centre C = R
According to the question we have,
\( \pi(8)^2 + \pi(15)^2 = \pi R^2 \)
\( \implies \) \( 64\pi + 225\pi = \pi R^2 \)
\( \implies \) \( 289\pi = \pi R^2 \)
\( \implies \) \( R^2 = 289 \)
\( \implies \) \( R = 17 \text{ cm} \)
Circumference of circle:
\( 2\pi r = 2\pi \times 17 \)
\( = 34\pi \text{ cm} \)
So, the circumference of the circle with centre C is \( 34\pi \text{ cm} \).

Question. The areas of two sectors of two different circles are equal. Is it necessary that their corresponding arc lengths are equal? Why? 
Answer: No. The given statement will be true for arcs of the same circle. But in different circles, it is not possible.
Area of 1st sector = \( \frac{1}{2} (r_1^2) \theta_1 \)
where, \( r_1 \) is the radius and \( \theta_1 \) is the angle.
Area of 2nd sector = \( \frac{1}{2} (r_2^2) \theta_2 \)
where, \( r_2 \) is the radius and \( \theta_2 \) is the angle subtended at the centre of the circle by the arc.
It is given that \( \frac{1}{2} r_1^2 \theta_1 = \frac{1}{2} r_2^2 \theta_2 \)
\( \implies \) \( r_1^2 \theta_1 = r_2^2 \theta_2 \)
Thus, it depends both on radius and angle subtended at the centre. But arc lengths depend only on radius.