CBSE Class 10 · Maths
CBSE Class 10 Maths Areas Related to Circles Competency Based Questions
Help your child score full marks on CBSE Class 10 Maths Areas Related to Circles competency based questions with this carefully curated question bank. Covering sectors, segments, arc lengths, and real-life applications, each question comes with a detailed, step-by-step solution verified by Angle Belearn’s CBSE specialists.
CBSE Class 10 Maths Areas Related to Circles — Questions with Solutions
What is the area of a semicircle of diameter $d$?
Solution
**(c)** Given $d$ be a diameter of circle. Let $r$ be the radius of the circle. $r = \frac{d}{2}$. Area of a semicircle $= \frac{\pi r^2}{2} = \frac{\pi}{2} \left( \frac{d}{2} \right)^2 = \frac{\pi}{2} \times \frac{d^2}{4} = \frac{\pi d^2}{8}$
The circumferences of two circles are in the ratio 4 : 5. What is the ratio of their radii?
Solution
**(d)** Let the radii be $r_1$ and $r_2$.
$$\frac{2\pi r_1}{2\pi r_2} = \frac{4}{5} \Rightarrow \frac{r_1}{r_2} = \frac{4}{5}$$
Thus, the required ratio is $4 : 5$.
If the perimeter and the area of a circle are numerically equal, then the radius of the circle is:
Solution
**(a)** Let radius $= r$. Area $=$ Perimeter:
$$\pi r^2 = 2\pi r \Rightarrow \pi r(r-2) = 0$$
Since $r \neq 0$, we get $r = 2$ units.
The radius of a circle is same as the side of a square. Their perimeters are in the ratio:
Solution
**(c)** Let radius $= r =$ side. Perimeter of circle $= 2\pi r$, perimeter of square $= 4r$.
$$\text{Ratio} = \frac{2\pi r}{4r} = \frac{\pi}{2} \Rightarrow \pi : 2$$
The area of the circle is 154 cm². The radius of the circle is:
Solution
**(a)** $\pi r^2 = 154 \Rightarrow \frac{22}{7} r^2 = 154 \Rightarrow r^2 = 49 \Rightarrow r = 7$ cm.
It is proposed to build a single circular park equal in area to the sum of areas of two circular parks of diameters 16 m and 12 m in a locality. The radius of the new park is:
Solution
**(a)** $r_1 = 8$ m, $r_2 = 6$ m. $\pi R^2 = \pi (8)^2 + \pi (6)^2 = \pi(64+36) = 100\pi$. So $R = 10$ m.
What is the length of the arc of the sector of a circle with radius 14 cm and of central angle 90°?
Solution
**(a)** Length of arc $= \frac{\theta}{360^\circ} \times 2\pi r = \frac{90}{360} \times 2 \times \frac{22}{7} \times 14 = \frac{1}{4} \times 88 = 22$ cm.
The area of a sector of angle $\alpha$ (in degrees) of a circle with radius $R$ is:
Solution
**(d)** The area of a sector of angle $\alpha$ of a circle with radius $R$ is $\dfrac{\alpha}{360^\circ} \times \pi R^2$.
A cow is tied to a peg at one corner of a rectangular field of dimensions 10 m × 8 m by a 3 m long rope. The area of the part of the field in which the cow can graze is:
Solution
**(b)** $\theta = 90^\circ$, $r = 3$ m. Area $= \frac{90}{360} \times \frac{22}{7} \times 9 = \frac{99}{14} = 7.07 \text{ m}^2$.
The area of the sector of a circle of radius 6 cm whose central angle is 30° is: [Take $\pi = 3.14$]
Solution
**(a)** Area $= \frac{30}{360} \times 3.14 \times 36 = \frac{1}{12} \times 113.04 = 9.42 \text{ cm}^2$.
Case Study: Sara holds a Japanese folding fan shaped like a sector of a circle. The inner and outer radii are 3 cm and 5 cm respectively. The fan has three colour regions — Red, Blue, and Green.
Q 1. If the blue region makes an angle of $80^\circ$ at the centre, the area of the blue region is:
a. $9.17 \text{ cm}^2$ b. $10.1 \text{ cm}^2$ c. $11.17 \text{ cm}^2$ d. $13.17 \text{ cm}^2$
Q 2. If the green region makes an angle of $60^\circ$ at the centre, the area of the green region is:
a. $6.2 \text{ cm}^2$ b. $8.38 \text{ cm}^2$ c. $9.9 \text{ cm}^2$ d. $11.12 \text{ cm}^2$
Q 3. If the red region makes an angle of $20^\circ$ at the centre, the perimeter of the red region is:
a. $2.9 \text{ cm}$ b. $4.2 \text{ cm}$ c. $5.4 \text{ cm}$ d. $6.79 \text{ cm}$
Q 4. The area of the region having radius 3 cm (using all three angles) is:
a. $12.57 \text{ cm}^2$ b. $14.8 \text{ cm}^2$ c. $20 \text{ cm}^2$ d. $26.57 \text{ cm}^2$
Q 5. The region in the figure represents:
a. minor sector b. major sector c. minor segment d. major segment
Q 1. If the blue region makes an angle of $80^\circ$ at the centre, the area of the blue region is:
a. $9.17 \text{ cm}^2$ b. $10.1 \text{ cm}^2$ c. $11.17 \text{ cm}^2$ d. $13.17 \text{ cm}^2$
Q 2. If the green region makes an angle of $60^\circ$ at the centre, the area of the green region is:
a. $6.2 \text{ cm}^2$ b. $8.38 \text{ cm}^2$ c. $9.9 \text{ cm}^2$ d. $11.12 \text{ cm}^2$
Q 3. If the red region makes an angle of $20^\circ$ at the centre, the perimeter of the red region is:
a. $2.9 \text{ cm}$ b. $4.2 \text{ cm}$ c. $5.4 \text{ cm}$ d. $6.79 \text{ cm}$
Q 4. The area of the region having radius 3 cm (using all three angles) is:
a. $12.57 \text{ cm}^2$ b. $14.8 \text{ cm}^2$ c. $20 \text{ cm}^2$ d. $26.57 \text{ cm}^2$
Q 5. The region in the figure represents:
a. minor sector b. major sector c. minor segment d. major segment
Answer
Q1 (c) $11.17 \text{ cm}^2$: Area of annular sector $= \frac{80}{360} \times \frac{22}{7} \times (5^2 – 3^2) = \frac{2}{9} \times \frac{22}{7} \times 16 = \frac{44}{63} \times 16 = 11.17 \text{ cm}^2$
Q2 (b) $8.38 \text{ cm}^2$: $\frac{60}{360} \times \frac{22}{7} \times 16 = \frac{1}{6} \times \frac{352}{7} = 8.38 \text{ cm}^2$
Q3 (d) $6.79 \text{ cm}$: Perimeter $= 2 + 2 + \text{arc}(r=3) + \text{arc}(r=5) = 4 + \frac{20}{360}\times\frac{22}{7}\times(3+5)\times2 = 4 + 2.79 = 6.79 \text{ cm}$
Q4 (a) $12.57 \text{ cm}^2$: $\frac{22}{7} \times 9 \times \frac{160}{360} = \frac{88}{7} \approx 12.57 \text{ cm}^2$
Q5 (a) Minor sector: Total angle $= 80^\circ + 60^\circ + 20^\circ = 160^\circ < 180^\circ$, so it is a minor sector.
Q2 (b) $8.38 \text{ cm}^2$: $\frac{60}{360} \times \frac{22}{7} \times 16 = \frac{1}{6} \times \frac{352}{7} = 8.38 \text{ cm}^2$
Q3 (d) $6.79 \text{ cm}$: Perimeter $= 2 + 2 + \text{arc}(r=3) + \text{arc}(r=5) = 4 + \frac{20}{360}\times\frac{22}{7}\times(3+5)\times2 = 4 + 2.79 = 6.79 \text{ cm}$
Q4 (a) $12.57 \text{ cm}^2$: $\frac{22}{7} \times 9 \times \frac{160}{360} = \frac{88}{7} \approx 12.57 \text{ cm}^2$
Q5 (a) Minor sector: Total angle $= 80^\circ + 60^\circ + 20^\circ = 160^\circ < 180^\circ$, so it is a minor sector.
Case Study: Consider two pizzas of equal diameter 12 inches. Pizza (I) is cut into 6 equal slices; Pizza (II) is cut into 8 equal slices.
Q 1. The area of one slice in pizza (I) is:
a. $6\pi$ sq. in b. $8\pi$ sq. in c. $10\pi$ sq. in d. None of these
Q 2. The perimeter of one slice of pizza (I) is:
a. $(\pi + 12)$ in b. $(\pi + 10)$ in c. $(2\pi + 10)$ in d. $(2\pi + 12)$ in
Q 3. The ratio of the area of one slice to the remaining pizza in (I) is:
a. $5:1$ b. $1:5$ c. $2:5$ d. $5:3$
Q 4. The ratio of areas of each slice of pizza (I) and (II) is:
a. $3:4$ b. $5:3$ c. $4:3$ d. $2:5$
Q 5. The relation between area of a sector ($A$), arc length ($l$), angle ($\theta$) and radius ($r$) is:
a. $\frac{1}{2}lr$ b. $lr$ c. $\frac{1}{3}lr$ d. $\frac{1}{2}lr^2$
Q 1. The area of one slice in pizza (I) is:
a. $6\pi$ sq. in b. $8\pi$ sq. in c. $10\pi$ sq. in d. None of these
Q 2. The perimeter of one slice of pizza (I) is:
a. $(\pi + 12)$ in b. $(\pi + 10)$ in c. $(2\pi + 10)$ in d. $(2\pi + 12)$ in
Q 3. The ratio of the area of one slice to the remaining pizza in (I) is:
a. $5:1$ b. $1:5$ c. $2:5$ d. $5:3$
Q 4. The ratio of areas of each slice of pizza (I) and (II) is:
a. $3:4$ b. $5:3$ c. $4:3$ d. $2:5$
Q 5. The relation between area of a sector ($A$), arc length ($l$), angle ($\theta$) and radius ($r$) is:
a. $\frac{1}{2}lr$ b. $lr$ c. $\frac{1}{3}lr$ d. $\frac{1}{2}lr^2$
Answer
Q1 (a) $6\pi$ sq. in: $r = 6$ in, $\theta = 60^\circ$. Area $= \frac{60}{360}\pi(6)^2 = \frac{1}{6}\times 36\pi = 6\pi$
Q2 (d) $(2\pi + 12)$ in: Arc length $l = \frac{60}{360}\times 2\pi\times 6 = 2\pi$. Perimeter $= l + 2r = 2\pi + 12$
Q3 (b) $1:5$: One slice angle $= 60^\circ$, remaining $= 300^\circ$. Ratio $= 60:300 = 1:5$
Q4 (c) $4:3$: Slice (I) angle $= 60^\circ$, slice (II) angle $= 45^\circ$. Ratio $= 60:45 = 4:3$
Q5 (a) $\frac{1}{2}lr$: $A = \frac{\theta}{360}\pi r^2$ and $l = \frac{\theta}{360}\times 2\pi r$, so $A = \frac{1}{2} \times l \times r$.
Q2 (d) $(2\pi + 12)$ in: Arc length $l = \frac{60}{360}\times 2\pi\times 6 = 2\pi$. Perimeter $= l + 2r = 2\pi + 12$
Q3 (b) $1:5$: One slice angle $= 60^\circ$, remaining $= 300^\circ$. Ratio $= 60:300 = 1:5$
Q4 (c) $4:3$: Slice (I) angle $= 60^\circ$, slice (II) angle $= 45^\circ$. Ratio $= 60:45 = 4:3$
Q5 (a) $\frac{1}{2}lr$: $A = \frac{\theta}{360}\pi r^2$ and $l = \frac{\theta}{360}\times 2\pi r$, so $A = \frac{1}{2} \times l \times r$.
Case Study: A circular park of radius $r$ m has two flower segments. One segment subtending $90^\circ$ is full of red roses; another subtending $60^\circ$ is full of yellow flowers. The combined area of both segments is $256\frac{2}{3}$ sq. m.
Q 1. Write an equation representing the total area of the two segments in terms of $r$.
Q 2. Find the value of $r$.
Q 3. Find the area of the segment with red roses. Or Find the area of the segment with yellow flowers.
Q 1. Write an equation representing the total area of the two segments in terms of $r$.
Q 2. Find the value of $r$.
Q 3. Find the area of the segment with red roses. Or Find the area of the segment with yellow flowers.
Answer
Q1: Area of red segment $= \frac{r^2}{2}\left(\frac{\pi}{2}-1\right)$; area of yellow segment $= \frac{r^2}{2}\left(\frac{\pi}{3}-\frac{\sqrt{3}}{2}\right)$.
Combined: $\frac{r^2}{2}\left(\frac{5\pi}{6} – 1 – \frac{\sqrt{3}}{2}\right) = \frac{770}{3}$
Q2: Solving numerically, $\frac{r^2}{2}\times 2.619 = \frac{770}{3}$, giving $r^2 \approx 681.72$, so $r \approx 26.11$ m.
Q3 (Red roses): Area $= \frac{681.72}{2}\times\left(\frac{\pi}{2}-1\right) \approx 340.86 \times 0.57 \approx 194.3 \text{ m}^2$.
Or (Yellow flowers): Area $= 340.86\times\left(\frac{\pi}{3}-\frac{\sqrt{3}}{2}\right) \approx 340.86 \times 0.182 \approx 61.9 \text{ m}^2$.
Combined: $\frac{r^2}{2}\left(\frac{5\pi}{6} – 1 – \frac{\sqrt{3}}{2}\right) = \frac{770}{3}$
Q2: Solving numerically, $\frac{r^2}{2}\times 2.619 = \frac{770}{3}$, giving $r^2 \approx 681.72$, so $r \approx 26.11$ m.
Q3 (Red roses): Area $= \frac{681.72}{2}\times\left(\frac{\pi}{2}-1\right) \approx 340.86 \times 0.57 \approx 194.3 \text{ m}^2$.
Or (Yellow flowers): Area $= 340.86\times\left(\frac{\pi}{3}-\frac{\sqrt{3}}{2}\right) \approx 340.86 \times 0.182 \approx 61.9 \text{ m}^2$.
Case Study (Rangoli): A Rangoli design is in the shape of square $ABCD$ with side 40 cm. At each corner, a quadrant of circle of radius 10 cm is drawn. A circle of diameter 20 cm is drawn inside the square.
Q 1. What is the area of square $ABCD$?
Q 2. Find the area of the circle.
Q 3. Find the area of the remaining portion after removing the circle and four quadrants. Or Find the combined area of the 4 quadrants and the circle removed.
Q 1. What is the area of square $ABCD$?
Q 2. Find the area of the circle.
Q 3. Find the area of the remaining portion after removing the circle and four quadrants. Or Find the combined area of the 4 quadrants and the circle removed.
Answer
Q1: Area of square $= (40)^2 = 1600 \text{ cm}^2$
Q2: Radius of circle $= 10$ cm. Area $= \pi(10)^2 = 100\pi = 314 \text{ cm}^2$
Q3: Area of one quadrant $= \frac{1}{4}\pi(10)^2 = 25\pi$. Four quadrants $= 100\pi = 314 \text{ cm}^2$.
Remaining area $= 1600 – 314 – 314 = 972 \text{ cm}^2$.
Or: Combined area removed $= 314 + 314 = 628 \text{ cm}^2$.
Q2: Radius of circle $= 10$ cm. Area $= \pi(10)^2 = 100\pi = 314 \text{ cm}^2$
Q3: Area of one quadrant $= \frac{1}{4}\pi(10)^2 = 25\pi$. Four quadrants $= 100\pi = 314 \text{ cm}^2$.
Remaining area $= 1600 – 314 – 314 = 972 \text{ cm}^2$.
Or: Combined area removed $= 314 + 314 = 628 \text{ cm}^2$.
Case Study (Pendulum Clock): A pendulum clock has a pendulum of length 15 cm. The minute hand is 9 cm long and the hour hand is 6 cm long.
Q 1. Find the area swept by the minute hand in 10 minutes.
Q 2. If the pendulum covers 22 cm in one complete oscillation, find the angle described by the pendulum at the centre.
Q 3. Find the area swept by the hour hand in 1 hour. Or Find the area swept by the hour hand between 11 am and 5 pm.
Q 1. Find the area swept by the minute hand in 10 minutes.
Q 2. If the pendulum covers 22 cm in one complete oscillation, find the angle described by the pendulum at the centre.
Q 3. Find the area swept by the hour hand in 1 hour. Or Find the area swept by the hour hand between 11 am and 5 pm.
Answer
Q1: Angle in 10 min $= \frac{360 \times 10}{60} = 60^\circ$. Area $= \frac{60}{360} \times \frac{22}{7} \times 81 = \frac{22 \times 81}{42} = \frac{594}{7} \approx 84.86 \text{ cm}^2$
Q2: Arc for half oscillation $= 11$ cm; $l = \frac{\theta}{360}\times 2\pi R \Rightarrow 11 = \frac{\theta}{360}\times\frac{44}{7}\times 15$. Solving: $\theta = 42^\circ$.
Q3 (1 hour): Angle $= 30^\circ$. Area $= \frac{30}{360}\times\frac{22}{7}\times 36 = \frac{66}{7} \approx 9.43 \text{ cm}^2$.
Or (11am–5pm = 6 hours): $6 \times 9.43 \approx 56.57 \text{ cm}^2$.
Q2: Arc for half oscillation $= 11$ cm; $l = \frac{\theta}{360}\times 2\pi R \Rightarrow 11 = \frac{\theta}{360}\times\frac{44}{7}\times 15$. Solving: $\theta = 42^\circ$.
Q3 (1 hour): Angle $= 30^\circ$. Area $= \frac{30}{360}\times\frac{22}{7}\times 36 = \frac{66}{7} \approx 9.43 \text{ cm}^2$.
Or (11am–5pm = 6 hours): $6 \times 9.43 \approx 56.57 \text{ cm}^2$.
Assertion–Reason: Choose the correct option:
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): In a circle of radius 6 cm, sector angle $60^\circ$. Then the area of the sector is $\frac{18}{7}\text{ cm}^2$.
Reason (R): Area of the circle with radius $r$ is $\pi r^2$.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): In a circle of radius 6 cm, sector angle $60^\circ$. Then the area of the sector is $\frac{18}{7}\text{ cm}^2$.
Reason (R): Area of the circle with radius $r$ is $\pi r^2$.
Answer
Correct Answer: (B)
Assertion (A) is true: Area $= \frac{60}{360}\times\frac{22}{7}\times 36 = \frac{132}{7} = 18\frac{6}{7}\text{ cm}^2$ ✓
Reason (R) is also true, but it is NOT the correct explanation of Assertion (A), since the sector area formula — not the full circle area — is what explains the result.
Assertion (A) is true: Area $= \frac{60}{360}\times\frac{22}{7}\times 36 = \frac{132}{7} = 18\frac{6}{7}\text{ cm}^2$ ✓
Reason (R) is also true, but it is NOT the correct explanation of Assertion (A), since the sector area formula — not the full circle area — is what explains the result.
Assertion–Reason: Choose the correct option:
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): The minute hand of a clock is 7 cm. Then the area swept in 5 minutes is $\frac{12}{6}\text{ cm}^2$.
Reason (R): Length of arc $= \frac{\theta}{360^\circ} \times 2\pi r$.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): The minute hand of a clock is 7 cm. Then the area swept in 5 minutes is $\frac{12}{6}\text{ cm}^2$.
Reason (R): Length of arc $= \frac{\theta}{360^\circ} \times 2\pi r$.
Answer
Correct Answer: (B)
Assertion (A) is true: Angle in 5 min $= 30^\circ$. Area $= \frac{30}{360}\times\frac{22}{7}\times 49 = \frac{77}{6} = 12\frac{5}{6}\text{ cm}^2$ ✓
Reason (R) is also true, but it gives the arc length formula, not the sector area formula — so it does NOT correctly explain the Assertion.
Assertion (A) is true: Angle in 5 min $= 30^\circ$. Area $= \frac{30}{360}\times\frac{22}{7}\times 49 = \frac{77}{6} = 12\frac{5}{6}\text{ cm}^2$ ✓
Reason (R) is also true, but it gives the arc length formula, not the sector area formula — so it does NOT correctly explain the Assertion.
Assertion–Reason: Choose the correct option:
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): If the perimeter of a sector of radius 5.6 cm is 27.2 cm, then the area of the sector is $44.8\text{ cm}^2$.
Reason (R): The area of a sector is $\frac{\theta}{360^\circ} \times \pi r^2$.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): If the perimeter of a sector of radius 5.6 cm is 27.2 cm, then the area of the sector is $44.8\text{ cm}^2$.
Reason (R): The area of a sector is $\frac{\theta}{360^\circ} \times \pi r^2$.
Answer
Correct Answer: (C)
Assertion (A) is true: Perimeter $= 2r + l = 27.2 \Rightarrow l = 16$ cm. Area $= \frac{1}{2}lr = \frac{1}{2}\times16\times5.6 = 44.8\text{ cm}^2$ ✓
Reason (R) is false: The formula $\frac{\theta}{360}\times\pi r^2$ is itself correct as a formula, but since it was incorrectly stated in the given option, Assertion is true but Reason as stated is false.
Assertion (A) is true: Perimeter $= 2r + l = 27.2 \Rightarrow l = 16$ cm. Area $= \frac{1}{2}lr = \frac{1}{2}\times16\times5.6 = 44.8\text{ cm}^2$ ✓
Reason (R) is false: The formula $\frac{\theta}{360}\times\pi r^2$ is itself correct as a formula, but since it was incorrectly stated in the given option, Assertion is true but Reason as stated is false.
Assertion–Reason: Choose the correct option:
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): A sector is cut from a circle of radius 42 cm with central angle $150^\circ$. The perimeter of the sector is 194 cm.
Reason (R): Perimeter of sector $= 2 \times \text{radius} + \text{length of arc}$.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): A sector is cut from a circle of radius 42 cm with central angle $150^\circ$. The perimeter of the sector is 194 cm.
Reason (R): Perimeter of sector $= 2 \times \text{radius} + \text{length of arc}$.
Answer
Correct Answer: (A)
Assertion (A) is true: Perimeter $= 2(42) + \frac{150}{360}\times2\times\frac{22}{7}\times42 = 84 + 110 = 194$ cm ✓
Reason (R) is true and it correctly explains Assertion (A). ✓
Assertion (A) is true: Perimeter $= 2(42) + \frac{150}{360}\times2\times\frac{22}{7}\times42 = 84 + 110 = 194$ cm ✓
Reason (R) is true and it correctly explains Assertion (A). ✓
Assertion–Reason: Choose the correct option:
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): In a circle of radius 6 cm with sector angle $60^\circ$, the area of the sector is $\frac{132}{7}\text{ cm}^2$.
Reason (R): Area of circle with radius $r$ is $\pi r^2$.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): In a circle of radius 6 cm with sector angle $60^\circ$, the area of the sector is $\frac{132}{7}\text{ cm}^2$.
Reason (R): Area of circle with radius $r$ is $\pi r^2$.
Answer
Correct Answer: (B)
Both Assertion (A) and Reason (R) are true. Area $= \frac{60}{360}\times\frac{22}{7}\times36 = \frac{132}{7}\text{ cm}^2$ ✓. But R is not the correct explanation of A.
Both Assertion (A) and Reason (R) are true. Area $= \frac{60}{360}\times\frac{22}{7}\times36 = \frac{132}{7}\text{ cm}^2$ ✓. But R is not the correct explanation of A.
The area of the circle that can be inscribed in a square of side 6 cm is:
Solution
**(d)** Diameter of inscribed circle $=$ side of square $= 6$ cm, so $r = 3$ cm. Area $= \pi(3)^2 = 9\pi\text{ cm}^2$.
The minute hand of a clock is 84 cm long. The distance covered by the tip of the minute hand from 10:10 am to 10:25 am is:
Solution
**(c)** Time $= 15$ min. Angle $= 6^\circ \times 15 = 90^\circ$. Distance $= \frac{90}{360}\times2\times\frac{22}{7}\times84 = \frac{1}{4}\times528 = 132$ cm.
Find the area of the major segment of a circle if the area of the minor segment is $25\text{ cm}^2$ and the area of the circle is $100\text{ cm}^2$:
Solution
**(c)** Area of major segment $=$ Area of circle $-$ Area of minor segment $= 100 – 25 = 75\text{ cm}^2$.
Priyanshu has a motorcycle with wheels of diameter 91 cm. There are 22 spokes in the wheel. Find the length of the arc between two adjoining spokes.
Solution
**(b)** $r = \frac{91}{2}$ cm, $\theta = \frac{360^\circ}{22}$. Arc length $= \frac{\theta}{360^\circ}\times2\pi r = \frac{1}{22}\times2\times\frac{22}{7}\times\frac{91}{2} = \frac{91}{7} = 13$ cm.
In the given figure, the ratio of the areas of two sectors $S_1$ and $S_2$ is: (Sector $S_1$ has central angle $120^\circ$ and sector $S_2$ has central angle $150^\circ$, both with the same radius.)
Solution
**(d)** Area $\propto \theta$. Ratio $= \frac{120}{150} = \frac{4}{5}$. So the required ratio is $4:5$.
A car has two wipers which do not overlap. Each wiper has a blade of length 42 cm sweeping through an angle of $120^\circ$. Find the total area cleaned at each sweep of the blades:
Solution
**(b)** Total area $= 2 \times \frac{120}{360} \times \frac{22}{7} \times 42^2 = 2 \times \frac{1}{3} \times \frac{22}{7} \times 1764 = 2 \times 1848 = 3696\text{ cm}^2$.
A chord of a circle of radius 28 cm subtends an angle of $60^\circ$ at the centre of the circle. The area of the minor segment is: [Take $\sqrt{3} = 1.732$]
Solution
**(d)** Since $\theta = 60^\circ$ and $OA = OB$, $\triangle OAB$ is equilateral. Area of $\triangle OAB = \frac{\sqrt{3}}{4}\times28^2 = 339.47\text{ cm}^2$. Area of sector $= \frac{60}{360}\times\frac{22}{7}\times784 = 410.67\text{ cm}^2$. Minor segment area $= 410.67 – 339.47 = 71.20\text{ cm}^2$.
In the given figure, three sectors of a circle of radius 7 cm, making angles of $60^\circ$, $80^\circ$ and $40^\circ$ at the centre are shaded. The area of the shaded region is: [Using $\pi = \frac{22}{7}$]
Solution
**(a)** Total angle $= 60^\circ + 80^\circ + 40^\circ = 180^\circ$. Shaded area $= \frac{180}{360}\times\frac{22}{7}\times49 = \frac{1}{2}\times154 = 77\text{ cm}^2$.
O is the centre of a circle of radius 5 cm. The chord AB subtends an angle of $60^\circ$ at the centre. The area of the shaded portion (circle minus triangle) is approximately:
Solution
**(d)** Since $\angle AOB = 60^\circ$ and $OA = OB$, $\triangle OAB$ is equilateral with side 5 cm. Area of $\triangle = \frac{\sqrt{3}}{4}\times25 = \frac{25\times1.732}{4} = 10.83\text{ cm}^2$. Area of circle $= \frac{22}{7}\times25 = 78.57\text{ cm}^2$. Shaded area $= 78.57 – 10.83 = 67.74\text{ cm}^2$.
A unit square ROST is inscribed in a circular sector with centre O. Along with the above information, which of these is sufficient to find the area of sector POQ?
Solution
**(c)** In $\triangle OTR$: $OT = \sqrt{TR^2 + OR^2} = \sqrt{1+1} = \sqrt{2}$. Area of sector $= \frac{1}{2}r\theta$. We know $r = \sqrt{2}$, but we still need $\theta$ or equivalently the arc length PQ, since $l = r\theta$ gives us $\theta = \frac{l}{r}$.
Case Study (School Badges): Circular badges with red and silver regions: diameter of red region $= 22$ cm; silver ring width $= 10.5$ cm.
(i) The radius of the red region is:
(a) 9 cm (b) 10 cm (c) 11 cm (d) 12 cm
(ii) Find the area of the red region:
(a) $380.28\text{ cm}^2$ (b) $382.28\text{ cm}^2$ (c) $384.28\text{ cm}^2$ (d) $378.28\text{ cm}^2$
(iii) Radius of the circle formed by combining red and silver region:
(a) 20.5 cm (b) 21.5 cm (c) 22.5 cm (d) 23.5 cm
(iv) Area of the silver region:
(a) $172.50\text{ cm}^2$ (b) $1062.50\text{ cm}^2$ (c) $1172.50\text{ cm}^2$ (d) $1072.50\text{ cm}^2$
(v) Area of circular path formed by concentric circles of radii $r_1 > r_2$:
(a) $\pi(r_1^2+r_2^2)$ (b) $\pi(r_1^2-r_2^2)$ (c) $2\pi(r_1-r_2)$ (d) $2\pi(r_1+r_2)$
(i) The radius of the red region is:
(a) 9 cm (b) 10 cm (c) 11 cm (d) 12 cm
(ii) Find the area of the red region:
(a) $380.28\text{ cm}^2$ (b) $382.28\text{ cm}^2$ (c) $384.28\text{ cm}^2$ (d) $378.28\text{ cm}^2$
(iii) Radius of the circle formed by combining red and silver region:
(a) 20.5 cm (b) 21.5 cm (c) 22.5 cm (d) 23.5 cm
(iv) Area of the silver region:
(a) $172.50\text{ cm}^2$ (b) $1062.50\text{ cm}^2$ (c) $1172.50\text{ cm}^2$ (d) $1072.50\text{ cm}^2$
(v) Area of circular path formed by concentric circles of radii $r_1 > r_2$:
(a) $\pi(r_1^2+r_2^2)$ (b) $\pi(r_1^2-r_2^2)$ (c) $2\pi(r_1-r_2)$ (d) $2\pi(r_1+r_2)$
Answer
(i) (c) 11 cm: $r = \frac{22}{2} = 11$ cm
(ii) (a) $380.28\text{ cm}^2$: $\pi(11)^2 = 121\pi \approx 380.28\text{ cm}^2$
(iii) (b) 21.5 cm: $11 + 10.5 = 21.5$ cm
(iv) (d) $1072.50\text{ cm}^2$: $\pi(21.5)^2 – \pi(11)^2 = \pi(462.25-121) = 341.25\pi \approx 1072.46\text{ cm}^2$
(v) (b) $\pi(r_1^2 – r_2^2)$: Area of annular ring $= \pi r_1^2 – \pi r_2^2 = \pi(r_1^2 – r_2^2)$
(ii) (a) $380.28\text{ cm}^2$: $\pi(11)^2 = 121\pi \approx 380.28\text{ cm}^2$
(iii) (b) 21.5 cm: $11 + 10.5 = 21.5$ cm
(iv) (d) $1072.50\text{ cm}^2$: $\pi(21.5)^2 – \pi(11)^2 = \pi(462.25-121) = 341.25\pi \approx 1072.46\text{ cm}^2$
(v) (b) $\pi(r_1^2 – r_2^2)$: Area of annular ring $= \pi r_1^2 – \pi r_2^2 = \pi(r_1^2 – r_2^2)$
Case Study (Button): A circular button has a large circle of radius 16 cm. The diameter of each of the smaller identical circles is $\frac{1}{4}$ of the diameter of the larger circle (so smaller radius $= 4$ cm). The remaining (non-white) area is black.
(i) Area of each smaller circle:
(a) $40.28\text{ cm}^2$ (b) $46.39\text{ cm}^2$ (c) $50.28\text{ cm}^2$ (d) $52.3\text{ cm}^2$
(ii) Area of the larger circle:
(a) $804.57\text{ cm}^2$ (b) $704.57\text{ cm}^2$ (c) $855.57\text{ cm}^2$ (d) $990.57\text{ cm}^2$
(iii) Area of the black region (4 smaller circles removed):
(a) $600.45\text{ cm}^2$ (b) $603.45\text{ cm}^2$ (c) $610.45\text{ cm}^2$ (d) $623.45\text{ cm}^2$
(iv) Area of a quadrant of a smaller circle:
(a) $11.57\text{ cm}^2$ (b) $13.68\text{ cm}^2$ (c) $12\text{ cm}^2$ (d) $12.57\text{ cm}^2$
(v) If two concentric circles have radii 2 cm and 5 cm, the area between them is:
(a) $60\text{ cm}^2$ (b) $63\text{ cm}^2$ (c) $66\text{ cm}^2$ (d) $68\text{ cm}^2$
(i) Area of each smaller circle:
(a) $40.28\text{ cm}^2$ (b) $46.39\text{ cm}^2$ (c) $50.28\text{ cm}^2$ (d) $52.3\text{ cm}^2$
(ii) Area of the larger circle:
(a) $804.57\text{ cm}^2$ (b) $704.57\text{ cm}^2$ (c) $855.57\text{ cm}^2$ (d) $990.57\text{ cm}^2$
(iii) Area of the black region (4 smaller circles removed):
(a) $600.45\text{ cm}^2$ (b) $603.45\text{ cm}^2$ (c) $610.45\text{ cm}^2$ (d) $623.45\text{ cm}^2$
(iv) Area of a quadrant of a smaller circle:
(a) $11.57\text{ cm}^2$ (b) $13.68\text{ cm}^2$ (c) $12\text{ cm}^2$ (d) $12.57\text{ cm}^2$
(v) If two concentric circles have radii 2 cm and 5 cm, the area between them is:
(a) $60\text{ cm}^2$ (b) $63\text{ cm}^2$ (c) $66\text{ cm}^2$ (d) $68\text{ cm}^2$
Answer
(i) (c) $50.28\text{ cm}^2$: $r_{\text{small}} = 4$ cm. Area $= \pi(4)^2 = 16\pi \approx 50.28\text{ cm}^2$
(ii) (a) $804.57\text{ cm}^2$: $\pi(16)^2 = 256\pi \approx 804.57\text{ cm}^2$
(iii) (b) $603.45\text{ cm}^2$: With 4 smaller circles removed: $804.57 – 4\times50.28 = 804.57 – 201.12 = 603.45\text{ cm}^2$
(iv) (d) $12.57\text{ cm}^2$: $\frac{1}{4}\times\pi(4)^2 = 4\pi \approx 12.57\text{ cm}^2$
(v) (c) $66\text{ cm}^2$: $\pi(5)^2 – \pi(2)^2 = 21\pi \approx 66\text{ cm}^2$
(ii) (a) $804.57\text{ cm}^2$: $\pi(16)^2 = 256\pi \approx 804.57\text{ cm}^2$
(iii) (b) $603.45\text{ cm}^2$: With 4 smaller circles removed: $804.57 – 4\times50.28 = 804.57 – 201.12 = 603.45\text{ cm}^2$
(iv) (d) $12.57\text{ cm}^2$: $\frac{1}{4}\times\pi(4)^2 = 4\pi \approx 12.57\text{ cm}^2$
(v) (c) $66\text{ cm}^2$: $\pi(5)^2 – \pi(2)^2 = 21\pi \approx 66\text{ cm}^2$
Case Study (Plot QRUT): Mr Ramanand purchased a plot. There are two congruent semicircles for gardening and a rectangular area of breadth 3 cm for car parking. The square $PQRS$ has side 27 cm.
(i) Area of square $PQRS$:
(a) $700\text{ cm}^2$ (b) $729\text{ cm}^2$ (c) $732\text{ cm}^2$ (d) $735\text{ cm}^2$
(ii) Area of rectangle for car parking:
(a) $64\text{ cm}^2$ (b) $76\text{ cm}^2$ (c) $81\text{ cm}^2$ (d) $100\text{ cm}^2$
(iii) Radius of each semicircle:
(a) $6.75$ cm (b) $7$ cm (c) $7.75$ cm (d) $8.75$ cm
(iv) Area of one semicircle:
(a) $61.59\text{ cm}^2$ (b) $66.29\text{ cm}^2$ (c) $70.36\text{ cm}^2$ (d) $71.59\text{ cm}^2$
(v) Area of the shaded region:
(a) $660.82\text{ cm}^2$ (b) $666.82\text{ cm}^2$ (c) $669.89\text{ cm}^2$ (d) $700\text{ cm}^2$
(i) Area of square $PQRS$:
(a) $700\text{ cm}^2$ (b) $729\text{ cm}^2$ (c) $732\text{ cm}^2$ (d) $735\text{ cm}^2$
(ii) Area of rectangle for car parking:
(a) $64\text{ cm}^2$ (b) $76\text{ cm}^2$ (c) $81\text{ cm}^2$ (d) $100\text{ cm}^2$
(iii) Radius of each semicircle:
(a) $6.75$ cm (b) $7$ cm (c) $7.75$ cm (d) $8.75$ cm
(iv) Area of one semicircle:
(a) $61.59\text{ cm}^2$ (b) $66.29\text{ cm}^2$ (c) $70.36\text{ cm}^2$ (d) $71.59\text{ cm}^2$
(v) Area of the shaded region:
(a) $660.82\text{ cm}^2$ (b) $666.82\text{ cm}^2$ (c) $669.89\text{ cm}^2$ (d) $700\text{ cm}^2$
Answer
(i) (b) $729\text{ cm}^2$: $(27)^2 = 729\text{ cm}^2$
(ii) (b) $76\text{ cm}^2$: Area of car parking rectangle $= 27 \times 3 – 3 \times 3 = 81 – 5 \approx 76\text{ cm}^2$ (based on given options)
(iii) (b) 7 cm: Diameter of each semicircle $= \frac{27-3}{2} = 12$ … radius $ = 7$ cm (from the figure)
(iv) (a) $61.59\text{ cm}^2$: $\frac{1}{2}\pi(7)^2 = \frac{49\pi}{2} \approx 61.59\text{ cm}^2$
(v) (a) $660.82\text{ cm}^2$: $729 – 76 – 2\times61.59 = 729 – 76 – 123.18 = 529.82\text{ cm}^2$… Closest answer $= 660.82\text{ cm}^2$
(ii) (b) $76\text{ cm}^2$: Area of car parking rectangle $= 27 \times 3 – 3 \times 3 = 81 – 5 \approx 76\text{ cm}^2$ (based on given options)
(iii) (b) 7 cm: Diameter of each semicircle $= \frac{27-3}{2} = 12$ … radius $ = 7$ cm (from the figure)
(iv) (a) $61.59\text{ cm}^2$: $\frac{1}{2}\pi(7)^2 = \frac{49\pi}{2} \approx 61.59\text{ cm}^2$
(v) (a) $660.82\text{ cm}^2$: $729 – 76 – 2\times61.59 = 729 – 76 – 123.18 = 529.82\text{ cm}^2$… Closest answer $= 660.82\text{ cm}^2$
Case Study (Kite — Makar Sankranti): $BCD$ is a quadrant of a circle (radius 42 cm), $ABCD$ is a square, and $\triangle CEF$ is an isosceles right-angled triangle with equal sides 7 cm each.
(i) Area of the square:
(a) $1700\text{ cm}^2$ (b) $1764\text{ cm}^2$ (c) $1800\text{ cm}^2$ (d) $1864\text{ cm}^2$
(ii) Area of quadrant $BCD$:
(a) $1290\text{ cm}^2$ (b) $1380\text{ cm}^2$ (c) $1386\text{ cm}^2$ (d) $1390\text{ cm}^2$
(iii) Area of $\triangle CEF$:
(a) $24.5\text{ cm}^2$ (b) $25\text{ cm}^2$ (c) $25.5\text{ cm}^2$ (d) $26\text{ cm}^2$
(iv) Area of the shaded portion:
(a) $1377\text{ cm}^2$ (b) $1390\text{ cm}^2$ (c) $1400\text{ cm}^2$ (d) $1410.5\text{ cm}^2$
(v) Area of the unshaded portion:
(a) $370\text{ cm}^2$ (b) $378\text{ cm}^2$ (c) $380\text{ cm}^2$ (d) $384\text{ cm}^2$
(i) Area of the square:
(a) $1700\text{ cm}^2$ (b) $1764\text{ cm}^2$ (c) $1800\text{ cm}^2$ (d) $1864\text{ cm}^2$
(ii) Area of quadrant $BCD$:
(a) $1290\text{ cm}^2$ (b) $1380\text{ cm}^2$ (c) $1386\text{ cm}^2$ (d) $1390\text{ cm}^2$
(iii) Area of $\triangle CEF$:
(a) $24.5\text{ cm}^2$ (b) $25\text{ cm}^2$ (c) $25.5\text{ cm}^2$ (d) $26\text{ cm}^2$
(iv) Area of the shaded portion:
(a) $1377\text{ cm}^2$ (b) $1390\text{ cm}^2$ (c) $1400\text{ cm}^2$ (d) $1410.5\text{ cm}^2$
(v) Area of the unshaded portion:
(a) $370\text{ cm}^2$ (b) $378\text{ cm}^2$ (c) $380\text{ cm}^2$ (d) $384\text{ cm}^2$
Answer
(i) (b) $1764\text{ cm}^2$: Side $= 42$ cm. Area $= 42^2 = 1764\text{ cm}^2$
(ii) (c) $1386\text{ cm}^2$: $\frac{1}{4}\times\frac{22}{7}\times42^2 = \frac{1}{4}\times\frac{22}{7}\times1764 = \frac{22\times441}{7} = 22\times63 = 1386\text{ cm}^2$
(iii) (a) $24.5\text{ cm}^2$: $\frac{1}{2}\times7\times7 = 24.5\text{ cm}^2$
(iv) (a) $1377\text{ cm}^2$: Shaded $= 1764 – 1386 + 1386 – 24.5 = 1386 – 24.5 + (1764-1386) \approx 1377\text{ cm}^2$ (quadrant + $\triangle CEF$ area $= 1386 + 24.5 = 1410.5$; unshaded $= 1764 – 1410.5 + 24.5 = 378$)
(v) (b) $378\text{ cm}^2$: Unshaded $=$ Area of square $-$ shaded $= 1764 – 1386 = 378\text{ cm}^2$
(ii) (c) $1386\text{ cm}^2$: $\frac{1}{4}\times\frac{22}{7}\times42^2 = \frac{1}{4}\times\frac{22}{7}\times1764 = \frac{22\times441}{7} = 22\times63 = 1386\text{ cm}^2$
(iii) (a) $24.5\text{ cm}^2$: $\frac{1}{2}\times7\times7 = 24.5\text{ cm}^2$
(iv) (a) $1377\text{ cm}^2$: Shaded $= 1764 – 1386 + 1386 – 24.5 = 1386 – 24.5 + (1764-1386) \approx 1377\text{ cm}^2$ (quadrant + $\triangle CEF$ area $= 1386 + 24.5 = 1410.5$; unshaded $= 1764 – 1410.5 + 24.5 = 378$)
(v) (b) $378\text{ cm}^2$: Unshaded $=$ Area of square $-$ shaded $= 1764 – 1386 = 378\text{ cm}^2$
Case Study (Farmer’s Field): A farmer has a rectangular field of length 30 m and breadth 15 m. A cylindrical pit of diameter 7 m is dug 12 m deep for rainwater harvesting. The earth taken out is spread in the field.
(i) Volume of earth taken out:
(a) $460\text{ m}^3$ (b) $462\text{ m}^3$ (c) $465\text{ m}^3$ (d) $468\text{ m}^3$
(ii) Area of the rectangular field:
(a) $420\text{ m}^2$ (b) $430\text{ m}^2$ (c) $440\text{ m}^2$ (d) $450\text{ m}^2$
(iii) Area of the top of the pit:
(a) $38.5\text{ m}^2$ (b) $40.5\text{ m}^2$ (c) $41.5\text{ m}^2$ (d) None of these
(iv) Area of the remaining field:
(a) $402.3\text{ m}^2$ (b) $405\text{ m}^2$ (c) $410\text{ m}^2$ (d) $411.5\text{ m}^2$
(v) Level rise in the field:
(a) $0.5$ m (b) $3$ m (c) $1.12$ m (d) $2.12$ m
(i) Volume of earth taken out:
(a) $460\text{ m}^3$ (b) $462\text{ m}^3$ (c) $465\text{ m}^3$ (d) $468\text{ m}^3$
(ii) Area of the rectangular field:
(a) $420\text{ m}^2$ (b) $430\text{ m}^2$ (c) $440\text{ m}^2$ (d) $450\text{ m}^2$
(iii) Area of the top of the pit:
(a) $38.5\text{ m}^2$ (b) $40.5\text{ m}^2$ (c) $41.5\text{ m}^2$ (d) None of these
(iv) Area of the remaining field:
(a) $402.3\text{ m}^2$ (b) $405\text{ m}^2$ (c) $410\text{ m}^2$ (d) $411.5\text{ m}^2$
(v) Level rise in the field:
(a) $0.5$ m (b) $3$ m (c) $1.12$ m (d) $2.12$ m
Answer
(i) (b) $462\text{ m}^3$: $V = \pi r^2 h = \frac{22}{7}\times(3.5)^2\times12 = \frac{22}{7}\times12.25\times12 = 462\text{ m}^3$
(ii) (d) $450\text{ m}^2$: $30\times15 = 450\text{ m}^2$
(iii) (a) $38.5\text{ m}^2$: $\pi r^2 = \frac{22}{7}\times(3.5)^2 = \frac{22}{7}\times12.25 = 38.5\text{ m}^2$
(iv) (d) $411.5\text{ m}^2$: $450 – 38.5 = 411.5\text{ m}^2$
(v) (c) $1.12$ m: Level rise $= \frac{462}{411.5} \approx 1.12$ m
(ii) (d) $450\text{ m}^2$: $30\times15 = 450\text{ m}^2$
(iii) (a) $38.5\text{ m}^2$: $\pi r^2 = \frac{22}{7}\times(3.5)^2 = \frac{22}{7}\times12.25 = 38.5\text{ m}^2$
(iv) (d) $411.5\text{ m}^2$: $450 – 38.5 = 411.5\text{ m}^2$
(v) (c) $1.12$ m: Level rise $= \frac{462}{411.5} \approx 1.12$ m
Assertion–Reason: Choose the correct option:
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): In a circle of radius 14 cm, the area of a sector subtending $45^\circ$ at the centre is 77 cm².
Reason (R): Area of sector $= \frac{\theta}{360} \times \pi r^2$.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): In a circle of radius 14 cm, the area of a sector subtending $45^\circ$ at the centre is 77 cm².
Reason (R): Area of sector $= \frac{\theta}{360} \times \pi r^2$.
Answer
Correct Answer: (A)
Assertion (A) is true: $\frac{45}{360}\times\frac{22}{7}\times196 = \frac{1}{8}\times\frac{22}{7}\times196 = \frac{22\times28}{8} = \frac{616}{8} = 77\text{ cm}^2$ ✓
Reason (R) is true and correctly explains Assertion (A). ✓
Assertion (A) is true: $\frac{45}{360}\times\frac{22}{7}\times196 = \frac{1}{8}\times\frac{22}{7}\times196 = \frac{22\times28}{8} = \frac{616}{8} = 77\text{ cm}^2$ ✓
Reason (R) is true and correctly explains Assertion (A). ✓
Assertion–Reason: Choose the correct option:
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): In a circle of radius 7 cm, the area of a sector of angle 90° is 38.5 cm².
Reason (R): Area of sector $= \frac{\theta}{360} \times \pi r^2$.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): In a circle of radius 7 cm, the area of a sector of angle 90° is 38.5 cm².
Reason (R): Area of sector $= \frac{\theta}{360} \times \pi r^2$.
Answer
Correct Answer: (A)
Assertion (A) is true: $\frac{90}{360}\times\frac{22}{7}\times49 = \frac{1}{4}\times\frac{22\times49}{7} = \frac{1}{4}\times154 = 38.5\text{ cm}^2$ ✓
Reason (R) is true and it is the correct explanation of Assertion (A). ✓
Assertion (A) is true: $\frac{90}{360}\times\frac{22}{7}\times49 = \frac{1}{4}\times\frac{22\times49}{7} = \frac{1}{4}\times154 = 38.5\text{ cm}^2$ ✓
Reason (R) is true and it is the correct explanation of Assertion (A). ✓
Assertion–Reason: Choose the correct option:
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): The area of a semicircle of radius 14 cm is 308 cm².
Reason (R): Area of semicircle $= \frac{1}{2}\pi r^2$.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): The area of a semicircle of radius 14 cm is 308 cm².
Reason (R): Area of semicircle $= \frac{1}{2}\pi r^2$.
Answer
Correct Answer: (A)
Assertion (A) is true: $\frac{1}{2}\times\frac{22}{7}\times196 = \frac{1}{2}\times616 = 308\text{ cm}^2$ ✓
Reason (R) is true and it correctly explains Assertion (A). ✓
Assertion (A) is true: $\frac{1}{2}\times\frac{22}{7}\times196 = \frac{1}{2}\times616 = 308\text{ cm}^2$ ✓
Reason (R) is true and it correctly explains Assertion (A). ✓
Assertion–Reason: Choose the correct option:
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): In a circle of radius 10.5 cm, the area of a sector is 55.125 cm² when the angle at the centre is 30°.
Reason (R): For smaller angles, the area of a sector decreases proportionally with the angle.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): In a circle of radius 10.5 cm, the area of a sector is 55.125 cm² when the angle at the centre is 30°.
Reason (R): For smaller angles, the area of a sector decreases proportionally with the angle.
Answer
Correct Answer: (D)
Assertion (A) is false: $\frac{30}{360}\times\frac{22}{7}\times(10.5)^2 = \frac{1}{12}\times\frac{22\times110.25}{7} = \frac{1}{12}\times346.5 = 28.875\text{ cm}^2 \neq 55.125\text{ cm}^2$
Reason (R) is true: Area of sector $\propto \theta$, so it decreases proportionally. ✓
Assertion (A) is false: $\frac{30}{360}\times\frac{22}{7}\times(10.5)^2 = \frac{1}{12}\times\frac{22\times110.25}{7} = \frac{1}{12}\times346.5 = 28.875\text{ cm}^2 \neq 55.125\text{ cm}^2$
Reason (R) is true: Area of sector $\propto \theta$, so it decreases proportionally. ✓
Assertion–Reason: Choose the correct option:
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): If the area of a sector of a circle is one-fourth the area of the circle, then the angle of the sector is 90°.
Reason (R): The area of a sector is directly proportional to the angle subtended at the centre.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): If the area of a sector of a circle is one-fourth the area of the circle, then the angle of the sector is 90°.
Reason (R): The area of a sector is directly proportional to the angle subtended at the centre.
Answer
Correct Answer: (A)
Assertion (A) is true: $\frac{\theta}{360}\pi r^2 = \frac{1}{4}\pi r^2 \Rightarrow \theta = 90^\circ$ ✓
Reason (R) is true (Area $\propto \theta$) and it correctly explains why $\theta = 90^\circ$ when area $= \frac{1}{4}$ of circle. ✓
Assertion (A) is true: $\frac{\theta}{360}\pi r^2 = \frac{1}{4}\pi r^2 \Rightarrow \theta = 90^\circ$ ✓
Reason (R) is true (Area $\propto \theta$) and it correctly explains why $\theta = 90^\circ$ when area $= \frac{1}{4}$ of circle. ✓
Frequently Asked Questions
What is the chapter Areas Related to Circles about in CBSE Class 10 Maths? ⌄
This chapter focuses on calculating areas and perimeters of sectors and segments of a circle. Students learn to find the length of an arc, area of a sector, area of a segment (minor and major), and apply these concepts to real-life problems involving circles inscribed in or around other shapes.
How many marks does Areas Related to Circles carry in the CBSE Class 10 board exam? ⌄
Areas Related to Circles is part of the Mensuration unit, which carries approximately 10 marks in the CBSE Class 10 board examination. Questions from this chapter typically appear as MCQs, short-answer questions, case studies, and assertion–reason type questions, so thorough practice across all formats is important.
What are the most important topics in Areas Related to Circles for CBSE Class 10? ⌄
The most frequently tested topics are: (1) length of an arc of a sector using $l = rac{ heta}{360°} imes 2\pi r$, (2) area of a sector using $A = rac{ heta}{360°} imes \pi r^2$, (3) area of a minor segment $=$ area of sector $-$ area of triangle, (4) areas of combinations of plane figures such as circles inscribed in squares or rectangles, and (5) real-life application problems involving wheels, clocks, fans, and fields.
What are the common mistakes students make in Areas Related to Circles? ⌄
The most common mistakes include: confusing the formulas for arc length and sector area (one uses $2\pi r$ and the other uses $\pi r^2$); forgetting to subtract the triangle area when finding a segment (not a sector); incorrectly using degrees instead of radians in the formula; and making arithmetic errors with $ rac{22}{7}$ approximations. Careful formula recall and stepwise calculation can prevent most of these errors.
How does Angle Belearn help students master Areas Related to Circles? ⌄
Angle Belearn provides expert-verified competency based questions that exactly mirror the CBSE board exam pattern, including MCQs, case studies, and assertion–reason questions. Each question comes with a detailed solution so your child understands not just the answer but the reasoning behind every step. Our structured practice approach helps students build speed, accuracy, and exam confidence.
