CBSE Class 10 · Maths
CBSE Class 10 Maths Areas Related to Circles Previous Year Questions
Help your child master CBSE Class 10 Maths Areas Related to Circles Previous Year Questions with this expert-verified collection sourced from real board papers spanning 2015–2024. Every question comes with a detailed step-by-step solution, so your child can confidently tackle arc lengths, sector areas, segment areas, and combined figure problems — concepts that appear consistently across all mark categories in the CBSE board exam.
CBSE Class 10 Maths Areas Related to Circles — Questions with Solutions
What is the area of a semi-circle of diameter ‘$d$’?
Solution
Answer: Option (C) is correct.
Explanation: The area of a semi-circle of diameter $d$ (so radius $r = \dfrac{d}{2}$) is: $$A = \frac{1}{2}\pi r^2 = \frac{1}{2}\pi\left(\frac{d}{2}\right)^2 = \frac{\pi d^{2}}{8}$$
Explanation: The area of a semi-circle of diameter $d$ (so radius $r = \dfrac{d}{2}$) is: $$A = \frac{1}{2}\pi r^2 = \frac{1}{2}\pi\left(\frac{d}{2}\right)^2 = \frac{\pi d^{2}}{8}$$
The circumferences of two circles are in the ratio $4:5$. What is the ratio of their radii?
Solution
Answer: Option (D) is correct.
Explanation: Circumference of a circle $= 2\pi r$. Therefore: $$\frac{2\pi r_1}{2\pi r_2} = \frac{4}{5} \Rightarrow \frac{r_1}{r_2} = \frac{4}{5}$$ Hence, the ratio of their radii $= 4:5$.
Explanation: Circumference of a circle $= 2\pi r$. Therefore: $$\frac{2\pi r_1}{2\pi r_2} = \frac{4}{5} \Rightarrow \frac{r_1}{r_2} = \frac{4}{5}$$ Hence, the ratio of their radii $= 4:5$.
A circle with radius 6 cm is shown below. The area of the shaded region in the circle is $\dfrac{5}{9}$ of the area of the circle. What is the length of the circle’s minor arc?
(Note: The figure is not to scale.)
(Note: The figure is not to scale.)
Solution
Answer: Option (A) is correct.
Explanation: Given radius $= 6$ cm. From the given condition: $$\frac{\pi r^{2}\theta}{360^{\circ}} = \frac{5}{9} \times \pi r^{2} \Rightarrow \theta = 200^{\circ}$$ Angle subtended at minor arc $= 360^{\circ} – 200^{\circ} = 160^{\circ}$
Length of minor arc $= \dfrac{2\pi r\theta_1}{360^{\circ}} = \dfrac{2 \times \pi \times 6 \times 160^{\circ}}{360^{\circ}} = \dfrac{16\pi}{3}$ cm
Explanation: Given radius $= 6$ cm. From the given condition: $$\frac{\pi r^{2}\theta}{360^{\circ}} = \frac{5}{9} \times \pi r^{2} \Rightarrow \theta = 200^{\circ}$$ Angle subtended at minor arc $= 360^{\circ} – 200^{\circ} = 160^{\circ}$
Length of minor arc $= \dfrac{2\pi r\theta_1}{360^{\circ}} = \dfrac{2 \times \pi \times 6 \times 160^{\circ}}{360^{\circ}} = \dfrac{16\pi}{3}$ cm
A regular pentagon is inscribed in a circle with centre O, of radius 5 cm, as shown below. What is the area of the shaded part of the circle?
Solution
Answer: Option (D) is correct.
Explanation: $$\text{Area of circle} = \pi r^2 = \pi \times (5)^2 = 25\pi \text{ cm}^2$$ The shaded region is $\dfrac{2}{5}$ of the circle (2 out of 5 equal sectors): $$\text{Area of shaded part} = \frac{2}{5} \times 25\pi = 10\pi \text{ cm}^2$$
Explanation: $$\text{Area of circle} = \pi r^2 = \pi \times (5)^2 = 25\pi \text{ cm}^2$$ The shaded region is $\dfrac{2}{5}$ of the circle (2 out of 5 equal sectors): $$\text{Area of shaded part} = \frac{2}{5} \times 25\pi = 10\pi \text{ cm}^2$$
In the given figure, the area of the segment ACB is
Solution
Answer: Option (A) is correct.
Explanation: Area of segment $= \dfrac{\pi r^{2}\theta}{360^{\circ}} – \dfrac{1}{2}r^{2}\sin\theta$
With $\theta = 90^{\circ}$: $$= \frac{\pi r^{2} \times 90^{\circ}}{360^{\circ}} – \frac{1}{2} \times r^{2} \times \sin 90^{\circ} = \frac{\pi r^2}{4} – \frac{r^2}{2} = \frac{r^{2}}{4}(\pi-2)$$
Explanation: Area of segment $= \dfrac{\pi r^{2}\theta}{360^{\circ}} – \dfrac{1}{2}r^{2}\sin\theta$
With $\theta = 90^{\circ}$: $$= \frac{\pi r^{2} \times 90^{\circ}}{360^{\circ}} – \frac{1}{2} \times r^{2} \times \sin 90^{\circ} = \frac{\pi r^2}{4} – \frac{r^2}{2} = \frac{r^{2}}{4}(\pi-2)$$
If the perimeter and the area of a circle are numerically equal, then the radius of the circle is:
Solution
Answer: Option (A) is correct.
Explanation: Let radius $= r$. Given perimeter $=$ area: $$2\pi r = \pi r^2 \Rightarrow 2 = r \Rightarrow r = 2 \text{ units}$$
Explanation: Let radius $= r$. Given perimeter $=$ area: $$2\pi r = \pi r^2 \Rightarrow 2 = r \Rightarrow r = 2 \text{ units}$$
The minute hand of a clock is 84 cm long. The distance covered by the tip of the minute hand from 10:10 a.m. to 10:25 a.m. is
Solution
Answer: Option (C) is correct.
Explanation: Length of minute hand (radius) $= 84$ cm.
In 1 minute, the minute hand turns $6^{\circ}$. In 15 minutes: $15 \times 6^{\circ} = 90^{\circ}$.
Distance covered by the tip: $$= \frac{\theta}{360^{\circ}} \times 2\pi r = \frac{90^{\circ}}{360^{\circ}} \times 2 \times \frac{22}{7} \times 84 = 132 \text{ cm}$$
Explanation: Length of minute hand (radius) $= 84$ cm.
In 1 minute, the minute hand turns $6^{\circ}$. In 15 minutes: $15 \times 6^{\circ} = 90^{\circ}$.
Distance covered by the tip: $$= \frac{\theta}{360^{\circ}} \times 2\pi r = \frac{90^{\circ}}{360^{\circ}} \times 2 \times \frac{22}{7} \times 84 = 132 \text{ cm}$$
If the perimeter of a circle is half that of a square, then the ratio of the area of the circle to the area of the square is
Solution
Answer: Option (D) is correct.
Explanation: Let radius of circle $= r$ and side of square $= a$. Given: $$2\pi r = \frac{1}{2}(4a) \Rightarrow r = \frac{a}{\pi} \Rightarrow \frac{r}{a} = \frac{1}{\pi}$$ $$\frac{\text{Area of circle}}{\text{Area of square}} = \frac{\pi r^2}{a^2} = \pi \times \frac{1}{\pi^2} = \frac{1}{\pi} = \frac{7}{22}$$
Explanation: Let radius of circle $= r$ and side of square $= a$. Given: $$2\pi r = \frac{1}{2}(4a) \Rightarrow r = \frac{a}{\pi} \Rightarrow \frac{r}{a} = \frac{1}{\pi}$$ $$\frac{\text{Area of circle}}{\text{Area of square}} = \frac{\pi r^2}{a^2} = \pi \times \frac{1}{\pi^2} = \frac{1}{\pi} = \frac{7}{22}$$
The number of revolutions made by a circular wheel of radius 0.25 m in rolling a distance of 11 km is
Solution
Answer: Option (D) is correct.
Explanation: Radius of wheel $(r) = 0.25$ m. Total distance $= 11$ km $= 11000$ m.
$$\text{No. of revolutions} \times 2\pi r = 11000$$ $$\Rightarrow \text{No. of revolutions} = \frac{11000 \times 7}{2 \times 22 \times 0.25} = 7000$$
Explanation: Radius of wheel $(r) = 0.25$ m. Total distance $= 11$ km $= 11000$ m.
$$\text{No. of revolutions} \times 2\pi r = 11000$$ $$\Rightarrow \text{No. of revolutions} = \frac{11000 \times 7}{2 \times 22 \times 0.25} = 7000$$
If the perimeter of a circle is equal to that of a square, then the ratio of their areas is
Solution
Answer: Option (B) is correct.
Explanation: Let radius of circle $= r$ and side of square $= a$. Given perimeters are equal: $$2\pi r = 4a \Rightarrow a = \frac{\pi r}{2}$$ $$\frac{\text{Area of circle}}{\text{Area of square}} = \frac{\pi r^2}{\left(\dfrac{\pi r}{2}\right)^2} = \frac{\pi r^2 \times 4}{\pi^2 r^2} = \frac{4}{\pi} = \frac{4 \times 7}{22} = \frac{14}{11}$$
Explanation: Let radius of circle $= r$ and side of square $= a$. Given perimeters are equal: $$2\pi r = 4a \Rightarrow a = \frac{\pi r}{2}$$ $$\frac{\text{Area of circle}}{\text{Area of square}} = \frac{\pi r^2}{\left(\dfrac{\pi r}{2}\right)^2} = \frac{\pi r^2 \times 4}{\pi^2 r^2} = \frac{4}{\pi} = \frac{4 \times 7}{22} = \frac{14}{11}$$
Shown below is a circle with centre O. Chord MN subtends an angle $x$ at O.
Which of these is true for the above circle?
I. $\dfrac{x}{360^{\circ}}=\dfrac{\text{length of arc MPN}}{\text{circumference of the circle}}$
II. $\dfrac{x}{360^{\circ}}=\dfrac{\text{minor sector area}}{\text{area of the circle}}$
Which of these is true for the above circle?
I. $\dfrac{x}{360^{\circ}}=\dfrac{\text{length of arc MPN}}{\text{circumference of the circle}}$
II. $\dfrac{x}{360^{\circ}}=\dfrac{\text{minor sector area}}{\text{area of the circle}}$
Solution
Answer: Option (C) is correct.
Explanation:
(I) Length of arc $= \dfrac{x}{360^{\circ}} \times 2\pi r$, so $\dfrac{x}{360^{\circ}} = \dfrac{\text{arc length}}{\text{circumference}}$ ✓
(II) Area of sector $= \dfrac{\pi r^2 \theta}{360^{\circ}}$, so $\dfrac{x}{360^{\circ}} = \dfrac{\text{sector area}}{\pi r^2}$ ✓
Both statements are correct.
Explanation:
(I) Length of arc $= \dfrac{x}{360^{\circ}} \times 2\pi r$, so $\dfrac{x}{360^{\circ}} = \dfrac{\text{arc length}}{\text{circumference}}$ ✓
(II) Area of sector $= \dfrac{\pi r^2 \theta}{360^{\circ}}$, so $\dfrac{x}{360^{\circ}} = \dfrac{\text{sector area}}{\pi r^2}$ ✓
Both statements are correct.
The ratio of the area of a quadrant of a circle to the area of the same circle is:
Solution
Answer: Option (C) is correct.
Explanation: A quadrant is one-fourth of the full circle. $$\frac{\text{Area of quadrant}}{\text{Area of circle}} = \frac{\frac{1}{4}\pi r^2}{\pi r^2} = \frac{1}{4}$$ Therefore the ratio is $1:4$.
Explanation: A quadrant is one-fourth of the full circle. $$\frac{\text{Area of quadrant}}{\text{Area of circle}} = \frac{\frac{1}{4}\pi r^2}{\pi r^2} = \frac{1}{4}$$ Therefore the ratio is $1:4$.
A piece of wire 20 cm long is bent into the form of an arc of a circle of radius $\dfrac{60}{\pi}$ cm. The angle subtended by the arc at the centre of the circle is:
Solution
Answer: Option (B) is correct.
Explanation: Using arc length formula $L = r\theta$ (radians): $$20 = \frac{60}{\pi} \times \theta \Rightarrow \theta = \frac{20\pi}{60} = \frac{\pi}{3} \text{ radians} = 60^{\circ}$$
Explanation: Using arc length formula $L = r\theta$ (radians): $$20 = \frac{60}{\pi} \times \theta \Rightarrow \theta = \frac{20\pi}{60} = \frac{\pi}{3} \text{ radians} = 60^{\circ}$$
What is the area of a semi-circle of diameter $d$?
Solution
Answer: Option (C) is correct.
Explanation: Radius $r = \dfrac{d}{2}$. $$A_{\text{semi-circle}} = \frac{1}{2}\pi r^2 = \frac{1}{2}\pi\left(\frac{d}{2}\right)^2 = \frac{1}{2} \times \frac{\pi d^2}{4} = \frac{\pi d^2}{8}$$
Explanation: Radius $r = \dfrac{d}{2}$. $$A_{\text{semi-circle}} = \frac{1}{2}\pi r^2 = \frac{1}{2}\pi\left(\frac{d}{2}\right)^2 = \frac{1}{2} \times \frac{\pi d^2}{4} = \frac{\pi d^2}{8}$$
In a right triangle $ABC$, right-angled at $B$, $BC = 12$ cm and $AB = 5$ cm. The radius of the circle inscribed in the triangle (in cm) is:
Solution
Answer: Option (C) is correct.
Explanation:
Hypotenuse $AC = \sqrt{5^2 + 12^2} = \sqrt{169} = 13$ cm
Semi-perimeter $s = \dfrac{5 + 12 + 13}{2} = 15$ cm
Area of triangle $= \dfrac{1}{2} \times 5 \times 12 = 30$ cm²
Inradius $r = \dfrac{\text{Area}}{s} = \dfrac{30}{15} = 2$ cm
Explanation:
Hypotenuse $AC = \sqrt{5^2 + 12^2} = \sqrt{169} = 13$ cm
Semi-perimeter $s = \dfrac{5 + 12 + 13}{2} = 15$ cm
Area of triangle $= \dfrac{1}{2} \times 5 \times 12 = 30$ cm²
Inradius $r = \dfrac{\text{Area}}{s} = \dfrac{30}{15} = 2$ cm
In the given figure, find the perimeter of the sector of a circle with radius 10.5 cm and of angle $60^{\circ}$. $\left(\text{Take } \pi = \dfrac{22}{7}\right)$
Answer
Given: radius $(r) = 10.5$ cm, angle $(\theta) = 60^{\circ}$
Length of arc APB: $$= \frac{\theta}{360^{\circ}} \times 2\pi r = \frac{60^{\circ}}{360^{\circ}} \times 2 \times \frac{22}{7} \times 10.5 = 11 \text{ cm}$$ Perimeter of sector OAPBO: $$= OA + \text{arc } APB + BO = 10.5 + 11 + 10.5 = \mathbf{32 \text{ cm}}$$
Length of arc APB: $$= \frac{\theta}{360^{\circ}} \times 2\pi r = \frac{60^{\circ}}{360^{\circ}} \times 2 \times \frac{22}{7} \times 10.5 = 11 \text{ cm}$$ Perimeter of sector OAPBO: $$= OA + \text{arc } APB + BO = 10.5 + 11 + 10.5 = \mathbf{32 \text{ cm}}$$
With vertices A, B and C of $\triangle ABC$ as centres, arcs are drawn with radii 14 cm and the three portions of the triangle so obtained are removed. Find the total area removed from the triangle.
Answer
The three removed portions are sectors at each vertex with radius $r = 14$ cm.
$$\text{Area of sector at A} = \frac{\angle A}{360^{\circ}}\pi r^2, \quad \text{Area at B} = \frac{\angle B}{360^{\circ}}\pi r^2, \quad \text{Area at C} = \frac{\angle C}{360^{\circ}}\pi r^2$$ Total removed area: $$= \frac{\pi r^2}{360^{\circ}}(\angle A + \angle B + \angle C) = \frac{\pi r^2}{360^{\circ}} \times 180^{\circ} = \frac{\pi r^2}{2}$$ $$= \frac{\frac{22}{7} \times 14 \times 14}{2} = \mathbf{308 \text{ cm}^2}$$
$$\text{Area of sector at A} = \frac{\angle A}{360^{\circ}}\pi r^2, \quad \text{Area at B} = \frac{\angle B}{360^{\circ}}\pi r^2, \quad \text{Area at C} = \frac{\angle C}{360^{\circ}}\pi r^2$$ Total removed area: $$= \frac{\pi r^2}{360^{\circ}}(\angle A + \angle B + \angle C) = \frac{\pi r^2}{360^{\circ}} \times 180^{\circ} = \frac{\pi r^2}{2}$$ $$= \frac{\frac{22}{7} \times 14 \times 14}{2} = \mathbf{308 \text{ cm}^2}$$
Find the area of the unshaded region shown in the given figure.
Answer
The side of the square $=$ Diameter of the semi-circle $= a$.
The horizontal/vertical extent of the white region $= 14 – 3 – 3 = 8$ cm
$\Rightarrow 2(\text{radius}) + \text{side of square} = 8$ cm $\Rightarrow 2a = 8$ cm $\Rightarrow a = 4$ cm
Area of unshaded region: $$= \text{Area of square of side 4 cm} + 4 \times \text{Area of semi-circle of diameter 4 cm}$$ $$= (4)^2 + 4 \times \frac{1}{2}\pi(2)^2 = 16 + 8\pi \text{ cm}^2$$
The horizontal/vertical extent of the white region $= 14 – 3 – 3 = 8$ cm
$\Rightarrow 2(\text{radius}) + \text{side of square} = 8$ cm $\Rightarrow 2a = 8$ cm $\Rightarrow a = 4$ cm
Area of unshaded region: $$= \text{Area of square of side 4 cm} + 4 \times \text{Area of semi-circle of diameter 4 cm}$$ $$= (4)^2 + 4 \times \frac{1}{2}\pi(2)^2 = 16 + 8\pi \text{ cm}^2$$
ABCD is a rhombus with side 3 cm. Two arcs are drawn from points A and C respectively such that the radius equals the side of the rhombus, as shown below.
If BD is a line of symmetry for the figure, then find the area of the shaded part of the figure in terms of $\pi$. Show your work.
If BD is a line of symmetry for the figure, then find the area of the shaded part of the figure in terms of $\pi$. Show your work.
Answer
Area of sector ABD $= \dfrac{60^{\circ}}{360^{\circ}} \times \pi \times 3^2 = \dfrac{3\pi}{2}$ cm²
Area of $\triangle ABD = \dfrac{\sqrt{3}}{4} \times 9 = \dfrac{9\sqrt{3}}{4}$ cm²
Required shaded area $= 2 \times (\text{sector ABD} – \triangle ABD)$: $$= 2 \times \left(\frac{3\pi}{2} – \frac{9\sqrt{3}}{4}\right) = \mathbf{3\pi – \frac{9\sqrt{3}}{2}} \text{ cm}^2$$
Area of $\triangle ABD = \dfrac{\sqrt{3}}{4} \times 9 = \dfrac{9\sqrt{3}}{4}$ cm²
Required shaded area $= 2 \times (\text{sector ABD} – \triangle ABD)$: $$= 2 \times \left(\frac{3\pi}{2} – \frac{9\sqrt{3}}{4}\right) = \mathbf{3\pi – \frac{9\sqrt{3}}{2}} \text{ cm}^2$$
Wasim made a model of Pac-Man, after playing the famous video game of the same name. The area of the model is $120\pi$ cm². Pac-Man’s mouth forms an angle of $60^{\circ}$ at the centre of the circle, as shown below.
Wasim wants to decorate the model by attaching a coloured ribbon to the entire boundary of the shape. What is the minimum length of the ribbon required in terms of $\pi$? Show your work.
Wasim wants to decorate the model by attaching a coloured ribbon to the entire boundary of the shape. What is the minimum length of the ribbon required in terms of $\pi$? Show your work.
Answer
Let radius $= r$ cm. The Pac-Man shape covers $\dfrac{300^{\circ}}{360^{\circ}}$ of the circle (since mouth angle $= 60^{\circ}$).
Area equation: $$120\pi = \frac{300}{360} \times \pi \times r^2 \Rightarrow r^2 = \frac{120 \times 360}{300} = 144 \Rightarrow r = 12 \text{ cm}$$ Length of ribbon (boundary = arc + 2 radii): $$= \frac{300}{360} \times 2\pi \times 12 + 2 \times 12 = 20\pi + 24 \text{ cm}$$ Minimum length of ribbon $= (20\pi + 24)$ cm
Area equation: $$120\pi = \frac{300}{360} \times \pi \times r^2 \Rightarrow r^2 = \frac{120 \times 360}{300} = 144 \Rightarrow r = 12 \text{ cm}$$ Length of ribbon (boundary = arc + 2 radii): $$= \frac{300}{360} \times 2\pi \times 12 + 2 \times 12 = 20\pi + 24 \text{ cm}$$ Minimum length of ribbon $= (20\pi + 24)$ cm
In a circle of radius 21 cm, an arc subtends an angle of $60^{\circ}$ at the centre. Find the area of the sector formed by the arc. Also, find the length of the arc.
Answer
Given: $\theta = 60^{\circ}$, $r = 21$ cm
(i) Area of sector: $$= \frac{\theta}{360^{\circ}} \times \pi r^2 = \frac{60^{\circ}}{360^{\circ}} \times \frac{22}{7} \times 21 \times 21 = \frac{1}{6} \times \frac{22}{7} \times 441 = \mathbf{231 \text{ cm}^2}$$ (ii) Length of arc: $$= \frac{\theta}{360^{\circ}} \times 2\pi r = \frac{60^{\circ}}{360^{\circ}} \times 2 \times \frac{22}{7} \times 21 = \mathbf{22 \text{ cm}}$$
(i) Area of sector: $$= \frac{\theta}{360^{\circ}} \times \pi r^2 = \frac{60^{\circ}}{360^{\circ}} \times \frac{22}{7} \times 21 \times 21 = \frac{1}{6} \times \frac{22}{7} \times 441 = \mathbf{231 \text{ cm}^2}$$ (ii) Length of arc: $$= \frac{\theta}{360^{\circ}} \times 2\pi r = \frac{60^{\circ}}{360^{\circ}} \times 2 \times \frac{22}{7} \times 21 = \mathbf{22 \text{ cm}}$$
Sides of a right triangular field are 25 m, 24 m and 7 m. At the three corners of the field, a cow, a buffalo and a horse are tied separately with ropes of 3.5 m each to graze in the field. Find the area of the field that cannot be grazed by these animals.
Answer
Radius of each grazing sector $(r) = 3.5$ m.
Sum of the three sectors’ areas: $$= \frac{\angle C + \angle B + \angle H}{360^{\circ}} \times \pi r^2 = \frac{180^{\circ}}{360^{\circ}} \times \frac{22}{7} \times (3.5)^2 = \frac{1}{2} \times 38.5 = 19.25 \text{ m}^2$$ Area of right triangle (legs 24 m and 7 m): $$= \frac{1}{2} \times 24 \times 7 = 84 \text{ m}^2$$ Area that cannot be grazed: $$= 84 – 19.25 = \mathbf{64.75 \text{ m}^2}$$
Sum of the three sectors’ areas: $$= \frac{\angle C + \angle B + \angle H}{360^{\circ}} \times \pi r^2 = \frac{180^{\circ}}{360^{\circ}} \times \frac{22}{7} \times (3.5)^2 = \frac{1}{2} \times 38.5 = 19.25 \text{ m}^2$$ Area of right triangle (legs 24 m and 7 m): $$= \frac{1}{2} \times 24 \times 7 = 84 \text{ m}^2$$ Area that cannot be grazed: $$= 84 – 19.25 = \mathbf{64.75 \text{ m}^2}$$
In the figure, a square OABC is inscribed in a quadrant OPBQ. If $OA = 15$ cm, find the area of the shaded region. (Use $\pi = 3.14$)
Answer
Since OABC is a square, $\angle OAB = 90^{\circ}$. In $\triangle OAB$:
Radius of quadrant $= OB = \sqrt{15^2 + 15^2} = 15\sqrt{2}$ cm
Area of quadrant OQBP: $$= \frac{90^{\circ}}{360^{\circ}} \times \pi \times (15\sqrt{2})^2 = \frac{1}{4} \times 3.14 \times 450 = 353.25 \text{ cm}^2$$ Area of square $= (15)^2 = 225$ cm²
Area of shaded region $= 353.25 – 225 = 128.25$ cm²
Radius of quadrant $= OB = \sqrt{15^2 + 15^2} = 15\sqrt{2}$ cm
Area of quadrant OQBP: $$= \frac{90^{\circ}}{360^{\circ}} \times \pi \times (15\sqrt{2})^2 = \frac{1}{4} \times 3.14 \times 450 = 353.25 \text{ cm}^2$$ Area of square $= (15)^2 = 225$ cm²
Area of shaded region $= 353.25 – 225 = 128.25$ cm²
ABDC is a quadrant of a circle of radius 28 cm and a semi-circle BEC is drawn with BC as diameter. Find the area of the shaded region. (Use $\pi = \dfrac{22}{7}$)
Answer
In $\triangle ABC$ (right-angled at A, with $AB = AC = 28$ cm):
$$BC^2 = 28^2 + 28^2 \Rightarrow BC = 28\sqrt{2} \text{ cm}$$
Radius of semi-circle on BC $= \dfrac{28\sqrt{2}}{2} = 14\sqrt{2}$ cm
Area of semi-circle: $$= \frac{1}{2} \times \frac{22}{7} \times (14\sqrt{2})^2 = \frac{1}{2} \times \frac{22}{7} \times 392 = 616 \text{ cm}^2$$ Area of $\triangle ABC = \dfrac{1}{2} \times 28 \times 28 = 392$ cm²
Area of quadrant $= \dfrac{1}{4} \times \dfrac{22}{7} \times 28^2 = 616$ cm²
Area of shaded region $=$ Semi-circle $+$ Triangle $-$ Quadrant $= 616 + 392 – 616 = 392$ cm²
Area of semi-circle: $$= \frac{1}{2} \times \frac{22}{7} \times (14\sqrt{2})^2 = \frac{1}{2} \times \frac{22}{7} \times 392 = 616 \text{ cm}^2$$ Area of $\triangle ABC = \dfrac{1}{2} \times 28 \times 28 = 392$ cm²
Area of quadrant $= \dfrac{1}{4} \times \dfrac{22}{7} \times 28^2 = 616$ cm²
Area of shaded region $=$ Semi-circle $+$ Triangle $-$ Quadrant $= 616 + 392 – 616 = 392$ cm²
A piece of wire 22 cm long is bent into the form of an arc of a circle subtending an angle of $60^{\circ}$ at its centre. Find the radius of the circle. $\left[\text{Use } \pi = \dfrac{22}{7}\right]$
Answer
Let radius $= r$ cm. Arc length $AB = 22$ cm, $\theta = 60^{\circ}$.
$$\text{Length of arc} = \frac{2\pi r\theta}{360^{\circ}}$$ $$22 = \frac{2 \times \frac{22}{7} \times r \times 60^{\circ}}{360^{\circ}} = \frac{22r}{21}$$ $$\Rightarrow r = 21 \text{ cm}$$ The radius of the circle is 21 cm.
$$\text{Length of arc} = \frac{2\pi r\theta}{360^{\circ}}$$ $$22 = \frac{2 \times \frac{22}{7} \times r \times 60^{\circ}}{360^{\circ}} = \frac{22r}{21}$$ $$\Rightarrow r = 21 \text{ cm}$$ The radius of the circle is 21 cm.
The length of the minute hand of a clock is 6 cm. Find the area swept by it when it moves from 7:05 p.m. to 7:40 p.m.
Answer
In 60 minutes, minute hand moves $360^{\circ}$, so in 1 minute it moves $6^{\circ}$.
From 7:05 to 7:40 $= 35$ minutes, angle swept $= 35 \times 6^{\circ} = 210^{\circ}$
Area swept: $$= \frac{210}{360} \times \pi \times 6^2 = \frac{7}{12} \times \frac{22}{7} \times 36 = \mathbf{66 \text{ cm}^2}$$
From 7:05 to 7:40 $= 35$ minutes, angle swept $= 35 \times 6^{\circ} = 210^{\circ}$
Area swept: $$= \frac{210}{360} \times \pi \times 6^2 = \frac{7}{12} \times \frac{22}{7} \times 36 = \mathbf{66 \text{ cm}^2}$$
In the given figure, arcs have been drawn of radius 7 cm each with vertices A, B, C and D of quadrilateral ABCD as centres. Find the area of the shaded region.
Answer
Let $\angle A = \theta_1,\ \angle B = \theta_2,\ \angle C = \theta_3,\ \angle D = \theta_4$.
Total shaded area: $$= \frac{\theta_1 + \theta_2 + \theta_3 + \theta_4}{360^{\circ}} \times \pi \times 7^2$$ By angle sum property of a quadrilateral: $\theta_1 + \theta_2 + \theta_3 + \theta_4 = 360^{\circ}$ $$= \frac{360^{\circ}}{360^{\circ}} \times \frac{22}{7} \times 49 = \mathbf{154 \text{ cm}^2}$$
Total shaded area: $$= \frac{\theta_1 + \theta_2 + \theta_3 + \theta_4}{360^{\circ}} \times \pi \times 7^2$$ By angle sum property of a quadrilateral: $\theta_1 + \theta_2 + \theta_3 + \theta_4 = 360^{\circ}$ $$= \frac{360^{\circ}}{360^{\circ}} \times \frac{22}{7} \times 49 = \mathbf{154 \text{ cm}^2}$$
Sneha had a rectangular tablecloth with one side measuring 30 cm which she wanted to keep on her circular table of radius 25 cm. After keeping it on the table, she realised that the corners of the tablecloth just touched the edge of the circular table, as shown in the figure.
(Note: The figure is not to scale.) Find the area of the table not covered by the tablecloth. (Use $\pi = 3.14$)
(Note: The figure is not to scale.) Find the area of the table not covered by the tablecloth. (Use $\pi = 3.14$)
Answer
Since the corners of the rectangle touch the circle, the diagonal of rectangle $=$ diameter of circle $= 50$ cm.
Using Pythagoras theorem: $DC^2 = 50^2 – 30^2 = 2500 – 900 = 1600 \Rightarrow DC = 40$ cm
Area of circle $= \pi r^2 = 3.14 \times 625 = 1962.5$ cm²
Area of tablecloth $= 30 \times 40 = 1200$ cm²
Area not covered $= 1962.5 – 1200 = 762.5$ cm²
Using Pythagoras theorem: $DC^2 = 50^2 – 30^2 = 2500 – 900 = 1600 \Rightarrow DC = 40$ cm
Area of circle $= \pi r^2 = 3.14 \times 625 = 1962.5$ cm²
Area of tablecloth $= 30 \times 40 = 1200$ cm²
Area not covered $= 1962.5 – 1200 = 762.5$ cm²
Sprinklers are crop irrigation equipment which rotate around a centre and spray water on the crops in the circular region. Two such high-power sprinklers, occupying negligible area, are installed in a straight line in a field such that they spray water on a common area. Points A and B are the sprinklers.
Both sprinklers spray over equal areas. Given $CD = 400$ m and $\angle CAD = \angle CBD = 90^{\circ}$.
(i) Find the radius of the circular region sprayed by the sprinkler.
(ii) Find the area of the overlapping region.
Both sprinklers spray over equal areas. Given $CD = 400$ m and $\angle CAD = \angle CBD = 90^{\circ}$.
(i) Find the radius of the circular region sprayed by the sprinkler.
(ii) Find the area of the overlapping region.
Answer
(i) Since $\angle CAD = 90^{\circ}$ and $CD = 400$ m, in right $\triangle ACD$:
$$CD^2 = AC^2 + AD^2 \Rightarrow 160000 = 2AC^2 \Rightarrow AC = 200\sqrt{2} \text{ m}$$ Radius $= 200\sqrt{2}$ m
(ii) Area of sector CAD $=$ Area of sector CBD: $$= \frac{90^{\circ}}{360^{\circ}} \times 3.14 \times (200\sqrt{2})^2 = \frac{1}{4} \times 3.14 \times 80000 = 62800 \text{ m}^2$$ Area of $\triangle CAD =$ Area of $\triangle CBD = \dfrac{1}{2} \times (200\sqrt{2})^2 = 40000$ m²
Area of overlapping region: $$= 2(62800) – 2(40000) = 125600 – 80000 = \mathbf{45600 \text{ m}^2}$$
(ii) Area of sector CAD $=$ Area of sector CBD: $$= \frac{90^{\circ}}{360^{\circ}} \times 3.14 \times (200\sqrt{2})^2 = \frac{1}{4} \times 3.14 \times 80000 = 62800 \text{ m}^2$$ Area of $\triangle CAD =$ Area of $\triangle CBD = \dfrac{1}{2} \times (200\sqrt{2})^2 = 40000$ m²
Area of overlapping region: $$= 2(62800) – 2(40000) = 125600 – 80000 = \mathbf{45600 \text{ m}^2}$$
Shown below is a circle with multiple chords. One of the chords is the diameter of the circle. $AB = 13.9$ cm, $MN = 14$ cm, $PQ = 14.1$ cm.
Find the measure of the angle subtended by a $4.7\pi$ cm arc at the circumference of the circle. Show your work and give valid reasons.
Find the measure of the angle subtended by a $4.7\pi$ cm arc at the circumference of the circle. Show your work and give valid reasons.
Answer
The diameter is the longest chord, so Diameter $= PQ = 14.1$ cm, Radius $= 7.05$ cm.
Using arc length formula: $$4.7\pi = \frac{2\pi r\theta}{360^{\circ}} \Rightarrow 4.7\pi = 2\pi \times \frac{14.1}{2} \times \frac{\theta}{360^{\circ}}$$ $$\Rightarrow \theta = \frac{4.7 \times 2 \times 360}{14.1 \times 2} = 120^{\circ} \quad (\text{central angle})$$ Angle subtended at circumference $= \dfrac{1}{2} \times$ central angle $= \dfrac{120^{\circ}}{2} = \mathbf{60^{\circ}}$
Using arc length formula: $$4.7\pi = \frac{2\pi r\theta}{360^{\circ}} \Rightarrow 4.7\pi = 2\pi \times \frac{14.1}{2} \times \frac{\theta}{360^{\circ}}$$ $$\Rightarrow \theta = \frac{4.7 \times 2 \times 360}{14.1 \times 2} = 120^{\circ} \quad (\text{central angle})$$ Angle subtended at circumference $= \dfrac{1}{2} \times$ central angle $= \dfrac{120^{\circ}}{2} = \mathbf{60^{\circ}}$
A park is of the shape of a circle of diameter 7 m. It is surrounded by a path of width 0.7 m. Find the expenditure of cementing the path, if its cost is ₹110 per sq. m.
Answer
Radius of park $= \dfrac{7}{2} = 3.5$ m
Radius of park with path $= 3.5 + 0.7 = 4.2$ m
Area of path (ring formula): $$= \pi(4.2)^2 – \pi(3.5)^2 = \frac{22}{7}(17.64 – 12.25) = \frac{22}{7} \times 5.39 = 22 \times 0.77 = 16.94 \text{ m}^2$$ Cost of cementing: $$= 16.94 \times 110 = \mathbf{₹\, 1863.40}$$
Radius of park with path $= 3.5 + 0.7 = 4.2$ m
Area of path (ring formula): $$= \pi(4.2)^2 – \pi(3.5)^2 = \frac{22}{7}(17.64 – 12.25) = \frac{22}{7} \times 5.39 = 22 \times 0.77 = 16.94 \text{ m}^2$$ Cost of cementing: $$= 16.94 \times 110 = \mathbf{₹\, 1863.40}$$
The given figure shows a sector OAP of a circle with centre O, containing $\angle\theta$. AB is perpendicular to the radius OA and meets OP produced at B. Prove that the perimeter of the shaded region is $r\left[\tan\theta + \sec\theta + \dfrac{\pi\theta}{180^{\circ}} – 1\right]$.
Answer
Perimeter of shaded region $= BP + AB + \text{arc } AP$
From the right-angled triangle OAB:
$\tan\theta = \dfrac{AB}{OA} = \dfrac{AB}{r} \Rightarrow AB = r\tan\theta$ …(1)
$\sec\theta = \dfrac{OB}{OA} = \dfrac{OB}{r} \Rightarrow OB = r\sec\theta$
$BP = OB – OP = r\sec\theta – r$ …(2)
Length of arc AP: $$= \frac{\theta}{360^{\circ}} \times 2\pi r = \frac{\pi r\theta}{180^{\circ}} \quad \ldots(3)$$ Adding (1), (2), and (3): $$\text{Perimeter} = r\tan\theta + (r\sec\theta – r) + \frac{\pi r\theta}{180^{\circ}} = r\left[\tan\theta + \sec\theta + \frac{\pi\theta}{180^{\circ}} – 1\right]$$ Hence proved.
From the right-angled triangle OAB:
$\tan\theta = \dfrac{AB}{OA} = \dfrac{AB}{r} \Rightarrow AB = r\tan\theta$ …(1)
$\sec\theta = \dfrac{OB}{OA} = \dfrac{OB}{r} \Rightarrow OB = r\sec\theta$
$BP = OB – OP = r\sec\theta – r$ …(2)
Length of arc AP: $$= \frac{\theta}{360^{\circ}} \times 2\pi r = \frac{\pi r\theta}{180^{\circ}} \quad \ldots(3)$$ Adding (1), (2), and (3): $$\text{Perimeter} = r\tan\theta + (r\sec\theta – r) + \frac{\pi r\theta}{180^{\circ}} = r\left[\tan\theta + \sec\theta + \frac{\pi\theta}{180^{\circ}} – 1\right]$$ Hence proved.
A chord of a circle of radius 14 cm subtends an angle of $60^{\circ}$ at the centre. Find the area of the corresponding minor segment of the circle. Also, find the area of the major segment of the circle.
Answer
Radius $r = 14$ cm, $\theta = 60^{\circ}$
Area of sector: $$A_{\text{sector}} = \frac{60^{\circ}}{360^{\circ}} \times \pi \times 14^2 = \frac{196\pi}{6} \approx 102.1 \text{ cm}^2$$ Area of triangle (two radii and chord): $$A_{\triangle} = \frac{1}{2}r^2\sin\theta = \frac{1}{2} \times 196 \times \sin 60^{\circ} = \frac{196\sqrt{3}}{4} \approx 84.9 \text{ cm}^2$$ Area of minor segment: $$= A_{\text{sector}} – A_{\triangle} \approx 102.1 – 84.9 = \mathbf{17.2 \text{ cm}^2}$$ Area of full circle: $$= \pi \times 14^2 = 196\pi \approx 615.8 \text{ cm}^2$$ Area of major segment: $$= 615.8 – 17.2 = \mathbf{598.6 \text{ cm}^2}$$
Area of sector: $$A_{\text{sector}} = \frac{60^{\circ}}{360^{\circ}} \times \pi \times 14^2 = \frac{196\pi}{6} \approx 102.1 \text{ cm}^2$$ Area of triangle (two radii and chord): $$A_{\triangle} = \frac{1}{2}r^2\sin\theta = \frac{1}{2} \times 196 \times \sin 60^{\circ} = \frac{196\sqrt{3}}{4} \approx 84.9 \text{ cm}^2$$ Area of minor segment: $$= A_{\text{sector}} – A_{\triangle} \approx 102.1 – 84.9 = \mathbf{17.2 \text{ cm}^2}$$ Area of full circle: $$= \pi \times 14^2 = 196\pi \approx 615.8 \text{ cm}^2$$ Area of major segment: $$= 615.8 – 17.2 = \mathbf{598.6 \text{ cm}^2}$$
A chord PQ of a circle of radius 10 cm subtends an angle of $60^{\circ}$ at the centre of the circle. Find the area of the major and minor segments of the circle.
Answer
Radius $r = 10$ cm, $\theta = 60^{\circ}$
Area of sector: $$A_{\text{sector}} = \frac{60^{\circ}}{360^{\circ}} \times \pi \times 100 = \frac{100\pi}{6} \approx 52.36 \text{ cm}^2$$ Area of triangle: $$A_{\triangle} = \frac{1}{2} \times 100 \times \sin 60^{\circ} = \frac{100\sqrt{3}}{4} \approx 43.3 \text{ cm}^2$$ Area of minor segment: $$= 52.36 – 43.3 \approx \mathbf{9.06 \text{ cm}^2}$$ Area of full circle $= 100\pi \approx 314.16$ cm²
Area of major segment: $$= 314.16 – 9.06 \approx \mathbf{305.1 \text{ cm}^2}$$
Area of sector: $$A_{\text{sector}} = \frac{60^{\circ}}{360^{\circ}} \times \pi \times 100 = \frac{100\pi}{6} \approx 52.36 \text{ cm}^2$$ Area of triangle: $$A_{\triangle} = \frac{1}{2} \times 100 \times \sin 60^{\circ} = \frac{100\sqrt{3}}{4} \approx 43.3 \text{ cm}^2$$ Area of minor segment: $$= 52.36 – 43.3 \approx \mathbf{9.06 \text{ cm}^2}$$ Area of full circle $= 100\pi \approx 314.16$ cm²
Area of major segment: $$= 314.16 – 9.06 \approx \mathbf{305.1 \text{ cm}^2}$$
The long and short hands of a clock are 6 cm and 4 cm long, respectively. Find the sum of the distances travelled by their tips in 24 hours.
Answer
Minute hand length $= 6$ cm; hour hand length $= 4$ cm.
Distance by minute hand tip: The minute hand completes 24 revolutions in 24 hours. $$= 24 \times 2\pi \times 6 = 288\pi \text{ cm}$$ Distance by hour hand tip: The hour hand completes 2 revolutions in 24 hours. $$= 2 \times 2\pi \times 4 = 16\pi \text{ cm}$$ Total distance: $$= 288\pi + 16\pi = 304\pi \approx \mathbf{954.93 \text{ cm}}$$
Distance by minute hand tip: The minute hand completes 24 revolutions in 24 hours. $$= 24 \times 2\pi \times 6 = 288\pi \text{ cm}$$ Distance by hour hand tip: The hour hand completes 2 revolutions in 24 hours. $$= 2 \times 2\pi \times 4 = 16\pi \text{ cm}$$ Total distance: $$= 288\pi + 16\pi = 304\pi \approx \mathbf{954.93 \text{ cm}}$$
Case Study — Read the following and answer any four questions from Q.1 to Q.5.
Jaya drew this rangoli design during a competition.
Circles $C_1$, $C_2$, $C_3$ and $C_4$ have common centre P. The radius of circle $C_1$ is 6 cm. The table below shows the radii of circles in terms of the radius of circle $C_1$:
1. The perimeter of circle $C_1$ is:
(A) $37\dfrac{5}{7}$ cm (B) $\dfrac{265}{7}$ cm (C) $37.81$ cm (D) $36\dfrac{5}{7}$ cm
2. The circumference of circle $C_2$ is:
(A) $12\pi$ cm (B) $24\pi$ cm (C) $15\pi$ cm (D) $20\pi$ cm
3. The area enclosed between $C_1$ and $C_2$, and between $C_3$ and $C_4$, has been painted. What area has been painted?
(A) $85\pi$ cm² (B) $108\pi$ cm² (C) $324\pi$ cm² (D) $846\pi$ cm²
4. Jaya wants to outline the boundaries of circles $C_2$ and $C_3$ with a ribbon. One roll of ribbon is 20 cm long. How many rolls would Jaya need? (Use $\pi = \dfrac{22}{7}$)
(A) 8 (B) 9 (C) 12 (D) 116
5. Since the radius of circle $C_4$ is 3.5 times that of $C_1$, the area occupied by circle $C_4$ is:
(A) 12.25 times of $C_1$ (B) 3.5 times of $C_1$ (C) 7.5 times of $C_1$ (D) 11.25 times of $C_1$
Jaya drew this rangoli design during a competition.
Circles $C_1$, $C_2$, $C_3$ and $C_4$ have common centre P. The radius of circle $C_1$ is 6 cm. The table below shows the radii of circles in terms of the radius of circle $C_1$:
| Radius of circle | Times the radius of $C_1$ |
|---|---|
| $C_2$ | 2 |
| $C_3$ | 2.5 |
| $C_4$ | 3.5 |
1. The perimeter of circle $C_1$ is:
(A) $37\dfrac{5}{7}$ cm (B) $\dfrac{265}{7}$ cm (C) $37.81$ cm (D) $36\dfrac{5}{7}$ cm
2. The circumference of circle $C_2$ is:
(A) $12\pi$ cm (B) $24\pi$ cm (C) $15\pi$ cm (D) $20\pi$ cm
3. The area enclosed between $C_1$ and $C_2$, and between $C_3$ and $C_4$, has been painted. What area has been painted?
(A) $85\pi$ cm² (B) $108\pi$ cm² (C) $324\pi$ cm² (D) $846\pi$ cm²
4. Jaya wants to outline the boundaries of circles $C_2$ and $C_3$ with a ribbon. One roll of ribbon is 20 cm long. How many rolls would Jaya need? (Use $\pi = \dfrac{22}{7}$)
(A) 8 (B) 9 (C) 12 (D) 116
5. Since the radius of circle $C_4$ is 3.5 times that of $C_1$, the area occupied by circle $C_4$ is:
(A) 12.25 times of $C_1$ (B) 3.5 times of $C_1$ (C) 7.5 times of $C_1$ (D) 11.25 times of $C_1$
Answer
1. Answer: (A)
Radius of $C_1 = 6$ cm. Perimeter $= 2 \times \dfrac{22}{7} \times 6 = \dfrac{264}{7} = 37\dfrac{5}{7}$ cm
2. Answer: (B)
Radius of $C_2 = 2 \times 6 = 12$ cm. Circumference $= 2\pi \times 12 = 24\pi$ cm
3. Answer: (C)
Area between $C_1$ and $C_2 = \pi(12^2 – 6^2) = \pi(144 – 36) = 108\pi$ cm²
Radius of $C_3 = 2.5 \times 6 = 15$ cm; $C_4 = 3.5 \times 6 = 21$ cm
Area between $C_3$ and $C_4 = \pi(21^2 – 15^2) = \pi(441 – 225) = 216\pi$ cm²
Total painted area $= 108\pi + 216\pi = 324\pi$ cm²
4. Answer: (B)
Circumference of $C_2 = \dfrac{528}{7}$ cm; Circumference of $C_3 = 2 \times \dfrac{22}{7} \times 15 = \dfrac{660}{7}$ cm
Total ribbon needed $= \dfrac{528 + 660}{7} = \dfrac{1188}{7} \approx 169.71$ cm
Number of rolls $= \dfrac{169.71}{20} \approx 8.49 \Rightarrow$ 9 rolls
5. Answer: (A)
Area of $C_1 = \pi(6)^2 = 36\pi$ cm²; Area of $C_4 = \pi(21)^2 = 441\pi$ cm²
$\dfrac{\text{Area of }C_4}{\text{Area of }C_1} = \dfrac{441\pi}{36\pi} = 12.25$ times
Radius of $C_1 = 6$ cm. Perimeter $= 2 \times \dfrac{22}{7} \times 6 = \dfrac{264}{7} = 37\dfrac{5}{7}$ cm
2. Answer: (B)
Radius of $C_2 = 2 \times 6 = 12$ cm. Circumference $= 2\pi \times 12 = 24\pi$ cm
3. Answer: (C)
Area between $C_1$ and $C_2 = \pi(12^2 – 6^2) = \pi(144 – 36) = 108\pi$ cm²
Radius of $C_3 = 2.5 \times 6 = 15$ cm; $C_4 = 3.5 \times 6 = 21$ cm
Area between $C_3$ and $C_4 = \pi(21^2 – 15^2) = \pi(441 – 225) = 216\pi$ cm²
Total painted area $= 108\pi + 216\pi = 324\pi$ cm²
4. Answer: (B)
Circumference of $C_2 = \dfrac{528}{7}$ cm; Circumference of $C_3 = 2 \times \dfrac{22}{7} \times 15 = \dfrac{660}{7}$ cm
Total ribbon needed $= \dfrac{528 + 660}{7} = \dfrac{1188}{7} \approx 169.71$ cm
Number of rolls $= \dfrac{169.71}{20} \approx 8.49 \Rightarrow$ 9 rolls
5. Answer: (A)
Area of $C_1 = \pi(6)^2 = 36\pi$ cm²; Area of $C_4 = \pi(21)^2 = 441\pi$ cm²
$\dfrac{\text{Area of }C_4}{\text{Area of }C_1} = \dfrac{441\pi}{36\pi} = 12.25$ times
Case Study — Read the following passage and answer the following questions.
The governing council of a local public development authority of Dehradun decided to build an adventurous playground on the top of a hill, which will have adequate space for parking.
After survey, it was decided to build a rectangular playground, with a semi-circular area allotted for parking at one end. The length and breadth of the rectangular playground are 14 units and 7 units respectively. There are two quadrants of radius 2 units on one side for special seats.
Based on the above information, answer the following questions:
(i) What is the total perimeter of the parking area?
(ii) What is the total area of parking and the two quadrants?
(iii) Find the cost of fencing the playground and parking area at the rate of ₹2 per unit.
The governing council of a local public development authority of Dehradun decided to build an adventurous playground on the top of a hill, which will have adequate space for parking.
After survey, it was decided to build a rectangular playground, with a semi-circular area allotted for parking at one end. The length and breadth of the rectangular playground are 14 units and 7 units respectively. There are two quadrants of radius 2 units on one side for special seats.
Based on the above information, answer the following questions:
(i) What is the total perimeter of the parking area?
(ii) What is the total area of parking and the two quadrants?
(iii) Find the cost of fencing the playground and parking area at the rate of ₹2 per unit.
Answer
The parking area is a semi-circle with diameter equal to the breadth of the playground $= 7$ units, so radius $r = 3.5$ units.
(i) Perimeter of parking area (semi-circle boundary $=$ arc $+$ diameter): $$= \pi r + 2r = \pi \times 3.5 + 7 = 3.5\pi + 7 \approx 18 \text{ units}$$ (ii) Total area of parking and two quadrants:
Area of parking (semi-circle): $$= \frac{\pi r^2}{2} = \frac{\pi \times 3.5^2}{2} = \frac{22}{7} \times \frac{12.25}{2} = 19.25 \text{ units}^2$$ Area of two quadrants (radius 2 units each): $$= 2 \times \frac{1}{4}\pi(2)^2 = 2\pi \approx 6.28 \text{ units}^2$$ Total $\approx 19.25 + 6.28 = 25.53$ units²
(iii) Cost of fencing:
Perimeter of playground $+ $ parking arc (diameter side is shared, not fenced): $$= 2(14 + 7) + \pi \times 3.5 = 42 + 3.5\pi \approx 42 + 11 = 53 \text{ units}$$ Cost $= 53 \times 2 = \mathbf{₹\,106}$
(i) Perimeter of parking area (semi-circle boundary $=$ arc $+$ diameter): $$= \pi r + 2r = \pi \times 3.5 + 7 = 3.5\pi + 7 \approx 18 \text{ units}$$ (ii) Total area of parking and two quadrants:
Area of parking (semi-circle): $$= \frac{\pi r^2}{2} = \frac{\pi \times 3.5^2}{2} = \frac{22}{7} \times \frac{12.25}{2} = 19.25 \text{ units}^2$$ Area of two quadrants (radius 2 units each): $$= 2 \times \frac{1}{4}\pi(2)^2 = 2\pi \approx 6.28 \text{ units}^2$$ Total $\approx 19.25 + 6.28 = 25.53$ units²
(iii) Cost of fencing:
Perimeter of playground $+ $ parking arc (diameter side is shared, not fenced): $$= 2(14 + 7) + \pi \times 3.5 = 42 + 3.5\pi \approx 42 + 11 = 53 \text{ units}$$ Cost $= 53 \times 2 = \mathbf{₹\,106}$
Frequently Asked Questions
What does the Areas Related to Circles chapter cover in CBSE Class 10 Maths? ⌄
The Areas Related to Circles chapter in CBSE Class 10 covers the perimeter (circumference) and area of a circle, length of an arc, area of a sector, area of a segment, and the area of combinations of plane figures involving circles. Students learn to apply these formulas to real-life scenarios involving sectors, segments, and composite shapes — all of which are regularly tested in board exams.
How many marks does Areas Related to Circles carry in the CBSE Class 10 board exam? ⌄
Areas Related to Circles falls under the Mensuration unit in CBSE Class 10 Maths, which carries approximately 10 marks in the board paper. Questions from this chapter typically appear as 1-mark MCQs, 2-mark short answers, 3-mark problems involving combinations of shapes, and occasionally as 5-mark case study questions — making it a high-value chapter across multiple question types.
What are the most important topics to focus on in Areas Related to Circles? ⌄
The most important topics are: the formula for arc length, area of a sector and area of a segment (sector minus triangle), areas of combined figures involving rectangles, triangles and semi-circles, and problems involving clock hands and rolling wheels. Questions that ask students to find ungrazed areas in triangular fields or areas of shaded regions in composite figures are especially frequent in previous year papers.
What common mistakes do students make when solving Areas Related to Circles questions? ⌄
A very common error is confusing the area of a sector with the area of a segment — students often forget to subtract the triangle’s area when finding the segment. Another frequent mistake is using diameter instead of radius in the formula $\pi r^2$. Students also sometimes use the wrong value of $\pi$ when the question specifies $\dfrac{22}{7}$ or $3.14$. Practising a variety of previous year questions helps your child avoid these errors under exam pressure.
How does Angle Belearn help students score well in Areas Related to Circles? ⌄
Angle Belearn’s CBSE specialists curate chapter-wise question banks drawn from real board papers, each paired with clear step-by-step solutions that show structured working — exactly what earns full marks in board exams. Regular practice with these verified questions builds both speed and accuracy in applying area and perimeter formulas, so your child walks into the exam confident and well-prepared.
