Vardaan Learning Institute
Chapter Solutions: Areas Related to Circles
Class: 10 (CBSE)
Subject: Mathematics
Max. Marks: 50
SECTION A: OBJECTIVE TYPE QUESTIONS (1 Mark Each)
1. Sum of areas equal to area of circle R, then:
$\pi R_1^2 + \pi R_2^2 = \pi R^2 \Rightarrow R_1^2 + R_2^2 = R^2$.
Correct Option: (b) $R_1^2 + R_2^2 = R^2$
2. Area of sector of angle p...
Area = $\frac{p}{360} \pi R^2$
$= \frac{p}{360 \times 2} 2\pi R^2 = \frac{p}{720} \times 2\pi R^2$.
$= \frac{p}{360 \times 2} 2\pi R^2 = \frac{p}{720} \times 2\pi R^2$.
Correct Option: (d) $\frac{p}{720} \times 2\pi R^2$
3. Area swept by minute hand (14 cm) in 5 min:
Angle = $30^\circ$ (since $360/60 \times 5 = 30$).
Area = $\frac{30}{360} \times \frac{22}{7} \times 14 \times 14$
$= \frac{1}{12} \times 22 \times 2 \times 14$
$= \frac{1}{12} \times 616 = \frac{154}{3}$ cm$^2$.
Area = $\frac{30}{360} \times \frac{22}{7} \times 14 \times 14$
$= \frac{1}{12} \times 22 \times 2 \times 14$
$= \frac{1}{12} \times 616 = \frac{154}{3}$ cm$^2$.
Correct Option: (a) $154/3$ cm$^2$
4. Perimeter of circle = Perimeter of square. Ratio of areas?
$2\pi r = 4a \Rightarrow a = \frac{\pi r}{2}$.
Ratio = $\pi r^2 : a^2 = \pi r^2 : (\frac{\pi r}{2})^2 = \pi r^2 : \frac{\pi^2 r^2}{4}$
$= 1 : \frac{\pi}{4} = 4 : \pi = 4 : \frac{22}{7} = 28 : 22 = 14 : 11$.
Ratio = $\pi r^2 : a^2 = \pi r^2 : (\frac{\pi r}{2})^2 = \pi r^2 : \frac{\pi^2 r^2}{4}$
$= 1 : \frac{\pi}{4} = 4 : \pi = 4 : \frac{22}{7} = 28 : 22 = 14 : 11$.
Correct Option: (b) 14:11
5. Wheel 1000 revs = 88km. Radius?
Distance = $n \times 2\pi r$. $88000 = 1000 \times 2 \times \frac{22}{7} \times r$.
$88 = 2 \times \frac{22}{7} \times r$.
$44/7 \times r = 88 \Rightarrow r = 14$ m.
$88 = 2 \times \frac{22}{7} \times r$.
$44/7 \times r = 88 \Rightarrow r = 14$ m.
Correct Option: (a) 14 m
6. Race track circumferences 437 and 503. Width?
$2\pi R - 2\pi r = 503 - 437 = 66$.
$2\pi(R-r) = 66$.
$R-r = \frac{66}{2\pi} = \frac{33}{22/7} = \frac{33 \times 7}{22} = \frac{3 \times 7}{2} = 10.5$ m.
$2\pi(R-r) = 66$.
$R-r = \frac{66}{2\pi} = \frac{33}{22/7} = \frac{33 \times 7}{22} = \frac{3 \times 7}{2} = 10.5$ m.
Correct Option: (a) 10.5 m
7. Area of circle inscribed in square of side 6 cm.
Diameter = Side = 6cm. Radius = 3cm.
Area = $\pi r^2 = \pi(3)^2 = 9\pi$ cm$^2$.
Area = $\pi r^2 = \pi(3)^2 = 9\pi$ cm$^2$.
Correct Option: (d) $9\pi$ cm$^2$
8. If C=P, relation between areas?
For same perimeter, circle has max area.
From Q4, Circle : Square = 14 : 11. So Circle > Square.
From Q4, Circle : Square = 14 : 11. So Circle > Square.
Correct Option: (b) Area of the circle > Area of the square
9. Area of sector (r=6, angle=60).
Area = $\frac{60}{360} \times \frac{22}{7} \times 6^2 = \frac{1}{6} \times \frac{22}{7} \times 36$
$= \frac{132}{7}$ cm$^2$.
$= \frac{132}{7}$ cm$^2$.
Correct Option: (a) 132/7 cm$^2$
10. Assertion: Minute hand area...
Reason is for arc length, but Assertion talks about Area. However, formula is correct for arc
length.
Let's check A. $r=7, t=30$. Angle = 180.
Area = $\frac{180}{360} \pi r^2 = \frac{1}{2} \times \frac{22}{7} \times 49 = 11 \times 7 = 77$.
A is True. R is True (formula for arc length is correct).
But R does not explain A (Area).
Let's check A. $r=7, t=30$. Angle = 180.
Area = $\frac{180}{360} \pi r^2 = \frac{1}{2} \times \frac{22}{7} \times 49 = 11 \times 7 = 77$.
A is True. R is True (formula for arc length is correct).
But R does not explain A (Area).
Correct Option: (b) Both A and R are true but R is not the correct explanation of A.
SECTION B: SHORT ANSWER TYPE QUESTIONS (2 Marks Each)
11. Area of quadrant circle with C=22 cm.
$2\pi r = 22 \Rightarrow r = 3.5$ cm.
Area = $\frac{1}{4} \pi r^2 = \frac{1}{4} \times \frac{22}{7} \times 3.5 \times 3.5 = 9.625$ cm$^2$.
Area = $\frac{1}{4} \pi r^2 = \frac{1}{4} \times \frac{22}{7} \times 3.5 \times 3.5 = 9.625$ cm$^2$.
9.625 cm$^2$
12. Area of minor segment (r=10m, angle=90).
Area of Sector = $\frac{90}{360} \times 3.14 \times 100 = 78.5$.
Area of $\triangle$ = $1/2 \times 10 \times 10 = 50$.
Seg Area = $78.5 - 50 = 28.5$ cm$^2$.
Area of $\triangle$ = $1/2 \times 10 \times 10 = 50$.
Seg Area = $78.5 - 50 = 28.5$ cm$^2$.
28.5 cm$^2$
13. Length of arc (r=21, angle=60).
Length = $\frac{60}{360} \times 2\pi r = \frac{1}{6} \times 2 \times \frac{22}{7} \times 21 = 22$ cm.
22 cm
14. Circle with C = C1 + C2. r1=19, r2=9.
$2\pi R = 2\pi r_1 + 2\pi r_2$.
$R = r_1 + r_2 = 19 + 9 = 28$ cm.
$R = r_1 + r_2 = 19 + 9 = 28$ cm.
28 cm
SECTION C: SHORT ANSWER TYPE II QUESTIONS (3 Marks Each)
15. Area of shaded region (Square - 2 Semicircles).
Area of Square = $14 \times 14 = 196$.
Area of 2 Semicircles = Area of 1 circle ($r=7$) = $\frac{22}{7} \times 7 \times 7 = 154$.
Shaded Area = $196 - 154 = 42$ cm$^2$.
Area of 2 Semicircles = Area of 1 circle ($r=7$) = $\frac{22}{7} \times 7 \times 7 = 154$.
Shaded Area = $196 - 154 = 42$ cm$^2$.
42 cm$^2$
16. Area of designed region (Leaf shape between quadrants).
Area = $2 \times (\text{Area of segment})$.
Area of 1 Quad = $\frac{1}{4} \times \frac{22}{7} \times 64 = 50.28$.
Area of $\triangle$ = $1/2 \times 8 \times 8 = 32$.
Area of segment = $50.28 - 32 = 18.28$.
Total Area = $2 \times 18.28 = 36.56 \approx 36.57$ cm$^2$.
Area of 1 Quad = $\frac{1}{4} \times \frac{22}{7} \times 64 = 50.28$.
Area of $\triangle$ = $1/2 \times 8 \times 8 = 32$.
Area of segment = $50.28 - 32 = 18.28$.
Total Area = $2 \times 18.28 = 36.56 \approx 36.57$ cm$^2$.
36.57 cm$^2$
17. Cost of designs (6 segments).
Area of 1 sector ($60^\circ$) = $\frac{60}{360} \times \frac{22}{7} \times 28^2 = 410.67$.
Area of eq. $\triangle$ = $\frac{\sqrt{3}}{4} \times 28^2 = 333.2$.
Area of 1 design = $410.67 - 333.2 = 77.47$.
Total Area = $6 \times 77.47 = 464.8$.
Cost = $464.8 \times 0.35 = 162.68$.
Area of eq. $\triangle$ = $\frac{\sqrt{3}}{4} \times 28^2 = 333.2$.
Area of 1 design = $410.67 - 333.2 = 77.47$.
Total Area = $6 \times 77.47 = 464.8$.
Cost = $464.8 \times 0.35 = 162.68$.
Rs. 162.68
18. Equilateral triangle OAB (side 12) + Circular arc (r=6) outside.
Area of circle = $\pi r^2 = \frac{22}{7} \times 36 = 113.14$.
Area of Equilat $\triangle$ = $\frac{\sqrt{3}}{4} \times 12^2 = 36\sqrt{3} = 62.35$.
We need shaded area. Is the arc inside or outside? Typically this question implies Total Area = Area Circle + Area Triangle - Area Sector (overlapped).
Angle = 60.
Area Sector = $\frac{60}{360} \times \frac{22}{7} \times 36 = \frac{132}{7} = 18.86$.
Total Area = $113.14 + 62.35 - 18.86 = 156.63$ cm$^2$.
Area of Equilat $\triangle$ = $\frac{\sqrt{3}}{4} \times 12^2 = 36\sqrt{3} = 62.35$.
We need shaded area. Is the arc inside or outside? Typically this question implies Total Area = Area Circle + Area Triangle - Area Sector (overlapped).
Angle = 60.
Area Sector = $\frac{60}{360} \times \frac{22}{7} \times 36 = \frac{132}{7} = 18.86$.
Total Area = $113.14 + 62.35 - 18.86 = 156.63$ cm$^2$.
156.63 cm$^2$
19. Square side 4cm, corners 1cm quad, center circle 2cm dia. Remaining area.
Area Square = $4^2 = 16$.
Area 4 Quads = Area 1 Circle ($r=1$) = $\pi (1)^2 = \pi$.
Area Center Circle ($r=1$) = $\pi (1)^2 = \pi$.
Remaining Area = $16 - 2\pi = 16 - 2(3.14) = 16 - 6.28 = 9.72$ cm$^2$.
Area 4 Quads = Area 1 Circle ($r=1$) = $\pi (1)^2 = \pi$.
Area Center Circle ($r=1$) = $\pi (1)^2 = \pi$.
Remaining Area = $16 - 2\pi = 16 - 2(3.14) = 16 - 6.28 = 9.72$ cm$^2$.
9.72 cm$^2$
20. Horse grazing. 15m square. (i) 5m rope (ii) Increase for 10m rope.
(i) Area = $\frac{90}{360} \pi (5)^2 = \frac{1}{4} \times 3.14 \times 25 = 19.625$ m$^2$.
(ii) New Area = $\frac{1}{4} \times 3.14 \times 10^2 = \frac{314}{4} = 78.5$.
Increase = $78.5 - 19.625 = 58.875$ m$^2$.
(ii) New Area = $\frac{1}{4} \times 3.14 \times 10^2 = \frac{314}{4} = 78.5$.
Increase = $78.5 - 19.625 = 58.875$ m$^2$.
(i) 19.625 m$^2$ (ii) 58.875 m$^2$
SECTION D: LONG ANSWER TYPE QUESTIONS (5 Marks Each)
21. Trapezium, arcs at A,B,C,D (r=7). Area shaded?
Area Trapezium = $\frac{1}{2}(AB+CD)h = \frac{1}{2}(18+32)(14) = \frac{1}{2}(50)(14) = 350$ cm$^2$.
Area of 4 sectors: $\angle A + \angle D = 180$, $\angle B + \angle C = 180$. Total angle 360.
Area sum = Area of 1 circle ($r=7$) = $\frac{22}{7} \times 7 \times 7 = 154$.
Shaded Area = $350 - 154 = 196$ cm$^2$.
Area of 4 sectors: $\angle A + \angle D = 180$, $\angle B + \angle C = 180$. Total angle 360.
Area sum = Area of 1 circle ($r=7$) = $\frac{22}{7} \times 7 \times 7 = 154$.
Shaded Area = $350 - 154 = 196$ cm$^2$.
196 cm$^2$
22. Quadrant OACB (r=3.5). OD=2. Area (i) Quadrant (ii) Shaded.
O is center. A, B on axes? Usually OACB implies a sector. Context suggests OD is on a radius.
Common problem: "OABC is a quadrant... OD=2cm... find area of shaded region" (Shaded is Quadrant - Triangle OBD?).
Assuming typical figure where $\triangle OBD$ is removed from quadrant.
(i) Area Quadrant = $\frac{1}{4} \times \frac{22}{7} \times 3.5 \times 3.5 = 9.625$ cm$^2$.
(ii) Area $\triangle OBD$ (assuming base OB=3.5, height OD=2) = $\frac{1}{2} \times 3.5 \times 2 = 3.5$.
Shaded = $9.625 - 3.5 = 6.125$ cm$^2$.
Common problem: "OABC is a quadrant... OD=2cm... find area of shaded region" (Shaded is Quadrant - Triangle OBD?).
Assuming typical figure where $\triangle OBD$ is removed from quadrant.
(i) Area Quadrant = $\frac{1}{4} \times \frac{22}{7} \times 3.5 \times 3.5 = 9.625$ cm$^2$.
(ii) Area $\triangle OBD$ (assuming base OB=3.5, height OD=2) = $\frac{1}{2} \times 3.5 \times 2 = 3.5$.
Shaded = $9.625 - 3.5 = 6.125$ cm$^2$.
(i) 9.625 cm$^2$ (ii) 6.125 cm$^2$
SECTION E: CASE STUDY ANSWERS
23. Case Study: Car Wipers
(i) Total Area = $2 \times \frac{115}{360} \times \frac{22}{7} \times 25 \times 25$
$= \frac{23}{36} \times \frac{22}{7} \times 625 \approx 1254.96$ cm$^2$.
(ii) Area of one sector = $1254.96 / 2 = 627.48$ cm$^2$.
(iii) If $90^\circ$: Area = $\frac{1}{4} \pi r^2 = 0.25 \times 3.14 \times 625$
$= 490.625$ cm$^2$.
$= \frac{23}{36} \times \frac{22}{7} \times 625 \approx 1254.96$ cm$^2$.
(ii) Area of one sector = $1254.96 / 2 = 627.48$ cm$^2$.
(iii) If $90^\circ$: Area = $\frac{1}{4} \pi r^2 = 0.25 \times 3.14 \times 625$
$= 490.625$ cm$^2$.
(i) $1254.96$ cm$^2$ (ii) $627.48$ cm$^2$ (iii) $490.625$ cm$^2$