CBSE Class 10 Mathematics Area Related To Circle Worksheet Set 06

Question. The diameter of the circle whose area is 301.84 cm\(^2\) is 
(a) 14.2 cm
(b) 12.8 cm
(c) 19.6 cm
(d) 15.6 cm
Answer: (c) 19.6 cm
Explanation: Let diameter of the circle be \( d \) cm.
\( \therefore \) Area = \( \pi r^2 = \pi \left( \frac{d}{2} \right)^2 \)

\( \implies 301.84 = \frac{22}{7} \times \frac{d^2}{4} \)

\( \implies d^2 = \frac{301.84 \times 4 \times 7}{22} \)

\( \implies d^2 = 384.16 \)

\( \implies d = 19.6 \) cm

Question. A piece of wire 20cm long is bent into the form of an arc of a circle subtending an angle of 60 at its centre. The radius of the circle is 
(a) \( \frac{20}{6 + \pi} \) cm
(b) \( \frac{30}{6 + \pi} \) cm
(c) \( \frac{60}{6 + \pi} \) cm
(d) \( \frac{15}{6 + \pi} \) cm
Answer: (c) \( \frac{60}{6 + \pi} \) cm
Explanation: Given: Length of arc + 2 \( \times \) Radius = 20 cm

\( \implies \frac{\theta}{360^\circ} \times 2\pi r + 2r = 20 \)

\( \implies \frac{60^\circ}{360^\circ} \times 2\pi r + 2r = 20 \)

\( \implies \frac{\pi r}{3} + 2r = 20 \)

\( \implies r \left( \frac{\pi}{3} + 2 \right) = 20 \)

\( \implies r \left( \frac{6 + \pi}{3} \right) = 20 \)

\( \implies r = \frac{60}{6 + \pi} \) cm

Question. The radius of a circle is 20 cm Three more concentric circles are drawn inside it in such a way that it is divided into four parts of equal area. The radius of the largest of the three concentric circles is 
(a) \( 14\sqrt{3} cm \)
(b) \( 10\sqrt{3} cm \)
(c) \( 8\sqrt{3} cm \)
(d) None of the options.
Answer: (b) \( 10\sqrt{3} cm \)
Explanation: Here, Radius of bigger circle (OA) = 20 cm
Let the radii of three concentric circle inscribe in the bigger circle be \( r, r_1 \text{ and } r_2 \) in decreasing order of length of radius respectively.
Then according to the question,
\( \pi (20)^2 - \pi r^2 = \pi r^2 - \pi r_1^2 \)

\( \implies 400 - r^2 = r^2 - r_1^2 \)

\( \implies 2r^2 = r_1^2 + 400 \) ……….(i)
Also \( \pi r^2 - \pi r_1^2 = \pi r_1^2 - \pi r_2^2 \)

\( \implies r^2 - r_1^2 = r_1^2 - r_2^2 \)

\( \implies r^2 = 2r_1^2 - r_2^2 \) ………..(ii)
And \( \pi r_1^2 - \pi r_2^2 = \pi r_2^2 \implies r_1^2 - r_2^2 = r_2^2 \)

\( \implies r_2^2 = \frac{1}{2} r_1^2 \) ……….(iii)
Putting the value of \( r_2^2 \) in eq. (ii), we get
\( r^2 = 2r_1^2 - \frac{1}{2} r_1^2 \)

\( \implies r^2 = \frac{3}{2} r_1^2 \)
Now, putting the value of \( r^2 \) in eq. (i), we get
\( 2 \times \frac{3}{2} r_1^2 = r_1^2 + 400 \)

\( \implies 3r_1^2 - r_1^2 = 400 \)

\( \implies 2r_1^2 = 400 \)

\( \implies r_1^2 = 200 \)
Actually, solving for the largest inner circle where the total area is divided into 4 equal parts:
Area of largest inner circle \( = \frac{3}{4} \times \text{Area of original circle} \)
\( \pi r^2 = \frac{3}{4} \pi (20)^2 \)

\( \implies r^2 = \frac{3}{4} \times 400 \)

\( \implies r^2 = 100 \times 3 \)

\( \implies r = 10\sqrt{3} \) cm or 17.32 m (approx)

Question. If the area of a circle is equal to the area of a square, then the ratio of their perimeters is 
(a) \( 2 : \pi \)
(b) 1 : 2
(c) \( \sqrt{\pi} : 2 \)
(d) \( \pi : 2 \)
Answer: (c) \( \sqrt{\pi} : 2 \)
Explanation: Let the radius of the circle be \( r \) and the side of the square be \( a \).
Then according to the question, \( \pi r^2 = a^2 \)

\( \implies a = r\sqrt{\pi} \) ……….(i)
Now, Ratio of their perimeters = \( \frac{2\pi r}{4a} \)

\( \implies \text{Ratio of their perimeters} = \frac{2\pi r}{4r\sqrt{\pi}} = \frac{\pi}{2\sqrt{\pi}} = \frac{\sqrt{\pi}}{2} \)
Ratio of their perimeters = \( \sqrt{\pi} : 2 \)

Question. A bicycle wheel makes 5000 revolutions in moving 11km. The diameter of the wheel is 
(a) 100cm
(b) 35cm
(c) 140cm
(d) 20cm
Answer: (d) 20cm
Explanation: Let diameter of the wheel be \( d \) cm.
Given: Distance = 11 km = 1100000 cm
No. of Revolutions = \( \frac{\text{Total distance}}{\text{Circumference of wheel}} \)

\( \implies 5000 = \frac{1100000 \times 7}{22 \times d} \)

\( \implies d = \frac{1100000 \times 2}{5000 \times 22} \)

\( \implies d = 100000 / 5000 \)
= 20 cm

Question. Find the area of a sector of angle p (in degrees) of a circle with radius R. 
Answer: Area of sector = \( \frac{\theta}{360} \times \pi r^2 \)
where \( \theta \)= angle, r = radius of circle
Here, we have \( \theta = p \) and radius = R
Putting these in formula,
Area of sector = \( \frac{p}{360} \times \pi R^2 \)

Question. A rope by which a cow is tethered is increased from 16m to 23m. How much additional ground does it have now to graze? 
Answer: Area grazed by the cow = \( \pi(16)^2 \)
If the length of the rope is increased, area grazed by cow = \( \pi(23)^2 \)
Hence, additional area grazed by cow
= \( \pi(23)^2 - \pi(16)^2 \)
= \( 858 \text{ m}^2 \)

Question. To warn ships for underwater rocks, a lighthouse spreads a red-coloured light over a sector of angle 72° to a distance of 15 km. Find the area of the sea over which the ships are warned. [Use \( \pi \) = 3.14.] 
Answer: Required area = area of the sector in which, r = 15 km and \( \theta = 72^\circ \)
\( = \left( \frac{\pi r^2 \theta}{360} \right) \)
\( = \left( 3.14 \times 15 \times 15 \times \frac{72}{360} \right) \text{ km}^2 \)
\( = \left( \frac{314 \times 45}{100} \right) \text{ km}^2 \)
\( = \frac{1413}{10} \text{ km}^2 \)
= 141.3 km\(^2\).

Question. Find the area of a quadrant of a circle whose circumference is 88 cm. 
Answer: We know that circumference of the circle = \( 2\pi r \).

\( \implies 88 = 2 \times \frac{22}{7} \times r \)

\( \implies 88 = \frac{44}{7} \times r \)

\( \implies 2 \times 7 = r \implies r = 14 \) cm
Now, Area of circle = \( \pi r^2 = \frac{22}{7} \times 14 \times 14 = 616 \text{ cm}^2 \)
Area of quadrant \( = \frac{1}{4} \times 616 = 154 \text{ cm}^2 \).

Question. If the perimeter of a protactor is 72 cm, then calculate its area. 
Answer: Given, Perimeter of semi-circular protactor = 72
Perimeter of a semi-circular protactor = Perimeter of semi-circle = \( 2r + \pi r = 72 \)

\( \implies r = \frac{72}{\pi + 2} = \frac{72 \times 7}{36} \)

\( \implies r = 14 \) cm
Area of protactor = \( \frac{1}{2} \pi r^2 = \frac{1}{2} \times \frac{22}{7} \times (14)^2 = 308 \text{ cm}^2 \)

Question. Find the area enclosed between two concentric circles of radii 3.5 cm and 7 cm. A third concentric circle is drawn outside the 7 cm circle, such that the area enclosed between it and the 7 cm circle is same as that between the two inner circles. Find the radius of the third circle correct to one decimal place. 
Answer: Radius of first circle = 3.5 cm
Radius of second circle = 7 cm
Let radius of third circle = x cm
Area between first and second circle
= \( \pi(7)^2 - \pi(3.5)^2 \)
= \( 36.75\pi \text{ cm}^2 \)
Area between second and third circle
= \( \pi(x)^2 - \pi(7)^2 \)
= \( \pi x^2 - 49\pi \)
According to question
Area between first and second circle = Area between second and third circle
\( 36.75\pi = \pi x^2 - 49\pi \)
\( 36.75\pi + 49\pi = \pi x^2 \)
\( 85.75\pi = \pi x^2 \)

\( \implies x^2 = 85.75 \)
\( x = \sqrt{85.75} \approx 9.26 \)cm

Question. The cost of fencing a circular field at the rate Rs 24 per metre is Rs 5280. The field is to be ploughed at the rate of Rs 0.50 per m\(^2\). Find the cost of ploughing the field. 
Answer: We have, Rate of fencing = Rs 24 per metre and, Total cost of fencing = Rs 5280
\( \therefore \) Length of the fence = \( \frac{\text{Total cost}}{\text{Rate}} = \frac{5280}{24} = 220 \) metre

\( \implies \) Circumference of the field = 220 metre

\( \implies 2\pi r = 220 \), where r is the radius of the field

\( \implies 2 \times \frac{22}{7} \times r = 220 \)

\( \implies r = \frac{220 \times 7}{22 \times 2} = 35 \)
Area of the field = \( \pi r^2 = \frac{22}{7} \times 35 \times 35 \text{m}^2 = 22 \times 5 \times 35 \text{m}^2 \)
It is given that the field is ploughed at the rate of Rs 0.50 per m\(^2\)
Cost of ploughing the field = Rs \( (22 \times 5 \times 35 \times 0.50) = \) Rs 1925

Question. A path of 4 m width runs round a semi-circular grassy plot whose circumference is \( 163 \frac{3}{7} \) m Find:
i. the area of the path
ii. the cost of gravelling the path at the rate of Rs 1.50 per square metre
iii. the cost of turfing the plot at the rate of 45 paise per m\(^2\). 
Answer: Let x be the radius of the semi-circular grassy plot.
Given, Circumference of grassy plot = \( 163 \frac{3}{7} \) m

\( \implies 2r + \pi r = 163 \frac{3}{7} m = \frac{1144}{7} m \)

\( \implies (2 + \pi) r = \frac{1144}{7} m \)

\( \implies \frac{36}{7} r = \frac{1144}{7} m \)

\( \implies r = \frac{1144}{36} = \frac{286}{9} \) m
\( \therefore \) Radius of semi-circular grassy plot = \( \frac{286}{9} \) m
Then, radius of outer semi-circle = \( \frac{286}{9} m + 4m = \frac{322}{9} \) m
i. Area of path = Area of outer semi-circle - Area of inner semi-circle
= \( \frac{1}{2} \pi \left( \frac{322}{9} \right)^2 - \frac{1}{2} \pi \left( \frac{286}{6} \right)^2 \)
= \( \frac{1}{2} \pi \left( \frac{322}{9} + \frac{286}{9} \right) \left( \frac{322}{9} - \frac{286}{9} \right) \)
= \( \frac{1}{2} \times \frac{22}{7} \times \frac{608}{9} \times \frac{36}{9} \)
= 424.63m\(^2\)
ii. Rate of gravelling the path = Rs.1.50 per m\(^2\)
\( \therefore \) Total cost = Rs. 1.50 \( \times \) 424.63
= Rs. 636.95
iii. Area of plot = \( \frac{1}{2} \pi r^2 \)
= \( \frac{1}{2} \times \frac{22}{7} \times \frac{286}{9} \times \frac{286}{9} \text{ m}^2 \)
= 1586.87m\(^2\)
Rate of turfing the plot = 45 paise per m\(^2\)
\( \therefore \) Total cost = 1586.87 \( \times \) 45paise
= 71409.15 paise
= Rs. 714.09

Question. A chord of a circle subtends an angle of \( \theta \) at the centre of the circle. The area of the minor segment cut off by the chord is one eighth of the area of the circle. Prove that \( 8 \sin \frac{\theta}{2} \cos \frac{\theta}{2} + \pi = \frac{\pi \theta}{45} \). 
Answer: Given Area of minor segment cut off by AB = \( \frac{1}{8} \times \) Area of circle

\( \implies \frac{\theta}{360^\circ} \times \pi r^2 - \frac{1}{2} r^2 \sin \theta = \frac{1}{8} \times \pi r^2 \)

\( \implies \frac{\theta}{360^\circ} \times \pi r^2 = \frac{1}{8} \pi r^2 + \frac{1}{2} r^2 \sin \theta \)

\( \implies \frac{\theta \pi}{360^\circ} = \frac{\pi}{8} + \frac{1}{2} \sin \theta \) [ Divide by \( r^2 \) ]

\( \implies \frac{\pi \theta}{360^\circ} = \frac{\pi + 4 \sin \theta}{8} \) [Multiply by 8]

\( \implies \frac{\pi \theta}{45^\circ} = \pi + 4 \times 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2} \) [\( \sin 2\theta = 2 \sin \theta \cos \theta \)]

\( \implies \frac{\pi \theta}{45^\circ} = \pi + 8 \sin \frac{\theta}{2} \cos \frac{\theta}{2} \)
Hence Proved