class 10 maths Areas Related to Circles questions with solutions

Q 1 :

A ‘circus’ is a company of performers who put on shows of acrobats, clowns, etc to entertain people. Started around 250 years back in open fields, now generally performed in tents. One such ‘Circus Tent’ is shown in figure. The tent is in the shape of a cylinder surmounted by a conical top. The height and diameter of cylindrical part are 9 m and 30 m respectively and height of conical part is 8 m with same diameter as that of the cylindrical part.

Based on the above information, answer the following questions:

(i) the area of the canvas used in making the tent.

$1650 m^{2}$
$- 1650 m^{2}$
$1550 m^{2}$
$1630 m^{2}$

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(1)

We have,
Radius of cylindrical portion = radius of conical part = 15 m $\Rightarrow r = 15 m$

and, height of cylindrical part , $h = 9 m$

Also, height of conical portion, $H = 8 m$

Slant height of cone

$l = \sqrt{(15)^{2} + (8)^{2}} = \sqrt{225 + 64} = \sqrt{289} = 17 m ∴ Area of canvas used in making tent$

$= CSA of cylinder + CSA of cone = 2 πrh + πrl$

$= 2 \times \frac{22}{7} \times 15 \times 9 + \frac{22}{7} \times 15 \times 17 = 848.57 + 801.43 = 1650 m^{2}$

Q 2 :

A ‘circus’ is a company of performers who put on shows of acrobats, clowns, etc to entertain people. Started around 250 years back in open fields, now generally performed in tents. One such ‘Circus Tent’ is shown in figure. The tent is in the shape of a cylinder surmounted by a conical top. The height and diameter of cylindrical part are 9 m and 30 m respectively and height of conical part is 8 m with same diameter as that of the cylindrical part.

Based on the above information, answer the following questions:

(ii) the cost of the canvas bought for the tent at the rate Rs 200 per sq m, if 30 sq m canvas was wasted during stitching.

$3, 34, 000$
$3, 36, 000$
$3, 30, 000$
$3, 32, 000$

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(2)

Now, area of canvas bought for the tent $= (1650 + 30) = 1680 m^{2}$

Total cost of the canvas $= R s (200 \times 1680) = R s 3, 36, 000$

Q 3 :

On a Sunday, your parents took you to a fair. You could see lot of toys displayed, and you wanted them to buy a RUBIK’S cube and strawberry ice-cream for you.

Observe the figures and answer the questions:

(i) Find the length of the diagonal of RUBIK’s cube if each edge measures 6 cm.

$6 \sqrt{2} cm$
$- 6 \sqrt{3} cm$
$5 \sqrt{3} cm$
$6 \sqrt{3} cm$

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(4)

We have length of the edge of a cube = 6 cm

$\Rightarrow a = 6 cm ∴ Length of the diagonal = a \sqrt{3}$

$= \sqrt{3} \times 6 = 6 \sqrt{3} cm$

Q 4 :

On a Sunday, your parents took you to a fair. You could see lot of toys displayed, and you wanted them to buy a RUBIK’S cube and strawberry ice-cream for you.

Observe the figures and answer the questions:

(ii) Find the volume of the RUBIK’s cube if the length of the edge is 7 cm.

$343 {cm}^{3}$
$345 {cm}^{3}$
$350 {cm}^{3}$
$342 {cm}^{3}$

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(1)

We have, edge of the cube, a = 7 cm

$∴ Volume of the cube = a^{3} = (7)^{3} = 343 {cm}^{3}$

Q 5 :

On a Sunday, your parents took you to a fair. You could see lot of toys displayed, and you wanted them to buy a RUBIK’S cube and strawberry ice-cream for you.

Observe the figures and answer the questions:

(iii) (a) What is the curved surface area of hemisphere (ice cream) if the base radius is 7 cm?

$305 {cm}^{2}$
$308 {cm}^{2}$
$258 {cm}^{2}$
$- 308 {cm}^{2}$

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(2)

Given base radius of hemisphere, r = 7 cm

$∴ Curved surface area of hemisphere (ice cream) = 2 π r^{2}$

$= 2 \times \frac{22}{7} \times (7)^{2} = 44 \times 7 = 308 {cm}^{2}$

Q 6 :

On a Sunday, your parents took you to a fair. You could see lot of toys displayed, and you wanted them to buy a RUBIK’S cube and strawberry ice-cream for you.

Observe the figures and answer the questions:

(iii) (b) If two cubes of edges 4 cm are joined end to end then find the surface area of the resulting cuboid.

$160 {cm}^{2}$
$150 {cm}^{2}$
$155 {cm}^{2}$
$152 {cm}^{2}$

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(1)

When two cubes of side (edge) 4 cm are joined end to end we get a cuboid of:

Length = 8 cm,
Breadth = 4 cm,
Height = 4 cm

$∴ Surface area of cuboid = 2 (l b + b h + l h)$

$= 2 (8 \times 4 + 4 \times 4 + 4 \times 8)$

$= 2 (32 + 16 + 32)$

$= 2 \times 80 = 160 {cm}^{2}$

Q 7 :

The Atal Tunnel, also known as the Rohtang Tunnel, is a highway tunnel built under the Rohtang Pass in the eastern Pir Panjal range of the Himalayas, located on the Leh–Manali highway in Himachal Pradesh. Spanning a length of 9.02 km, it is the world’s longest tunnel situated above 10,000 feet and is named after the former Prime Minister of India, Atal Bihari Vajpayee. The cross-section of a tunnel is shown in figure. The radius of circular part is $5 \sqrt{2} m e t r e s a n d ∠ A O B = 90 ° .$

Based on the above information, answer the following questions:

(i) Find the width of the tunnel (same as the width of the road).

10 m
15 m
20 m
30 m

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(1)

$I n r i g h t ∆ A O B, u s i n g P y t h a g o r a s T h e o r e m : A B^{2} = O A^{2} + O B^{2}$

$A B^{2} = (5 \sqrt{2})^{2} + (5 \sqrt{2})^{2} A B^{2} = 50 + 50 = 100$

$A B = 10 m The width of the tunnel is 10 m .$

Q 8 :

The Atal Tunnel, also known as the Rohtang Tunnel, is a highway tunnel built under the Rohtang Pass in the eastern Pir Panjal range of the Himalayas, located on the Leh–Manali highway in Himachal Pradesh. Spanning a length of 9.02 km, it is the world’s longest tunnel situated above 10,000 feet and is named after the former Prime Minister of India, Atal Bihari Vajpayee. The cross-section of a tunnel is shown in figure. The radius of circular part is $5 square root of 2 space end root m e t r e s space a n d angle A O B equals 90 degree.$

Based on the above information, answer the following questions:

(ii) Find the height of the tunnel.

$5 (\sqrt{2} + 3) m$
$4 (\sqrt{2} + 1) m$
$5 (\sqrt{2} + 1) m$
$5 (\sqrt{- 2} + 1) m$

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(3)

$T h e h e i g h t o f t h e t u n n e l w i l l b e (O M + O C) . I n ∆ O A M a n d ∆ O B M$

$∠ O A M = ∠ O B M (∵ O A = O B) ∠ O M A = ∠ O M B = 90 °$

$B y A A s i m i l a r i t y, ∆ O A M ~ ∆ O B M A M = M B = \frac{A B}{2} = 5 m$

$In right ∆ OMB : {OB}^{2} = {OM}^{2} + {MB}^{2}$

$(5 \sqrt{2})^{2} = {OM}^{2} + (5)^{2} 50 = {OM}^{2} + 25 OM = 5 m$

$OC = OA = 5 \sqrt{2} m ∴ Height of the tunnel = OM + OC = 5 (\sqrt{2} + 1) m$

Q 9 :

The Atal Tunnel, also known as the Rohtang Tunnel, is a highway tunnel built under the Rohtang Pass in the eastern Pir Panjal range of the Himalayas, located on the Leh–Manali highway in Himachal Pradesh. Spanning a length of 9.02 km, it is the world’s longest tunnel situated above 10,000 feet and is named after the former Prime Minister of India, Atal Bihari Vajpayee. The cross-section of a tunnel is shown in figure. The radius of circular part is $5 square root of 2 space end root m e t r e s space a n d angle A O B equals 90 degree.$

Based on the above information, answer the following questions:

(iii) (a) What is the perimeter of cross-section of the tunnel?

$(\frac{15}{\sqrt{2}} π + 10) m$
$(\frac{- 15}{\sqrt{2}} π + 10) m$
$(\frac{14}{\sqrt{2}} π + 10) m$
$(\frac{15}{\sqrt{3}} π + 10) m$

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(1)

Perimeter of cross-section of the tunnel

$= Major arc A B + A B = \frac{270 °}{360 °} \times 2 π r + 10$

$= (\frac{3}{4} \times 2 π \times 5 \sqrt{2} + 10) = (\frac{15}{\sqrt{2}} π + 10) m$

Q 10 :

The Atal Tunnel, also known as the Rohtang Tunnel, is a highway tunnel built under the Rohtang Pass in the eastern Pir Panjal range of the Himalayas, located on the Leh–Manali highway in Himachal Pradesh. Spanning a length of 9.02 km, it is the world’s longest tunnel situated above 10,000 feet and is named after the former Prime Minister of India, Atal Bihari Vajpayee. The cross-section of a tunnel is shown in figure. The radius of circular part is $5 square root of 2 space end root m e t r e s space a n d angle A O B equals 90 degree.$

Based on the above information, answer the following questions:

(iii) (b) What is the area of cross-section of the tunnel?

$\frac{70 π}{2} + 25 m^{2}$
$\frac{75 π}{2} + 25 m^{2}$
$\frac{75 π}{2} + 20 m^{2}$
$\frac{75 π}{2} + 19 m^{2}$

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(2)

Area of cross-section of the tunnel

$= \frac{270 °}{360 °} \times π r^{2} + ar △ A O B$

$= \frac{3}{4} π r^{2} + \frac{1}{2} \times O A \times O B$

$= \frac{3}{4} π (5 \sqrt{2})^{2} + \frac{1}{2} \times 5 \sqrt{2} \times 5 \sqrt{2}$

$= \frac{75 π}{2} + 25 m^{2}$

Topic Question Set

Q 3 :

On a Sunday, your parents took you to a fair. You could see lot of toys displayed, and you wanted them to buy a RUBIK’S cube and strawberry ice-cream for you.

Observe the figures and answer the questions:

(i) Find the length of the diagonal of RUBIK’s cube if each edge measures 6 cm.

Q 4 :

On a Sunday, your parents took you to a fair. You could see lot of toys displayed, and you wanted them to buy a RUBIK’S cube and strawberry ice-cream for you.

Observe the figures and answer the questions:

(ii) Find the volume of the RUBIK’s cube if the length of the edge is 7 cm.

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