Class 10 maths Areas Related to Circles Hots Questions

Q 1 :

Raju installed a fence around a circular field, with the total cost amounting to Rs 6,000 at a rate of Rs 30 per m. He now plans to plough the field.

Based on the above information answer the following questions:

(i) What is the perimeter of the field?

200 m
400 m
600 m
300 m

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(1)

Let r m be the radius of the circular field.

$∴ L e n g t h o f f e n c e = P e r i m e t e r o f c i r c l e = 2 π r T o t a l c o s t o f f e n c i n g = 2 π r \times 30 6000 = 60 π r \Rightarrow 100 = π r \Rightarrow r = \frac{100}{π} (i) P e r i m e t e r o f t h e f i e l d 2 π r = 2 π \times \frac{100}{π} = 200 m$

Q 2 :

Raju installed a fence around a circular field, with the total cost amounting to Rs 6,000 at a rate of Rs 30 per m. He now plans to plough the field.

Based on the above information answer the following questions:

(ii) What is the radius of the field?

$1 \frac{8}{11} m$
$1 \frac{9}{11} m$
$1 \frac{9}{10} m$
$2 \frac{9}{11} m$

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(2)

Radius of the field

$r = \frac{100}{π} = \frac{700}{22} = \frac{350}{11} = 31 \frac{9}{11} m$

Q 3 :

Raju installed a fence around a circular field, with the total cost amounting to Rs 6,000 at a rate of Rs 30 per m. He now plans to plough the field.

Based on the above information answer the following questions:

(iii) (a) What is the area of the field?

$2181.81 m^{2}$
$3181.82 m^{2}$
$3180.82 m^{2}$
$3181.81 m^{2}$

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(4)

Area of the field

$π r^{2} = π \times {(\frac{100}{π})}^{2} = π \times \frac{100 \times 100}{π^{2}} = \frac{10000}{π} = \frac{10000}{\frac{22}{7}} = \frac{70000}{22} \approx 3181.81 m^{2}$

Q 4 :

A brooch is a small piece of jewelry that features a pin on the back, allowing it to be fastened to a dress, blouse, or coat. Below are designs for some brooches.

Design A: Brooch A is crafted from silver wire in the shape of a circle with a diameter of 28 mm. The wire is also used to create 4 diameters, dividing the circle into 8 equal parts.

Design B: Brooch B is made with two colors, gold and silver. The outer portion is made of gold. The circumference of the silver section is 44 mm, and the gold section is uniformly 3 mm wide.

Based on the above information answer the following questions:

(i) In design A, what is the total length of the wire required?

200 m
400 m
600 m
500 m

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(1)

In design A, diameter of the circle is 28 mm.

$∴ Circumference of the circle = π \times 28 mm = \frac{22}{7} \times 28 mm = 88 mm$

$Length of the wire used in 4 diameters$

$= 4 \times 28 mm = 112 mm$

$∴ Total length of the required wire = (112 + 88) mm = 200 mm$

Q 5 :

Raju installed a fence around a circular field, with the total cost amounting to Rs 6,000 at a rate of Rs 30 per m. He now plans to plough the field.

Based on the above information answer the following questions:

(iii) (b) Find the cost of ploughing the field at the rate of $R s 0.50 p e r m ² .$

$1591.90$
$- 1590.90$
$1490.90$
$1590.90$

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(4)

Total cost of ploughing the field

$3181.81 \times 0.50 = R s 1590.90$

Q 6 :

A brooch is a small piece of jewelry that features a pin on the back, allowing it to be fastened to a dress, blouse, or coat. Below are designs for some brooches.

Design A: Brooch A is crafted from silver wire in the shape of a circle with a diameter of 28 mm. The wire is also used to create 4 diameters, dividing the circle into 8 equal parts.

Design B: Brooch B is made with two colors, gold and silver. The outer portion is made of gold. The circumference of the silver section is 44 mm, and the gold section is uniformly 3 mm wide.

Based on the above information answer the following questions:

(ii) In design A, what is the area of each part of the brooch?

$77 m m^{2}$
$55 m m^{2}$
$70 m m^{2}$
$72 m m^{2}$

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(1)

Area of each part of brooch

$= \frac{1}{8} (Area of the circle) = (\frac{1}{8} \times \frac{22}{7} \times 14^{2}) {mm}^{2} = 77 m m^{2}$

Q 7 :

A brooch is a small piece of jewelry that features a pin on the back, allowing it to be fastened to a dress, blouse, or coat. Below are designs for some brooches.

Design A: Brooch A is crafted from silver wire in the shape of a circle with a diameter of 28 mm. The wire is also used to create 4 diameters, dividing the circle into 8 equal parts.

Design B: Brooch B is made with two colors, gold and silver. The outer portion is made of gold. The circumference of the silver section is 44 mm, and the gold section is uniformly 3 mm wide.

Based on the above information answer the following questions:

(iii) (a) In design B, find the circumference of the outer part (golden).

7 mm
6 mm
5 mm
4 mm

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(1)

Let r mm be the radius of the silver part. Then,

$2 \times \frac{22}{7} \times r = 44 \Rightarrow r = 7 mm .$

Q 8 :

A brooch is a small piece of jewelry that features a pin on the back, allowing it to be fastened to a dress, blouse, or coat. Below are designs for some brooches.

Design A: Brooch A is crafted from silver wire in the shape of a circle with a diameter of 28 mm. The wire is also used to create 4 diameters, dividing the circle into 8 equal parts.

Design B: Brooch B is made with two colors, gold and silver. The outer portion is made of gold. The circumference of the silver section is 44 mm, and the gold section is uniformly 3 mm wide.

Based on the above information answer the following questions:

(iii) (b) A boy is playing with brooch B. He makes revolutions with it along its edge. How many complete revolutions must it take it cover $80 π m m ?$

n = 6
n = 4
n = 3
n = 2

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(2)

Outer circumference of Brooch B:

$= 2 π \times 10 mm = 20 π mm Suppose to cover 80 π mm brooch B takes ncomplete revolutions . Then, 20 π \times n = 80 π \Rightarrow n = 4$

Q 9 :

A stable owner has four horses. He usually ties these horses with 7 m long rope to pegs at each corner of a square shaped grass field of 20 m length, to graze in his farm. But tying with rope sometimes results in injuries to his horses, so he decided to build fence around the area so that each horse can graze.

Based on the above information answer the following questions:

(i) Find the area of the square shaped grass field.

$200 m^{2}$
$400 m^{2}$
$100 m^{2}$
$250 m^{2}$

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(2)

Area of the square shaped grass field $= 20 m \times 20 m = 400 m^{2}$

Q 10 :

A stable owner has four horses. He usually ties these horses with 7 m long rope to pegs at each corner of a square shaped grass field of 20 m length, to graze in his farm. But tying with rope sometimes results in injuries to his horses, so he decided to build fence around the area so that each horse can graze.

Based on the above information answer the following questions:

(ii) (a) Find the area of the total field in which these horses can graze.

$150 m^{2}$
$152 m^{2}$
$145 m^{2}$
$154 m^{2}$

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(4)

Area of the total field in which these horses can graze

$= 4 \times \frac{90 °}{360 °} \times π r^{2} = π r^{2} = \frac{22}{7} \times (7)^{2} = 154 m^{2}$

Q 11 :

A stable owner has four horses. He usually ties these horses with 7 m long rope to pegs at each corner of a square shaped grass field of 20 m length, to graze in his farm. But tying with rope sometimes results in injuries to his horses, so he decided to build fence around the area so that each horse can graze.

Based on the above information answer the following questions:

(ii) (b) If the length of the rope of each horse is increased from 7 m to 10 m, find the area grazed by one horse. $(U s e π = 3.14)$

$78.5 m^{2}$
$68.5 m^{2}$
$73.5 m^{2}$
$69.5 m^{2}$

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(1)

When length of rope is increased, we have r = 10 m

Area grazed by one horse

$= \frac{90}{360} \times π r^{2} = \frac{1}{4} \times 3.14 \times (10)^{2} = \frac{1}{4} \times 3.14 \times 100 = \frac{314}{4} = 78.5 m^{2}$

Q 12 :

A stable owner has four horses. He usually ties these horses with 7 m long rope to pegs at each corner of a square shaped grass field of 20 m length, to graze in his farm. But tying with rope sometimes results in injuries to his horses, so he decided to build fence around the area so that each horse can graze.

Based on the above information answer the following questions:

(iii) What is area of the field that is left ungrazed, if the length of the rope of each horse is 7 m?

$246 m^{2}$
$245 m^{2}$
$235 m^{2}$
$- 246 m^{2}$

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(1)

Area of the field left ungrazed

$= Area of square field - 4 \times area of field in which horses can graze = (20)^{2} - 4 \times \frac{90}{360} \times \frac{22}{7} \times (7)^{2} = 400 - 154 = 246 m^{2}$

Q 13 :

A brooch is crafted from silver wire in the shape of a circle with a diameter of 35 mm. The wire is also used to create 5 diameters, dividing the circle into 10 equal sectors, as shown in the figure.

Based on the above information, answer the following questions:

(i) What is the radius of the circle?

$17.5 mm$
$16.5 mm$
$13.5 mm$
$10.5 mm$

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(1)

Radius of the circle $= \frac{35}{2} mm = 17.5 mm$

Q 14 :

A brooch is crafted from silver wire in the shape of a circle with a diameter of 35 mm. The wire is also used to create 5 diameters, dividing the circle into 10 equal sectors, as shown in the figure.

Based on the above information, answer the following questions:

(ii) What is the circumference of the brooch?

$110 mm$
$120 mm$
$115 mm$
$129 mm$

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(1)

Circumference of the brooch $= 2 π r$

$= 2 \times \frac{22}{7} \times 17.5 = 44 \times 2.5 = 110 mm$

Q 15 :

A brooch is crafted from silver wire in the shape of a circle with a diameter of 35 mm. The wire is also used to create 5 diameters, dividing the circle into 10 equal sectors, as shown in the figure.

Based on the above information, answer the following questions:

(iii) (a) What is the total length of silver wire required?

$250 mm$
$282 mm$
$285 mm$
$200 mm$

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(3)

Total length of the required wire

$= 2 π r + 5 d = 110 + 5 \times 35 = 110 + 175 = 285 mm$

Q 16 :

A brooch is crafted from silver wire in the shape of a circle with a diameter of 35 mm. The wire is also used to create 5 diameters, dividing the circle into 10 equal sectors, as shown in the figure.

Based on the above information, answer the following questions:

(iii) (b) What is the area of each sector of the brooch?

$96.25 {mm}^{2}$
$86.25 {mm}^{2}$
$94.25 {mm}^{2}$
$93.25 {mm}^{2}$

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(1)

Angle subtended by arc at each sector $= \frac{360 °}{10} = 36 °$

∴ Area of each sector $= \frac{36 °}{360 °} \times π \times {(\frac{35}{2})}^{2} = \frac{1}{10} \times \frac{22}{7} \times \frac{35}{2} \times \frac{35}{2} = 96.25 {mm}^{2}$

Q 17 :

A pendulum clock uses a swinging weight as its timekeeping element. Invented in 1656 by Christiaan Huygens, the pendulum clock was the most precise timekeeper in the world, which is why it became so widely used. Its accuracy helped support the faster pace of life necessary for the Industrial Revolution. By the 1930s and 40s, home pendulum clocks were largely replaced by more affordable synchronous electric clocks. Today, pendulum clocks are mostly valued for their decorative and antique qualities.

Dhaani bought a pendulum clock for her living room. The clock features a small pendulum with a length of 45 cm. The minute hand is 14 cm long, while the hour hand is 6 cm long.

Based on the above information, answer the following questions

(i) What is the area swept by the minute hand in 14 minutes?

$143.73 c m^{2}$
$143.72 c m^{2}$
$140.73 c m^{2}$
$139.73 c m^{2}$

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(1)

Angle described by a minute hand in 14 minutes $= 6 ° \times 14 = 84 °$

$R a d i u s r = 14 c m (l e n g t h o f m i n u t e h a n d) ∴ Area swept by the minute hand = \frac{84 °}{360 °} \times \frac{22}{7} \times (14)^{2} = \frac{2156}{15} \approx 143.73 c m^{2}$

Q 18 :

A pendulum clock uses a swinging weight as its timekeeping element. Invented in 1656 by Christiaan Huygens, the pendulum clock was the most precise timekeeper in the world, which is why it became so widely used. Its accuracy helped support the faster pace of life necessary for the Industrial Revolution. By the 1930s and 40s, home pendulum clocks were largely replaced by more affordable synchronous electric clocks. Today, pendulum clocks are mostly valued for their decorative and antique qualities.

Dhaani bought a pendulum clock for her living room. The clock features a small pendulum with a length of 45 cm. The minute hand is 14 cm long, while the hour hand is 6 cm long.

Based on the above information, answer the following questions

(ii) What is the angle described by hour hand in 10 minutes?

$2 °$
$5 °$
$3 °$
$1 °$

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(2)

Since in 12 hours, a hour hand describes $360 °,$

$In 1 hour, angle described = \frac{360 °}{12} = 30 ° T h e r e f o r e, a n g l e d e s c r i b e d b y h o u r h a n d i n 10 m i n u t e s = (30 \times \frac{10}{60}) ° = 5 °$

Q 19 :

A pendulum clock uses a swinging weight as its timekeeping element. Invented in 1656 by Christiaan Huygens, the pendulum clock was the most precise timekeeper in the world, which is why it became so widely used. Its accuracy helped support the faster pace of life necessary for the Industrial Revolution. By the 1930s and 40s, home pendulum clocks were largely replaced by more affordable synchronous electric clocks. Today, pendulum clocks are mostly valued for their decorative and antique qualities.

Dhaani bought a pendulum clock for her living room. The clock features a small pendulum with a length of 45 cm. The minute hand is 14 cm long, while the hour hand is 6 cm long.

Based on the above information, answer the following questions

(iii) (a) What is the distance covered by the tip of hour hand in 3.5 hours?

11 cm
13 cm
12 cm
5 cm

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(1)

Angle described by the tip of hour hand in 3.5 hours

$= 30 ° \times 3.5 = 105 ° Distance covered = \frac{105 °}{360 °} \times 2 \times \frac{22}{7} \times 6 = 11 cm$

Q 20 :

A pendulum clock uses a swinging weight as its timekeeping element. Invented in 1656 by Christiaan Huygens, the pendulum clock was the most precise timekeeper in the world, which is why it became so widely used. Its accuracy helped support the faster pace of life necessary for the Industrial Revolution. By the 1930s and 40s, home pendulum clocks were largely replaced by more affordable synchronous electric clocks. Today, pendulum clocks are mostly valued for their decorative and antique qualities.

Dhaani bought a pendulum clock for her living room. The clock features a small pendulum with a length of 45 cm. The minute hand is 14 cm long, while the hour hand is 6 cm long.

Based on the above information, answer the following questions

(iii) (b) If the tip of pendulum covers a distance of 66 cm in complete oscillation, what is the angle described by pendulum at the centre?

$44 °$
$39 °$
$42 °$
$40 °$

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(3)

Length of the pendulum

$r = 45 c m L e t θ b e t h e a n g l e d e s c r i b e d b y p e n d u l u m a t t h e c e n t r e D i s \tan c e c o v e r e d b y a p e n d u l u m i n c o m p l e t e o s c i l l a t i o n = 2 \times \frac{θ}{360 °} \times 2 π r G i v e n d i s \tan c e = 66 c m 66 = 2 \times \frac{θ}{360 °} \times 2 \times \frac{22}{7} \times 45 66 = \frac{88 \times 45}{7 \times 360 °} \times θ 66 = \frac{11}{7} θ θ = \frac{66 \times 7}{11} = 42 °$

Q 21 :

Pragati made a paper flower using 4 identical circles and a dotted square. The front view and back view of the flower is as shown below.

The diameter of each circle is the same as the length of the side of the square, 42 m.

Based on the above information, answer the following questions:

(i) Find the perimeter of the flower in the back view. Show your work.

564 m
555 m
552 m
550 m

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(1)

We have,
Length of the side of the square = 42 m
Radius of the circle = 21 m

Arc length of four sectors of circle (overlapped by square)

$= 4 \times \frac{90 °}{360 °} \times 2 π (21) = 2 \times \frac{22}{7} \times 21 = 132 m ∴ Arc length of four circles without square portion = 4 \times 2 πr - 132 = 8 \times \frac{22}{7} \times 21 - 132 = 528 - 132 = 396 m Also, perimeter of square = 4 \times 42 = 168 m ∴ Required perimeter of the flower = 396 + 168 = 564 m$

Q 22 :

Pragati made a paper flower using 4 identical circles and a dotted square. The front view and back view of the flower is as shown below.

The diameter of each circle is the same as the length of the side of the square, 42 m.

Based on the above information, answer the following questions:

(ii) Find the area of the dotted region from the front view. Show your work.

$375 m^{2}$
$378 m^{2}$
$373 m^{2}$
$370 m^{2}$

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(2)

Area of dotted region (front view)

= Area of square - Area of four sectors

$= (42)^{2} - 4 \times \frac{90 °}{360 °} \times \frac{22}{7} \times (21)^{2} = 1764 - \frac{22}{7} \times 21 \times 21 = 1764 - 1386 = 378 m^{2}$

Q 23 :

Pragati made a paper flower using 4 identical circles and a dotted square. The front view and back view of the flower is as shown below.

The diameter of each circle is the same as the length of the side of the square, 42 m.

Based on the above information, answer the following questions:

(iii) (a) Is the area of the flower the same in the front and back view? Justify your answer with proper working. $(T a k e π = 22 / 7)$

View are equal
View are not equal
none of these
Only Front

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(1)

We have,

Area of the flower from front view

=Area of the four circles + Area of dotted region

$= 4 \times π r^{2} + 378 = 4 \times \frac{22}{7} \times (21)^{2} + 378 = \frac{88}{7} \times 21 \times 21 + 378 = 5544 + 378 = 5922 m^{2}$

$A r e a o f t h e f l o w e r f r o m b a c k v i e w$

$= 4 \times \frac{3}{4} \times Area of a circle + Area of a square = 3 \times {πr}^{2} + (42)^{2}$

$= 3 \times \frac{22}{7} \times 21 \times 21 + 1764 = 4158 + 1764 = 5922 m^{2}$

$Yes, area of the flower from front view and back view are equal .$

Q 24 :

Pragati made a paper flower using 4 identical circles and a dotted square. The front view and back view of the flower is as shown below.

The diameter of each circle is the same as the length of the side of the square, 42 m.

Based on the above information, answer the following questions:

(iii) (b) Find the total area of dotted and non-dotted region in back view.

$5920 m^{2}$
$5910 m^{2}$
$5820 m^{2}$
$5922 m^{2}$

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(4)

Total area of back view

=Area of square +4 × Area of one longer sector of circle

Each longer sector angle:

$(360 ° - 90 °) = 270 ° = (42)^{2} + 4 \times \frac{270 °}{360 °} \times \frac{22}{7} \times 21^{2} = 1764 + 4 \times 4158 = (1764 + 4158) = 5922 m^{2}$

Q 25 :

The diagram alongside shows a trapezium enclosed within a circle with a radius of 3 cm.

The length of AD is 5.5 cm and length of BC is 4 cm.

Based on the above information, answer the following questions:

(i) What is the length of RP?

$\sqrt{5} cm$
$\sqrt{4} cm$
$\sqrt{1} cm$
$\sqrt{2} cm$

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(1)

We have radius of the circle PC = 3 cm, r = 3 cm

$A l s o, A B C D i s a t r a p e z i u m s u c h t h a t A D ∥ B C a n d B C = 4 cm$

$PR bisects the chord BC of the circle because PR ⟂ BC .$

$BR = RC = 2 cm In right triangle ∆ PRC, we have : {PC}^{2} = {PR}^{2} + {RC}^{2}$

$(3)^{2} = {PR}^{2} + (2)^{2} 9 = {PR}^{2} + 4 {PR}^{2} = 5 \Rightarrow PR = \sqrt{5} cm$

Q 26 :

The diagram alongside shows a trapezium enclosed within a circle with a radius of 3 cm.

The length of AD is 5.5 cm and length of BC is 4 cm.

Based on the above information, answer the following questions:

(ii) Find area of trapezium ABCD.

$16.624 {cm}^{2}$
$16.615 {cm}^{2}$
$16.625 {cm}^{2}$
$15.625 {cm}^{2}$

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(3)

We have distance between parallel lines AD and BC = 3.5 cm, h = 3.5 cm

Also, $A D = 5.5 cm, B C = 4 cm Area of trapezium : = \frac{1}{2} \times (AD + BC) \times h = \frac{1}{2} \times (5.5 + 4) \times 3.5 = \frac{1}{2} \times 9.5 \times 3.5 = 16.625 {cm}^{2}$

Q 27 :

The diagram alongside shows a trapezium enclosed within a circle with a radius of 3 cm.

The length of AD is 5.5 cm and length of BC is 4 cm.

Based on the above information, answer the following questions:

(iii) The distance between AD and BC is 3.5 cm. What is the area of the circle outside the trapezium?

$(U s e π = \frac{22}{7})$

$11.60 {cm}^{2}$
$11.66 {cm}^{2}$
$11.35 {cm}^{2}$
$11.56 {cm}^{2}$

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(2)

Area of the circle:

$π \times (3)^{2} = \frac{22}{7} \times 9 = 28.285 c m^{2}$

$A r e a o f t h e c i r c l e o u t s i d e t h e t r a p e z i u m :$

$= Area of circle - Area of trapezium$

$= 28.285 - 16.625 = 11.66 {cm}^{2}$

Topic Question Set

Q 1 :

Raju installed a fence around a circular field, with the total cost amounting to Rs 6,000 at a rate of Rs 30 per m. He now plans to plough the field.

Based on the above information answer the following questions:

(i) What is the perimeter of the field?

Q 2 :

Raju installed a fence around a circular field, with the total cost amounting to Rs 6,000 at a rate of Rs 30 per m. He now plans to plough the field.

Based on the above information answer the following questions:

(ii) What is the radius of the field?

Q 3 :

Raju installed a fence around a circular field, with the total cost amounting to Rs 6,000 at a rate of Rs 30 per m. He now plans to plough the field.

Based on the above information answer the following questions:

(iii) (a) What is the area of the field?

Q 25 :

The diagram alongside shows a trapezium enclosed within a circle with a radius of 3 cm.

The length of AD is 5.5 cm and length of BC is 4 cm.

Based on the above information, answer the following questions:

(i) What is the length of RP?

Q 26 :

The diagram alongside shows a trapezium enclosed within a circle with a radius of 3 cm.

The length of AD is 5.5 cm and length of BC is 4 cm.

Based on the above information, answer the following questions:

(ii) Find area of trapezium ABCD.

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Topic Question Set

Q 1 : Raju installed a fence around a circular field, with the total cost amounting to Rs 6,000 at a rate of Rs 30 per m. He now plans to plough the field. Based on the above information answer the following questions: (i) What is the perimeter of the field?

Q 2 : Raju installed a fence around a circular field, with the total cost amounting to Rs 6,000 at a rate of Rs 30 per m. He now plans to plough the field. Based on the above information answer the following questions: (ii) What is the radius of the field?

Q 3 : Raju installed a fence around a circular field, with the total cost amounting to Rs 6,000 at a rate of Rs 30 per m. He now plans to plough the field. Based on the above information answer the following questions: (iii) (a) What is the area of the field?

Q 5 : Raju installed a fence around a circular field, with the total cost amounting to Rs 6,000 at a rate of Rs 30 per m. He now plans to plough the field. Based on the above information answer the following questions: (iii) (b) Find the cost of ploughing the field at the rate of Rs 0.50 per m².

Q 25 : The diagram alongside shows a trapezium enclosed within a circle with a radius of 3 cm. The length of AD is 5.5 cm and length of BC is 4 cm. Based on the above information, answer the following questions: (i) What is the length of RP?

Q 26 : The diagram alongside shows a trapezium enclosed within a circle with a radius of 3 cm. The length of AD is 5.5 cm and length of BC is 4 cm. Based on the above information, answer the following questions: (ii) Find area of trapezium ABCD.

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Q 1 :

Raju installed a fence around a circular field, with the total cost amounting to Rs 6,000 at a rate of Rs 30 per m. He now plans to plough the field.

Based on the above information answer the following questions:

(i) What is the perimeter of the field?

Q 2 :

Raju installed a fence around a circular field, with the total cost amounting to Rs 6,000 at a rate of Rs 30 per m. He now plans to plough the field.

Based on the above information answer the following questions:

(ii) What is the radius of the field?

Q 3 :

Raju installed a fence around a circular field, with the total cost amounting to Rs 6,000 at a rate of Rs 30 per m. He now plans to plough the field.

Based on the above information answer the following questions:

(iii) (a) What is the area of the field?

Q 25 :

The diagram alongside shows a trapezium enclosed within a circle with a radius of 3 cm.

The length of AD is 5.5 cm and length of BC is 4 cm.

Based on the above information, answer the following questions:

(i) What is the length of RP?

Q 26 :

The diagram alongside shows a trapezium enclosed within a circle with a radius of 3 cm.

The length of AD is 5.5 cm and length of BC is 4 cm.

Based on the above information, answer the following questions:

(ii) Find area of trapezium ABCD.