3 Marks Question

Question 13 Marks

A brooch is made with silver wire in the form of a circle with diameter $35\ mm.$ The wire is also used in making $5$ diameters which divide the circle into $10$ equal sectors as shown in figure. Find:

the total length of the silver wire required.
the area of each sector of the brooch.

Answer

$\because $ Diameter $= 35\ mm$
$\therefore $ Radius = $\frac { 35 } { 2 } \mathrm { mm }$
$\therefore $ Circum ference = $2\pi r$
$= 2 \times \frac { 22 } { 7 } \times \frac { 35 } { 2 } = 110\ mm ..... (1)$
Length of $5$ diameters
$= 35 \times 5 = 175\ mm ...... ( 2 )$
$\therefore $ The total length of the silver wire required
$= 110 + 175 = 285\ mm $
$r = \frac { 35 } { 2 } m m , \theta = \frac { 360 ^ { \circ } } { 10 } = 36 ^ { \circ }$
$\therefore $ The area of each sector of the brooch
$= \frac { \theta } { 360 } \times \pi r ^ { 2 }$
$= \frac { 36 } { 360 } \times \frac { 22 } { 7 } \times \frac { 35 } { 2 } \times \frac { 35 } { 2 } = \frac { 385 } { 4 } \mathrm { mm } ^ { 2 }$

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Question 23 Marks

A horse is tied to a peg at one corner of a square shaped grass field of side $15\ m$ by means of a $5\ m$ long rope. Find

the area of that part of the field in which the horse can graze.
the increase in the grazing area if the rope were $10\ m$ long instead of $5\ m ($Use $\pi = 3.14)$

Answer

The area of that part of the field in which the horse can graze if the length of the rope is $5\ cm$
$= \frac { 1 } { 4 } \pi r ^ { 2 } = \frac { 1 } { 4 } \times 3.14 \times ( 5 ) ^ { 2 } = \frac { 1 } { 4 } \times 78.5 = 19.625 \mathrm { m } ^ { 2 }$
The area of that part of the field in which the horse can graze if the length of the rope is $10\ m$
$= \frac { 1 } { 4 } \pi r ^ { 2 } = \frac { 1 } { 4 } \times 3.14 \times ( 10 ) ^ { 2 } = 78.5 \mathrm { m } ^ { 2 }$
$\therefore $ The increase in the grazing area
$= 78.5 - 19.625 = 58.875\ cm^2$

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Question 33 Marks

A chord of a circle of radius $10\ cm$ subtends a right angle at the centre. Find the area of the corresponding:

minor segment
major sector.

Answer

$r = 10\ cm, \theta = 90 ^ { \circ }$
Area of minor sector $ = \frac { \theta } { 360 } \times \pi r ^ { 2 }$
$= \frac { 90 } { 360 } \times 3.14 \times 10 \times 10 = 78.5 \mathrm { cm } ^ { 2 }$
Area of $\triangle O A B = \frac { O A \times O B } { 2 }$
$= \frac { 10 \times 10 } { 2 } = 50 \mathrm { cm } ^ { 2 }$
$\therefore $ Area of the minor segment
$=$ Area of minor sector $-$ Area of $\triangle O A B$
$= 78.5 \mathrm { cm } ^ { 2 } - 50 \mathrm { cm } ^ { 2 } = 28.5 \mathrm { cm } ^ { 2 }$
Area of major sector $= \pi x ^ { 2 } - 28.5$
$= 3.14 \times 10 \times 10 - 28.5$
$= 314 - 28.5 = 285.5 \mathrm { cm } ^ { 2 }$

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Question 43 Marks

Find the area of a quadrant of a circle whose circumference is $22\ cm.$

Answer

Let the radius of the circle be $r \ cm.$
Then, circumference of the circle $= 2\pi r \ cm$
According to the question,
$2\pi r = 22$
$\Rightarrow 2 \times \frac { 22 } { 7 } \times \mathrm { r } = 22$
$\Rightarrow \mathrm { r } = \frac { 22 \times 7 } { 2 \times 22 } \Rightarrow \mathrm { r } = \frac { 7 } { 2 } \mathrm { cm }$
For a quadrant of a circle,
Area $ =\frac{1}{4} \pi r^2$
$= \frac {1 } { 4 } \times \frac { 22 } { 7 } \times \left( \frac { 7 } { 2 } \right) ^ { 2 }$
$= \frac { 1 } { 4 } \times \frac { 22 } { 7 } \times \frac { 7 } { 2 } \times \frac { 7 } { 2 } = \frac { 77 } { 8 } \mathrm { cm } ^ { 2 }$

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Question 53 Marks

Find the area of the sector of a circle with radius $4 \ cm$ and of angle $30^\circ .$ Also, find the area of the corresponding major sector.$($use $ \pi =3.14 )$

Answer

Here, $\theta = 30^\circ $ and $r = 4 \ cm$
Area of sector $OAPB =\frac { \theta } { 360 } \times \pi r ^ { 2 }$
Let $'A'$ be the area of corresponding major sector.
Then, $A =$ Area of sector $OAQB$
$\Rightarrow A =$ Area of the circle $-$ Area of the corresponding minor sector
$\Rightarrow A = \pi r ^ { 2 } - \frac { \theta } { 360 } \times \pi r ^ { 2 }$
$\Rightarrow A = \pi r ^ { 2 } \left( 1 - \frac { \theta } { 360 } \right)$
$\Rightarrow A = 3.14 \times 4 \times 4 \left( 1 - \frac { 30 } { 360 } \right) \mathrm { cm } ^ { 2 }$
$\Rightarrow A = 3.14 \times 4 \times 4 \times \frac { 11 } { 12 } \mathrm { cm } ^ { 2 }$$= \frac { 3.14 \times 44 } { 3 } \mathrm { cm } ^ { 2 }= 46.05 cm^2$

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