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 = $2\pi r$
$= 2 \times \frac { 22 } { 7 } \times \frac { 35 } { 2 } = 110mm ..... (1)$
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
$= 78.5 - 19.625 = 58.875cm^2$

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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 $= 2 \pi rcm$
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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Question 63 Marks

A chord of a circle having length 10 cm subtends a right angle at the centre. Find the area of the corresponding:
(i) Minor segment
(ii) Major sector
(Use $\pi=3.14$ )

Answer

Let the chord be AB, and the center of the circle be O. The length of the chord is 10 cm, and it subtends a right angle at the center, so $\angle A O B=90^{\circ}$.. In the right-angled triangle AOB, the sides OA and OB are the radii (r) of the circle.
Using the Pythagorean theorem for $\triangle A O B:$
$\begin{array}{l}O A^2+O B^2=A B^2 \\ r^2+r^2=10^2 \\ 2 r^2=100 \\ r^2=50 \\ r=\sqrt{50}=5 \sqrt{2} cm\end{array}$
(i) The area of the minor segment is the area of the sector OAB minus the area of the triangle OAB.
Area of Sector OAB:
$\begin{array}{l}\text { Area }=\frac{\theta}{360^{\circ}} \times \pi r^2=\frac{ g }{360^{\circ}} \times 3.14 \times 50 \\ \text { Area }=\frac{1}{4} \times 3.14 \times 50=3.14 \times 12.5=39.25 sq . cm \end{array}$
Area of Triangle OAB:
$\begin{array}{l}\text { Area }=\frac{1}{2} \times \text { base } \times \text { height }=\frac{1}{2} \times O A \times O B=\frac{1}{2} \times r \times r=\frac{1}{2} r^2 \\ \text { Area }=\frac{1}{2} \times 50=25 \text { sq.cm }\end{array}$
Area of Minor Segment:
Area of Sector - Area of Triangle = 39.25−25=14.25 sq.cm
(ii) The area of the major sector is the area of the circle minus the area of the minor sector.
Area$=\pi r^2=3.14 \times 50=157 sq . cm$
Area of Circle - Area of Minor Sector = 157−39.25=117.75 sq.cm

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