Maths 3 Marks Question

1. $\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 ​​​​​
2. $r = \frac { 35 } { 2 } m m , \theta = \frac { 360 ^ { \circ } } { 10 } = 36 ^ { \circ }$
   $\therefore $ The area of each sector of the brooch
   $= \frac { \theta } { 360 } \times \pi r ^ { 2 }$
   $= \frac { 36 } { 360 } \times \frac { 22 } { 7 } \times \frac { 35 } { 2 } \times \frac { 35 } { 2 } = \frac { 385 } { 4 } \mathrm { mm } ^ { 2 }$​​​​​​​

1. The area of that part of the field in which the horse can graze if the length of the rope is 5cm
   $= \frac { 1 } { 4 } \pi r ^ { 2 } = \frac { 1 } { 4 } \times 3.14 \times ( 5 ) ^ { 2 } = \frac { 1 } { 4 } \times 78.5 = 19.625 \mathrm { m } ^ { 2 }$
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
3. $= 78.5 - 19.625 = 58.875cm^2$

1. 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 $\triangle O A B$
   $= 78.5 \mathrm { cm } ^ { 2 } - 50 \mathrm { cm } ^ { 2 } = 28.5 \mathrm { cm } ^ { 2 }$
2. Area of major sector = $\pi x ^ { 2 } - 28.5$
   $= 3.14 \times 10 \times 10 - 28.5$
   $= 314 - 28.5 = 285.5 \mathrm { cm } ^ { 2 }$
   

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 = $\frac{1}{4} \pi r^2$
$= \frac {1 } { 4 } \times \frac { 22 } { 7 } \times \left( \frac { 7 } { 2 } \right) ^ { 2 }$
$= \frac { 1 } { 4 } \times \frac { 22 } { 7 } \times \frac { 7 } { 2 } \times \frac { 7 } { 2 } = \frac { 77 } { 8 } \mathrm { cm } ^ { 2 }$