### Solved Examples

**Question 1: **A square is inscribed in a circle of radius 5 cm. Calculate the area of the shaded region shown in the figure.

** Solution: **

We are given that the radius of the circle = 5 cm

Therefore, the diameter = 2 r

= 2 ( 5 )

= 10 cm

Area of the shaded region = Area of the circle - Area of the square

Area of the Circle = $\pi$ $r^{2}$

= $\pi$ 5^{2}

= 25 $\pi$

Let us assume that the side of the square = a cm

and the diagonal ( diameter of the circle ) of the square = d cm

Since each angle of a square measure 90^{o} ,

According to Pythagoras Theorem,

a^{2} + a^{2} = d^{2} = 10^{2}

=> 2 a^{2} = 100

=> a^{2} = $\frac{100}{2}$

= 50

Therefore, the area of the square = 50 cm^{2}

Area of the shaded region = 25 $\pi$ - 50

= 25 x 3.14 - 50

= 78.5 - 50

= 28.5 cm^{2}

**Question 2: **A wire is bent in the form of a square encloses an area of 484 cm

^{2}. The same wire is bent in the form of a circle. Find the area enclosed by the circle.

** Solution: **

We are given that the area of the square = 484 cm^{2}

Therefore, side of the square = $\sqrt{484}$

= 22 cm

Perimeter of the square = 4 a

= 4 x 22

= 88 cm

The length of 88 cm is bent to form a circle of radius r cm

Therefore, Circumference of the circle = Perimeter of the square

=> 2 $\pi$ r = 88

=> $\pi$ r = $\frac{88}{2}$

= 44

=> 3.14 r = 44

=> r = $\frac{44}{\pi}$

Area of the circle = $\pi$ r^{2}

= $\pi$ $\left (\frac{44}{\pi } \right )^{2}$

= $\pi$ $\frac{44\times 44}{\pi \times \pi }$

= $\frac{44\times 44}{\pi}$

= $\frac{1936}{\pi}$

= 616.25 cm^{2}

**Therefore, the Area of the circle = 616.25 cm**^{2}

**Question 3: **The diameter of a circular park is 60 m. A 3m wide road runs on the
outside around it. Calculate the cost of constructing the road at $\$$ 7
per square meter.

** Solution: **

We are given that the Diameter of the park = 60 m

Therefore, the radius of the park = $\frac{60}{2}$

= 30 m

Therefore, inner

Width of the road = 3 m

Therefore, the outer radius of the circle ( R ) = 30 + 3 = 33 m

Area of the road = Area of the outer circle - Area of inner circle

= $\pi$ R^{2} - $\pi$ r^{2}

= $\pi$ ( R^{2} - r^{2} ) = $\pi$ ( 33^{2} - 30^{2} )

= $\pi$ ( 33 + 30 ) ( 33 - 30 )

= $\pi$ ( 63 ) ( 3 ) = 593.76 m^{2}

Cost of constructing the road per sq. m = $\$$ 7

Cost of constructing the road of area 593.76 m^{2} = 593.76 x 7

= $\$$ 4156.32

Therefore, **cost of constructing the road = $\$$ 4156.32 **

**Question 4: **TRUE is a square of side 20 cm with centers T, R, U and E. Four
circles are drawn such that each circle touches externally two of the
remaining three circles. Find the area of the shaded region.

** Solution: **

We are given that the side of the square = 20 cm

Let us observe the diagram closely to find the area of the shaded region.

The shaded region is inside the square TRUE and the square has $\frac{1}{4}$ circle( quadrants ) at each corner.

Therefore area of the shaded region = Area of the square - Area of the four quadrants

Area of the square = 20^{2}

= 20 x 20

= 400 cm^{2}

Radius of each circle = $\frac{20}{2}$

= 10 cm

Area of the 4 quadrants = 4 x ( $\frac{1}{4}$ $\pi$ r^{2} )

= $\pi$ r^{2}

= $\pi$ 10^{2}

= 3.14 x 100

= 314 cm^{2}

** Area of the shaded region = 400 - 314**

= 86 cm^{2}