# Application Problems on Area of a Circle

We will discuss here about the Application problems on Area of a circle.

1. The minute hand of a clock is 7 cm long. Find the area traced out by the minute hand of the clock between 4.15 PM to 4.35 PM on a day.

Solution:

The angle through which the minute hand rotates in 20 minutes (i.e., 4:35 PM – 4:15 PM) is $$\frac{20}{60}$$ × 360°, i.e., 120°

Therefore, the required area = The area of the sector of central angle 120°

= $$\frac{θ}{360}$$ × πr2

= $$\frac{120}{360}$$ × $$\frac{22}{7}$$  × 72 cm2, [Since, θ = 120, r = 7 cm]

= $$\frac{1}{3}$$ × 22 × 7 cm2.

= $$\frac{154}{3}$$ cm2.

= 51$$\frac{1}{3}$$ cm2.

2. The cross section of a tunnel is in the shape of a semicircle surmounted on the longer side of a rectangle whose shorter side measures 6 m. If the perimeter of the cross section is 66 m, find the breadth and height of the tunnel.

Solution:

Let the radius of the secicircle be r m.

Then, the perimeter of the cross section

= PQ + QR +PS + Semicircle STR

= (2r + 6 + 6 + πr) m

= (2r + 12 + $$\frac{22}{7}$$ r) m

= (12 + 2r + $$\frac{22}{7}$$ r) m

= (12 + $$\frac{36}{7}$$ r) m

Therefore, 66m = (12 + $$\frac{36}{7}$$ r) m

⟹ 66 = 12 + $$\frac{36}{7}$$ r

⟹ 12 + $$\frac{36}{7}$$ r = 66

⟹ $$\frac{36}{7}$$ r = 66 - 12

⟹ $$\frac{36}{7}$$ r = 54

⟹ r = 54 × $$\frac{7}{36}$$

⟹ r = $$\frac{21}{2}$$.

Therefore, PQ = Breadth of the tunnel = 2r m = 2 × $$\frac{21}{2}$$ = 21 m.

And height of the tunnel = r m + 6 m

= $$\frac{21}{2}$$ m + 6 m

= $$\frac{21}{2}$$ m + 6 m

= $$\frac{33}{2}$$ m

= 16.5 m.