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.
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