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.

10th Grade Math

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