Not related to the Umbrella Corporation

Geometry Level 3

If the area of the shaded region equals 7 x cos x 7x - \cos x , what is the value of π x ? \frac \pi x ?


The answer is 6.

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3 solutions

The area of the shaded region

A = Area of the radius-2 semicircle 2 × Area of 1 3 of radius-1 circle Area of 1 6 of radius-1 circle 2 × Area of side length-1 equilateral triangle = 1 2 π ( 2 2 ) 2 × 1 3 π 1 6 π 2 × 1 2 ( 1 ) ( 3 2 ) = ( 2 2 3 1 6 ) π 3 2 = 7 π 6 3 2 = 7 π 6 cos ( π 6 ) \begin{aligned} A & = \text{Area of the radius-2 semicircle } \\ & \quad - 2\times \text{Area of } \frac{1}{3} \text{ of radius-1 circle } \\ & \quad - \text{Area of } \frac{1}{6} \text{ of radius-1 circle } \\ & \quad - 2\times \text{Area of side length-1 equilateral triangle } \\ & = \frac{1}{2} \pi \left(2^2 \right) - 2 \times \frac{1}{3} \pi - \frac{1}{6} \pi - 2 \times \frac{1}{2}\left(1 \right)\left(\frac{\sqrt{3}}{2} \right)\\ & = \left(2-\frac{2}{3}-\frac{1}{6} \right) \pi - \frac{\sqrt{3}}{2} = \frac{7\pi}{6} - \frac{\sqrt{3}}{2} = \frac{7\pi}{6} - \cos{\left( \frac{\pi}{6} \right)} \end{aligned}

x = π 6 π x = 6 \Rightarrow x = \dfrac{\pi}{6} \quad \Rightarrow \dfrac {\pi}{x} = \boxed{6}

Yay! Correct! Doesn't this look like an umbrella?

Pi Han Goh - 6 years, 1 month ago

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Yes, it does and it reminded me of the recent Hong Kong Umbrella Revolution. Sorry to be political.

Chew-Seong Cheong - 6 years, 1 month ago

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I solved it using integration.........

rajat kharbanda - 6 years, 1 month ago

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@Rajat Kharbanda Yes, post solution. I am not very fond of calculus and integration.

Chew-Seong Cheong - 6 years, 1 month ago

Lol! Umbrella Corp . haha

A Former Brilliant Member - 6 years, 1 month ago

truly a fantastic solution

rajat kharbanda - 6 years, 1 month ago

Sir , how did u get that the 2 parts of circle with radius 1cm occupies 1/3 area of the original circle i.e. third line of ur answer and also that part of circle that occupies 1/6 area of another circle(i.e. fourth line) ??

Chirayu Bhardwaj - 5 years, 5 months ago

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I have changed the figure. π 3 \frac{\pi}{3} and π 6 \frac{\pi}{6} indicate the one-third and one-sixth of a radius-1 circle respectively. Hope it helps.

Chew-Seong Cheong - 5 years, 5 months ago

Total required area = 2*Pink area=All area -2( green+blue+red) areas. R e q u i r e d a r e a = 2 π 2 ( 1 2 π 6 3 4 1 2 2 π 3 ) = { 7 π 6 3 2 } = 7 π 6 C o s π 6 = X C o s X π 6 = X . . S o π X = 6. \text{Total required area = 2*Pink area=All area -2( green+blue+red) areas.}\\ Required ~~area=2*\pi ~-~2*\left (~~\dfrac 1 2 *\dfrac \pi 6 ~-\dfrac{\sqrt 3}{4}-\dfrac 1 2 *\dfrac{2* \pi}{ 3} \right )\\ =\left \{~~\dfrac{7*\pi}{6}-\dfrac{\sqrt3}{2} \right \} =\dfrac{7*\pi}{6}-Cos{\dfrac{\pi}{6}}=X-Cos{X}\\\implies~\dfrac{\pi}{6}=X..So~~\dfrac{\pi}{X}=\huge 6.

Gabi Codrin
Oct 9, 2015

Solution by Rusu Bogdan.

  1. Thank you!

Bogdan Rusu - 5 years, 8 months ago

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