Subtract Out The Hole

Geometry Level 1

Circles of radii 2 and 3 are externally tangent and are circumscribed by a third circle, as shown in the figure. Find the area of the shaded region.

6 π 6\pi 4 π 4\pi 9 π 9\pi 3 π 3\pi 12 π 12\pi

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Anish Harsha by Anish Harsha , 20, India

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2 solutions

Rishabh Jain
Feb 5, 2016

Radius of the biggest circle = A B 2 = 3 + 3 + 2 + 2 2 = 5 \frac{AB}{2}=\dfrac{3+3+2+2}{2}=5
Hence shaded area= Area of biggest circle - (Area of two smaller circles) \color{#302B94}{\text{Area of biggest circle - (Area of two smaller circles)}}
= π ( 5 ) 2 − ( π ( 3 ) 2 + π ( 2 ) 2 ) \pi (5)^2- (\pi(3)^2+\pi(2)^2) = 12 π \Large 12\pi

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Why you take AB as diameter of bigger circle?

Yashkrit Gupta - 5 years, 4 months ago

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because they are touching externally

Modo 942000 - 5 years, 4 months ago

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Can i get proof of this?

Yashkrit Gupta - 5 years, 4 months ago

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@Yashkrit Gupta – "Circles of radii 2 and 3 are externally tangent" copied from the text

Modo 942000 - 5 years, 3 months ago

Area of shaded region = Area of big circle − Area of two smaller circles. \color{#D61F06}{\text{Area of shaded region}}=\color{#3D99F6}{\text{Area of big circle}}-\color{#20A900}{\text{Area of two smaller circles.}}

Radius of big circle = 3 + 3 + 2 + 2 2 = 5 \color{#624F41}{\text{Radius of big circle}}=\frac{\color{magenta}{3}+\color{magenta}{3}+\color{#EC7300}{2}+\color{#EC7300}{2}}{2}=\color{grey}{5}

Now, Area of shaded region.
⇒ π ( 5 ) 2 − [ π ( 3 ) 2 + π ( 2 ) 2 ] = 12 π \Rightarrow \pi (\color{grey}{5})^2-[\pi (\color{magenta}{3})^2+\pi (\color{#EC7300}{2})^2]=\boxed{\color{#302B94}{12} \pi}

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