Triangle and Semicircle

Geometry Level 2

In the image bellow $AC=CB=10\text{cm}$ , $AB=6\text{cm}$ and $AM=MB$ . The height $BH$ is tangent to the gray semi-circle with center in $M$ in the point $D$ . What is the radius of the gray semi-circle?

The answer is 0.9.

1 solution

A Former Brilliant Member
Dec 22, 2017

Relevant wiki: Cosine Rule (Law of Cosines)

By cosine rule, we have

$10^2=10^2+6^2-2(10)(6)(\cos A)$ $\color{#D61F06}\implies$ $A=\cos^{-1} 0.3$

$\angle ABH = 90 - \angle A = 90 - \cos^{-1} 0.3$

$\sin \angle ABH = \dfrac{MD}{3}$ $\color{#D61F06}\implies$ $MD=3[\sin (90-\cos^{-1}0.3)] = \boxed{0.9}$

Triangle and Semicircle

The answer is 0.9.

1 solution

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