A peculiar Area

Geometry Level 4

In a $\triangle ABC$ , $a = 11$ and $\sin A = \dfrac{3}{7}$ where $a$ is the side opposite $\angle A$ and $0 < A < \dfrac{\pi}{2}$ . If the side lengths of $\triangle ABC$ are $11,b$ and $c$ , where $b$ is the largest possible side of $\triangle ABC$ , find the area of $\triangle ABC$ . If the area of this triangle is of the form $\dfrac{k\sqrt{m}}{n},$ where $k$ and $n$ are prime numbers, find $k+m+n.$

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The answer is 1224.

1 solution

Mark Hennings
Jan 25, 2017

The side $b$ is maximised when $b$ is a diameter of the circumcircle of $ABC$ . Thus $b \; = \; 2R \; = \; \frac{a}{\sin A} \; = \; \frac{77}{3}$ and $ABC$ is right-angled, with $\angle B = 90^\circ$ . By Pythagoras, $c = \frac{22\sqrt{10}}{3}$ , and so the triangle has area $\tfrac12 \times 11 \times \frac{22\sqrt{10}}{3} \; = \; \frac{121\sqrt{10}}{3} \; = \; \frac{11\sqrt{1210}}{3}$ making the answer $1210 + 11 + 3 = \boxed{1224}$ .

A peculiar Area

Try a similar problem here.

The answer is 1224.

1 solution

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