Triangles in Semicircle

Treating the area of the triangle as 1/2 X base X height,
we know that the maximum base is the entire diameter, which has length 2,
and the maximum height is to the top of the semicircle, which is the entire radius of length 1.

Hence, the maximum area is $\frac{1}{2} \times 2 \times 1$ , which is achieved by the triangle with vertices $(-1, 0), (0, 1), (1, 0)$ .

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Now, the reason for creating this problem, is that it allows us to easily solve the following question

What is the area of the largest rectangle that we can inscribe in a circle of radius 1?

It is very well-known that the largest rectangle is a square. The goal is to obtain a proof that follows as a corollary of my question. If you want a hint of how to prove that, check out the related problem .

Let $AP = a$ , $BP=b$ and $\angle PAB = \theta$ . We know that angle extended in a semicircle $\angle APB = \frac \pi 2$ . Then the area of the triangle inscribed in the semicircle is given by $A = \frac 12 ab$ $= \frac 12 (2\cos \theta)(2 \sin \theta)$ $= \sin 2\theta$ . $A$ is maximum, when $\sin 2\theta$ is maximum, that is $\sin 2\theta = 1$ , and $A_{max} = \boxed{1}$ .

Here,

A= .5* base* height

A = .5 * 2 * 1

A=1.00

Base = 2 * radius

Height = radius