Question

I am floored. For the love of sweet Jesus someone please help me.

The question is to try and find the largest possible area of a right triangle that has a 28 cm long hypotenuse.

Units are square centimeters.

I've tried a really hard to figure it out but I keep getting it wrong. Please help!

Answer 1

The area of a right triangle of legs a,b is

A= - ab 2

The hypotenuse is c² = a² + b²

We also know that if a,b are both real value then the square of is difference is non-negative: (a−b)²≥0. (Strictly positive unless a=b.)

If we solve: (a−b)² = a²+2ab+b²,  where 2ab=4A. So we can replace and

$0\leq{(a-b)^{2}}$

$0 \leq {a^{2}}+2ab+b^{2}$

$0 \leq {c^{2}}-4A$

4A<

$A\leq\frac{1}{4}c^{2}$

So we already have this identity: $A\leq\frac{1}{4}c^{2}$

In this problem c=28 so our answer is:

So our answer is:

$A\leq\frac{1}{4}28^{2} = \frac{1}{2} * 784 = 196$

$A\leq196$

Largest possible area = 196 square centimetres.