Thread: Solving Lengths & Areas

1. Solving Lengths & Areas

I have a right-angled isosceles triangle with hypotenuse c = 100mm. I'm being asked to calculate:
1. length of AB in mm
2. area of triangle ABD in sq mm
Workings including answer would be appreciated. I've spent too much time reading up on Pythagorean Theorem, and getting way too confused.
Cheers

2. Since it's an isoceles triangle, AB = AD. So, by Pythagoras' Theorem:
$AB^{2} + BD^{2} = c^{2}$
$AB^{2} + AB^{2} = 100^{2}$
$2AB^{2} = 100^{2}$
$AB^{2} = \frac{100^{2}}{2}$
etc. etc.

Area of a triangle is given by $\frac{1}{2}(\text{base} \times \text{height})$. Here, your height and base are the same length so:
$A = \frac{1}{2}(AB)(BD) = \frac{1}{2}(AB)(AB) = ...$