1. ## Perimeter Problem

AE = 5 cm = base of a triangle. The other 2 sides are equal to 4 cm. Show, by calculation, that FX = 3.12 cm (correct to 2 decimal places), where FX is the height of a triangle (and the angle is 90 degrees).

AE = 5 cm = base of a triangle. Show, by calculation, that FX = 3.12 cm (correct to 2 decimal places), where FX is the height of a triangle.

This is clearly incomplete. Can we have the whole question please?

3. ## Height of a Triangle

Well, I know the height of a triangle is found by: a^2 = b^2 + c^2.
How can I show, by calculation, that FX = 3.12 cm (correct to 2 decimal places), where FX is the height of a triangle?

4. So, if the base is 5 cm, and each of the other two sides are 4cm, then you would draw a vertical line from the centre of the base to the vertex. Then using Pythagorus you would have $\displaystyle c=\sqrt{4^2-2.5^2}$ which $\displaystyle \approx 3.12$

When $\displaystyle c$ = FX, $\displaystyle a$ = 4cm, and $\displaystyle b$ = 2.5cm (half the base).

5. Ah! How silly of me! I forgot to halve the 5!

6. Haha. An easy mistake to make