JAMB 1985 · UME · Q33

In $\triangle XYZ$ , $\angle XKZ = 90^\circ$ , $XK = 15$ cm, $XZ = 25$ cm and $YK = 8$ cm. Find the area of $\triangle XYZ$ .

Worked solution (try it first)

In the right-angled triangle $XKZ$ , by Pythagoras, $KZ = \sqrt{25^2 - 15^2}$ , which is $\sqrt{400} = 20$ cm.
The base of $\triangle XYZ$ is $YZ = 8 + 20 = 28$ cm, and its height is $XK = 15$ cm.
Area $= \frac12 \times 28 \times 15 = 210$ sq. cm, option B.