JAMB 1987 · UME · Q34

In the figure, $\angle XYZ = \angle YTZ = 90^\circ$ , $XT = 9$ cm and $TZ = 16$ cm. Find $YZ$ .

Worked solution (try it first)

The whole hypotenuse is $XZ = 9 + 16 = 25$ cm.
Triangles $YTZ$ and $XYZ$ are similar (they share the angle at $Z$ and each has a right angle), so $\dfrac{YZ}{XZ} = \dfrac{TZ}{YZ}$ , which gives $YZ^2 = TZ \times XZ$ .
So $YZ^2 = 16 \times 25 = 400$ .
So $YZ = 20$ cm, option B.