waecmaths question:
In the diagram $\angle QPR={{90}^{\circ }}$ , if ${{q}^{2}}=25-{{r}^{2}}$ find the value of p
Option A:
3
Option B:
4
Option C:
5
Option D:
6
waecmaths solution:
$\begin{align} & \text{Using pythagoras theorem} \\ & {{p}^{2}}={{r}^{2}}+{{q}^{2}} \\ & {{p}^{2}}=(25-{{r}^{2}})+{{r}^{2}} \\ & {{p}^{2}}=25 \\ & p=\sqrt{25}=5 \\\end{align}$
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