Question

In $△PQR,\angle PQR={90}^o,QS\perp PR$. If $PQ=a$, $QR=b$, $RP=c$ and $QS=p$, show that $pc=ab$.

Answer 1

It is given that $\angle PQR={90}^0$, $QS\perp PR$, $PQ=a$, $QR=b$, $RP=c$ and $QS=p$

Consider,

$P{Q}^2=RP\times PS\Rightarrow a^2=c\times PS\Rightarrow PS=\frac{a^2}{c}.......\left(1\right)$

Similarly, we have,

$Q{R}^2=RP\times RS\Rightarrow b^2=c\times RS\Rightarrow RS=\frac{b^2}{c}.......\left(2\right)$

And finally,

$Q{S}^2=RS\times PS\Rightarrow p^2=RS\times PS.......\left(3\right)$

Substitute equations $\left(1\right)$ and $\left(2\right)$ in equation $\left(3\right)$, we get

$Q{S}^2=RS\times PS{p}^2=\frac{b^2}{c}\times \frac{a^2}{c}\Rightarrow p^2=\frac{a^2{b}^2}{c^2}\Rightarrow p=\sqrt{\frac{a^2{b}^2}{c^2}}\Rightarrow p=\frac{ab}{c}\Rightarrow pc=ab$

Hence, $pc=ab$.