Question

In a right triangle with sides $a$ and $b$, hypotenuse $H$, the altitude $p$ drawn on hypotenuse is, then

• A. $a+b=H+p$
• B. $ab=H.p$
• C. $ap=bH$
• D. $a^2+b^2=H^2+p^2$

Answer 1

The correct option is B $ab=H.p$

In $\Delta ABC$

$a^2+b^2=H^2$

In $\Delta BCD$

$\sin (90-\theta )=\frac{P}{Q}$

$⟹\cos \theta =\frac{P}{Q}$

In $\Delta ACD$

$\sin \theta =\frac{p}{b}$

As we know that

${\sin }^{2}x+{\cos }^{2}x=1$

$\frac{P^2}{a^2}+\frac{P^2}{b^2}=1$

$P^2(\frac{1}{a^2}+\frac{1}{b^2})=1$

$⟹P^2=\frac{a^2{b}^2}{a^2+b^2}$

$⟹P^2=\frac{a^2{b}^2}{H^2}$

$⟹H^2{P}^2=a^2{b}^2$

$⟹PH=ab$