Question

In $\Delta ABC$, $\angle ACB={90}^{\circ }$, AB = C units, BC = a units, AC = b units, CD is perpendicular to AB and CD = p units. Then $\frac{1}{p^2}=$

• A. Scalene
• B. Isosceles but not right angled
• C. Equilateral
• D. Right Ðd & isosceles

Answer 1

The correct option is C

Equilateral

$\angle ACB={90}^{\circ }$

$A{B}^2=A{C}^2+B{C}^2$

$\Rightarrow c^2=b^2+a^2$ ....... (1)

Area $\left(\Delta ABC\right)=\frac{1}{2}\times AB\times CD$

$=\frac{1}{2}\times BC\times AC$

$=\frac{1}{2}\times c\times p=\frac{1}{2}\times a\times b$

$\Rightarrow C=\frac{ab}{p}$

Put this in (1), we get

$\frac{a^2{b}^2}{p^2}=b^2+a^2$

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

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

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