Question

In a $△ABC,AB=AC=37$. Let $D$ be a point on $BC$ such that $BD=7,AD=33$. The length of $CD$ is

• A. 7
• B. 11
• C. 40
• D. not determinable

Answer 1

Answer: (C)

Given, $AB=AC=37$
$AD=33$
$BD=7$

$A{B}^2=A{E}^2+B{E}^2$ ...(i)
In $△ADE$,
$A{D}^2=A{E}^2+D{E}^2$ ...(ii)
$\Rightarrow A{B}^2-A{D}^2=B{E}^2-D{E}^2$
$\Rightarrow A{B}^2-A{D}^2=(BE+DE)(BE-DE)$
$\Rightarrow A{B}^2-A{D}^2=(CE+DE)(BD)[\because BE=CE]$
$\Rightarrow A{B}^2-A{D}^2=CD\cdot BD$
$\Rightarrow CD=\frac{A{B}^2-A{D}^2}{BD}$
$\Rightarrow CD=\frac{37^2-33^2}{7}$ [given]
$\Rightarrow CD=\frac{(37+33)(37-33)}{7}$
$\Rightarrow CD=\frac{70\times 4}{7}=40$