In an equilateral triangle ABC, AD is drawn perpendicular to BC meeting BC in A. If $(A D)^{2} =$ x $(B D)^{2}$ . Find $x$ .

Open in App

Solution

$∵ E q u i l a t e r a l t r i a n g l e,$
$⟹ ∠ A = ∠ B = ∠ C = 60^{\circ}$
$I n △ A B D :$
$⟹ tan (∠ B) = \frac{A D}{B D}$
$⟹ tan (60) = \frac{A D}{B D}$
$⟹ \sqrt{3} = \frac{A D}{B D}$
$s q u a r i n g$
${A D}^{2} = 3 {B D}^{2}$
$∴ x = 3$

Suggest Corrections

Similar questions

△

⊥

{A D}^{2} = 3 {B D}^{2}

If in an equilateral triangle $△ A B C, A D$ is perpendicular to $B C,$ then $A D^{2} =$

If $△ A B C$ is an equilateral triangle such that $A D$ is perpendicular to $B C$ , then $A D^{2} =$

Join BYJU'S Learning Program