Question

Find the length of altitude AD of an isosceles triangle ABC in which AB = AC = 2a units and BC = a units.​

Answer 1

✬ Altitude = a√15/2 units ✬

Step-by-step explanation:

Given:

• An isosceles triangle ABC.
• Two sides are equal i.e AB = AC = 2a.
• Length of BC is a units.

To Find:

• What is the length of altitude AD ?

Solution: Let the length of altitude be x units. We have

• AB = AC = 2a (hypotenuse)

• AD = x (perpendicular)

• BC = a

• BD = DC = a/2 (base)

Now applying Pythagoras Theorem in ∆ADB

★ Hypotenuse² = P² + B² ★

$\implies{\rm }$ AB² = AD² + BD²

$\implies{\rm }$ (2a)² = x² + (a/2)²

$\implies{\rm }$ 4a² = x² + a²/4

$\implies{\rm }$ 4a² – a²/4 = x²

$\implies{\rm }$ 16a² – a²/4 = x²

$\implies{\rm }$ √15a²/4 = x

$\implies{\rm }$ a√15/2 = x

Hence, the length of altitude of isosceles triangle ABC is a√15/2 units.

Answer 2

Answer:

Please refer the attachment for the diagram.

Question :-

• Find the length of altitude AD of an isosceles triangle ABC in which AB = AC = 2a units and BC = a units..

Find Out :-

• Find the length of altitude AD of an isosceles triangle.

Given in the question :

• An isosceles triangle ABC.
• Two sides are equal i.e AB = AC = 2a.
• Length of BC is a units.

Solution :-

Let the length of altitude be x units.

• AB = AC (Perpendicular) = 2a
• AD = (perpendicular) = x
• BC = a
• BD = DC = (base) = $\dfrac{a}{2}$

Now applying Pythagoras Theorem in ∆ADB

$\red{ \boxed{\sf{(Hypotenuse)^2 = (Perpendicular)^2 + (Base)^2}}}$

$\curlyeqsucc$ AB² = AD² + BD²

$\curlyeqsucc$ 2a² = x² + $\sf \bigg(\dfrac{a}{2}\bigg)^2$

$\curlyeqsucc$ 4a² = x² + $\sf\dfrac{a^2}{4}$

$\curlyeqsucc$ 4a² – $\sf\dfrac{a^2}{4}$ = x²

$\curlyeqsucc$ 16a² – $\sf\dfrac{a^2}{4}$ = x²

$\curlyeqsucc$ $\sf \sqrt{\dfrac{15a^2}{4}} = x$

$\curlyeqsucc$ $\sf \dfrac{a\sqrt{15}}{2} = x$

Therefore, the length of altitude of isosceles triangle ABC is a√15/2 units.