#### Question

In an isosceles triangle ABC, AB = AC = 25 cm, BC = 14 cm. Calculate the altitude from A on BC.

#### Solution

We have

AB = AC = 25 cm and BC = 14 cm

In ΔABD and ΔACD

∠ADB = ∠ADC [Each 90°]

AB = AC [Each 25 cm]

AD = AD [Common]

Then, ΔABD ≅ ΔACD [By RHS condition]

∴ BD = CD = 7 cm [By c.p.c.t]

In ΔADB, by Pythagoras theorem

AD^{2} + BD^{2} = AB^{2}

⇒ AD^{2} + 7^{2} = 25^{2}

⇒ AD^{2} = 625 − 49 = 576

⇒ AD = `sqrt576` = 24 cm

Is there an error in this question or solution?

#### APPEARS IN

Solution In an Isosceles Triangle Abc, Ab = Ac = 25 Cm, Bc = 14 Cm. Calculate the Altitude from a on Bc. Concept: Application of Pythagoras Theorem in Acute Angle and Obtuse Angle.