#### Question

In an isosceles triangle ABC, if AB = AC = 13 cm and the altitude from A on BC is 5 cm, find BC.

#### Solution

In ΔADB, by Pythagoras theorem

AD^{2} + BD^{2} = 13^{2}

⇒ 25 + BD^{2} = 169

⇒ BD^{2} = 169 − 2 = 144

⇒ BD = `sqrt144` = 12 cm

In ΔADB and ΔADC

∠ADB = ∠ADC [Each 90°]

AB = AC [Each 13 cm]

AD = AD [Common]

Then, ΔADB ≅ ΔADC [By RHS condition]

∴ BD = CD = 12 cm [By c.p.c.t]

Hence, BC = 12 + 12 = 24 cm

Is there an error in this question or solution?

#### APPEARS IN

Solution In an Isosceles Triangle Abc, If Ab = Ac = 13 Cm and the Altitude from a on Bc is 5 Cm, Find Bc. Concept: Application of Pythagoras Theorem in Acute Angle and Obtuse Angle.