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

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Question

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

Sum
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Solution

In ΔADB, by Pythagoras theorem

AD2 + BD2 = 132

⇒ 25 + BD2 = 169

⇒ BD2 = 169 − 25 = 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

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Chapter 7: Triangles - EXERCISE 7.6 [Page 7.97]

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R.D. Sharma Mathematics [English] Class 10
Chapter 7 Triangles
EXERCISE 7.6 | Q 9. | Page 7.97
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