Question

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

Answer 1

SOLUTION :  

Given :  

AB = AC = 25 cm and BC = 14 cm

In ∆ABD and ∆ACD,

∠ADB = ∠ADC   [Each 90°]

AB = AC      [Each 25 cm]

AD = AD      [Common]

∆ABD ≅ ∆ACD     [By RHS condition]

∴ BD = CD = 7 cm     [By c.p.c.t]

In ∆ADB,

AB² = AD² + BD²

[By  using Pythagoras theorem ]

25² = AD² + 7²

AD² = 625 − 49  

AD² = 576

AD = √576  

AD = 24 cm  

Hence, the altitude AD on BC  is 24 cm.

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Answer 2

Answer:

Step-by-step explanation:

Given :-

AB = AC = 25 cm and BC = 14 cm

To Find :-

In this triangle altitude from A to BC is AD, we know that isosceles altitude on non equal sides also median.

BD = CD = 14/2 = 7 cm

In ∆ADB,

By applying Pythagoras theorem, we get

AB² = AD² + BD²

⇒  25² = AD² + 7²

⇒  AD² = 625 − 49  

⇒  AD² = 576

⇒  AD = √576  

⇒  AD = 24 cm  

Hence, the altitude AD on BC  is 24 cm.