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Base ‘BC’ of an isosceles ∆ABC is 20 cm and equal sides is 26 cm, then the length of altitude through the vertex ‘A’ is

(A)23 cm (B)24 cm (C)25 cm (D)26 cm


Draw perpendicular AD In isosceles triangle ABC ,BD =CD=10 cm In right angled triangle ABD use Pythagoras theorem ,h²=p² +b², therefore AD(p)=√676-100=√576=24
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Draw a perpendicular from A on to BC.  It will meet BC at the center D of BC.  This can be show n  from the symmetry.

Apply Pythagoras formula in  triangle  ADB or ADC,
         26² = AD² + (BC/2)² =  AD² + 10²

           AD = 24 cm

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