the length of isoceles triangle is 10cm if its altitude is 1cm shorter than its equal sides find area of triangle

**Given, **the base of an isosceles triangle = BC = 10 cm

Let the equal sides AB and AC of isosceles triangle be x cm.

∴ AB = AC = x cm

Now, altitude of triangle AD = (x - 1) cm

We know that, in an isosceles triangle altitude from the opposite vertex will always bisect the base.

So, BD = DC =

Now, In Δ ABD by pythagoras theorem,

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

⇒ x^{2} = (x-1)^{2}+ 5^{2}

⇒ x^{2} = x^{2}+ 1 - 2x + 25

⇒ -26 = -2x

⇒ x = 13 cm

So, AD = x - 1 = 13 - 1 = 12 cm

Now area of ABC =

=

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