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Question 105
The perimeter of an isosceles, right angled triangle is 2p unit. The area of the same traingle is :
Solution
lets assume the sides to be (a,b,c) . (In isosceles a=b ; also as it is right angled $$c= a\times\sqrt{2}$$)(c is the hypotenuse)
a+b+c = 2p
a+a+√2 a = 2p
a = $$\frac{2p}{(2 + \sqrt{2})}$$
Now area of triangle (A) = $$\frac{1}{2} \times ab$$
A= $$(1/2) \times a^2$$
A= (1/2) $$\frac{2p}{(2 + \sqrt{2})^2}$$
= 2p² ⁄ (4 + 4√ 2 + 2)
= 2p² ⁄ (6 + 4√ 2)
= p² ⁄ (3 + 2√ 2)
A= $$(3 - 2 \sqrt{2} ) p^2$$ sq.unit
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