Question 112

The perimeter of a certain isosceles right triangle is 10 + $$10 \sqrt{2}\ $$ cm. What is the length of the hypotenuse of the triangle ?

Solution

ABC is an isosceles right angled triangle, where $$\angle$$ B = $$90^\circ$$

Let AB = BC = $$x$$

=> $$(AC)^2=x^2+x^2=2x^2$$

=> $$AC=\sqrt{2x^2}=\sqrt2x$$

Perimeter = $$x+x+\sqrt2x=10+10\sqrt2$$

=> $$2x+\sqrt2x=10+10\sqrt2$$

=> $$\sqrt2x(\sqrt2+1)=10(\sqrt2+1)$$

=> $$\sqrt2x=10$$

$$\therefore$$ Length of hypotenuse = 10 cm

=> Ans - (B)

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