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  • An isosceles right triangle has area 8 cm^{2} . The length of its hypotenuse is (A) sqrt{32} cm (B) sqrt{16} cm (C) sqrt{48} cm (D) sqrt{24} cm

An isosceles right triangle has area 8\; cm^{^{2}} . The length of its hypotenuse is

(A) \sqrt{32}\; cm

(B) \sqrt{16}\; cm

(C) \sqrt{48}\; cm

(D) \sqrt{24}\; cm

Answers (1)

Solution

An isosceles right triangle is given.

According to definition of right triangle, one angle should be 90o

According to definition of isosceles triangle any two sides equal.

i.e., AB = BC

Suppose equal sides of triangle be = x cm

[AB = BC = x]

Area of isosceles triangle = \frac{1}{2} × base × height

\Rightarrow 8 cm2 = \frac{1}{2} × AB × BC

\Rightarrow 8 × 2 = x × x   where  [AB = BC = x]

\Rightarrow 16 = x^{2}

\Rightarrow x = \sqrt{16}

\Rightarrow x = 4 cm

So AB = BC = 4 cm

In \DeltaABC, using Pythagoras theorem

(AC)2 = (AB)2 + (BC)2

(AC)2 = (4)2 + (4)2

(AC)2 = 16 + 16

AC = \sqrt{32}\; cm

Hence hypotenuse of \DeltaABC is \sqrt{32}\; cm.

Hence option (A) is correct.

Posted by

infoexpert21

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