Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • 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)

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                      [\Theta 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 DABC is \sqrt{32}\; cm.

Hence option (A) is correct.

Posted by

infoexpert21

View full answer

Similar Questions

Latest Question