Question

An isosceles right triangle has an area of 8 cm square find the length of the hypotenuse

Answer 1

(a) Given, area of an isosceles right triangle = 8 cm2
Area of an isosceles triangle = 1/2 (Base x Height)
⇒  8 = 1/2 (Base x Base)
[∴ base = height, as triangle is an isosceles triangle]
⇒ (Base)2 =16 ⇒  Base= 4 cm



In ΔABC, using Pythagoras theorem

AC2 = AB2 + BC2 = 42 + 42 = 16 + 16
⇒ AC2 = 32 ⇒  AC = √32 cm
[taking positive square root because length is always positive]
Hence, the length of its hypotenuse is √32 cm.

Answer 2

$\underline{\Huge\mathfrak{Question-}}$

An isosceles right triangle has an area of 8 cm square. Find the length of the hypotenuse.

$\underline{\Huge\mathfrak{Answer-}}$

Here ,
Both the perpendicular and base are equal in length.

Than,
The Area of triangle = 1/2 x b x h

Let b = h = "X"

Now ,

$= > \frac{1}{2} \times x \times x = 8$

$= > x^{2} = 16$

$= > x = \frac{16}{4} = 4$

$= > x = 4$

So ,
Perpendicular = Base = 4 cm

Now ,
By Pythagoras theorem ;-

$= > 4^{2} + 4^{2} = \sqrt{16 + 16} = \sqrt{32}$

$= > 4 \sqrt{2} \: cm$