Question

An isosceles right triangle has an area 8 cm^2 .The length of it's hypotenuse is
(a) √32 cm
(b) √16 cm
(c) √48cm
(d) √24cm​

Answer 1

Answer:

Hypotenuse of given isosceles triangle is root 32 cm

Step-by-step explanation:

Let the isosceles right triangle be triangle ABC

where angle B is 90° and base = height=x

Given area of triangle ABC= 8cm^2

so , area(ABC) = 8 cm^2

1/2 × base × height = 8cm^2

1/2 (x)(x) = 8

1/2 x^2 = 8

x^2 = 8×2

x^2 = 16

x = 4cm

so, base = height =4cm

Using Pythagoras Theorem,

(Hypotenuse)^2 = (Base)^2 + (Perpendicular)^2

AC ^2 = AB ^2 + BC ^2

AC ^2 = 4^2 + 4^2

AC ^2 = 16 + 16

AC ^2 = 32

AC = root 32 cm

Hence, Hypotenuse of given isosceles triangle is root 32 cm

Answer 2

$\blue{\bold{\underline{\underline{Answer:}}}}$

${\green{\therefore AC=\sqrt{32}\:cm}}$

$</p><p>\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{ \underline \mathfrak{Given : }} \\ \implies \text{Area \: of \: issoceles \: right \: triangle = 8 cm}^{2} \\ \\\red{ \underline \mathfrak{To \: Find : }} \\ \implies \text{Length \: of \: hypotenuse = ?}$

• According to given question :

$\text{Consider \: ABC \: be \: a \: right \: angled \: issoceles \: triangle} \\\circ \: \: \angle\text{B = 90}\degree \\\circ \: \: \: \text{AC = hypotenuse} \\\circ \: \: \: \text{AB = BC \: \: \: \: \: (issoceles \: triangle)} \\ \\\implies Area \: of \: triangle = \frac{1}{2} \times Base \times Height\\\\\implies 8 = \frac{1}{2} \times AB \times BC \\ \\ \implies 8 = \frac{1}{2} \times AB \times AB \: \: \: \: \: \: \: \: \: (AB = BC) \\ \\ \implies 8 \times 2 = {AB}^{2} \\ \\ \implies AB = \sqrt{16} \\ \\ \green{\implies \text{AB = 4 \: cm}} \\ \\ \text{In \: right \:angled } \triangle \: abc \\ \implies {h}^{2} = {p}^{2} + {b}^{2} \\ \\ \implies {h}^{2} = {4}^{2} + {4}^{2} \\ \\ \implies {h}^{2} = 16 + 16 \\ \\ \implies {h}^{2} = 32 \\ \\ \green{\implies h = \sqrt{32} \: cm }$