Question

An iscoceles right angled triangle has an area of 8 cm square. Find the hypotenuse by Heron's formula.....
Please.

Answer 1

AnswEr :

Given that,

• Area of the traingle is 8 cm²

• The triangle is a right iscoceles triangle

Let the similar sides of the triangle be 'a'

The hypotenuse in terms of side would be :

$\sf \: h {}^{2} = {a}^{2} + {a}^{2} \\ \\ \longmapsto \: \sf \: h = a \sqrt{2}$

The sides of the traingle are a,a and a√2

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If the semi perimeter of the triangle is s :

$\sf \: \longmapsto \: s = \dfrac{2a + a \sqrt{2} }{2} \\ \\ \sf \: \longmapsto \: s = \dfrac{a(2 + \sqrt{2} )}{2}$

Applying the heron's formula,

$\sf \: Area = \sqrt{s(s - x)(s - y)(s - z)} \\ \\ \longrightarrow \: \sf \: 8 = \sqrt{s(s - a)(s - a)(s - a \sqrt{2} )} \\ \\ \longrightarrow \: \sf \: 64 = s(s - a) {}^{2} (s - a \sqrt{2} ) \\ \\ \longrightarrow \: \sf \: 64 = \dfrac{a(2 + \sqrt{2} )}{2} \times \bigg( \dfrac{a(2 + \sqrt{2}) }{2} - a \bigg) {}^{2} \times ( \dfrac{a(2 + \sqrt{2} )}{2} -a \sqrt{2} ) \\ \\ \longrightarrow \: \sf \: 64 = \dfrac{a(2 + \sqrt{2}) }{2} \times \bigg( \dfrac{2a + a \sqrt{2} - 2a}{2} \bigg) {}^{2} \times ( \dfrac{2a + a \sqrt{2} - 2a \sqrt{2} }{2} ) \\ \\ \longrightarrow \: \sf \: 64 = \dfrac{a(2 + \sqrt{2}) }{2} \times ( \dfrac{a \sqrt{2} }{2} ) {}^{2} \times \dfrac{a(2 - \sqrt{2} )}{2} \\ \\ \longrightarrow \sf \: 64 \times 8 = a {}^{4} (2 + \sqrt{2} )(2 - \sqrt{2} ) \\ \\ \longrightarrow \sf \: 64 \times 8 = 2{a}^{4} \\ \\ \longrightarrow \: \sf \: {a}^{4} = 256 \\ \\ \longrightarrow \: \boxed{ \boxed{ \sf a = 4 \: cm}}$

Now,

The hypotenuse of the triangle is of the form a√2

Thus,

Hypotenuse of the triangle is  4√2 cm

Answer 2

Given: An isosceles right-angled triangle having an area of 8 cm².

To find: The hypotenuse.

Answer:

Let's assume one of the triangle's sides to be 'a' cm.

Since it's isosceles, another one of its sides will be 'a' cm as well.

Given that it's a right-angled triangle, let's find the third side using the Pythagorus formula.

Hypotenuse² = Base² + Height²

Hypotenuse² = a² + a²

Hypotenuse² = 2a²

Hypotenuse = √2a²

Hypotenuse = √2 a cm.

Since we now have all it's sides, let's use Heron's formula to find the value of each side.

$\bf Heron's\ formula:\ \sf \sqrt{s(s\ -\ x)(s\ -\ y)(s\ -\ z)}$

Where 's', is the semi-perimeter and x, y, and z are the sides of the triangle.

s = (a + a + √2 a)/2

s = (2a + √2 a)/2

$\sf s\ =\ \dfrac{2a\ +\ \sqrt{2}\ a}{2}$

Now, from the values we have,

• s = (2a + √2 a)/2
• x = a
• y = a
• z = √2 a

Using them in the formula,

$\sf Area\ [8]\ =\ \sqrt{\bigg(\dfrac{2a\ +\ \sqrt{2}\ a}{2}\bigg)\ \times\ \bigg(\dfrac{2a\ +\ \sqrt{2}\ a}{2}\ -\ a\bigg)^2\ \times\ \bigg(\dfrac{2a\ +\ \sqrt{2}\ a}{2}\ -\ \sqrt{2}\ a\bigg)}$

Squaring both sides,

$\sf 64\ =\ \bigg(\dfrac{2a\ +\ \sqrt{2}\ a}{2}\bigg)\ \times\ \ \bigg(\dfrac{2a\ +\ \sqrt{2}\ a\ -\ 2a}{2}\bigg)^2\ \times\ \bigg(\dfrac{2a\ +\ \sqrt{2}\ a\ -\ 2\sqrt{2}\ a}{2}\bigg)$

$\sf 64\ =\ \bigg(\dfrac{2a\ +\ \sqrt{2}\ a}{2}\bigg)\ \times\ \ \bigg(\dfrac{\sqrt{2a}}{2}\bigg)^2\ \times\ \bigg(\dfrac{2a\ -\sqrt{2}\ a}{2}\bigg)\\\\\\\\64\ =\ \dfrac{\bigg(2a\ +\ \sqrt{2}\ a\bigg)\ \times\ \bigg(\sqrt{2}\ a\bigg)^2\ \times\ \bigg(2a\ -\ \sqrt{2}\ a\bigg)}{8}\\\\\\\\64\ \times\ 8\ =\ \bigg(2a\ +\ \sqrt{2}\ a\bigg)\ \times\ \bigg(\sqrt{2}\ a\bigg)^2\ \times\ \bigg(2a\ -\ \sqrt{2}\ a\bigg)\\\\\\512\ =\ 2a^4\\\\\\\dfrac{512}{2}\ =\ a^4\\\\\\256\ =\ a^4\\\\\\\bf a\ =\ 4\ cm$

Since a = 4 cm, its sides are 4 cm, 4 cm and 4√2 cm [hypotenuse].