Question

if the perimeter of isosceles right angles triangle is root 2+1 find the area

Answer 1

let a , a and b be the sides of isosceles right angle triangle

in isosceles right Triangle ABC,

base = perpendicular= a, hypotenuse
= b

by Pythagoras theorem,

b^2 = (a)^2 + (a)^2

b^2 = 2a^2 => b = √2a

if perimeter = (√2 + 1) units

2a + b = √2 + 1

2a + √2a = √2 + 1

a( 2 +√2) = (√2 + 1)

a = (√2+1)/ (2 +√2)

a = (√2 +1) ×(2 -√2)/ (2 +√2)×(2 - √2)

a = (2√2 + 2 - 2 - √2)/ (4 - 2)

a = √2/2 units

a =base=perpendicular= √2/2units

area = (base × perpendicular)/ 2

= (√2/2 ) × (√2/2) / 2

= 2/ 4 /2 = (2/ 4) × (1/ 2)

= 2/ 8= 1/ 4 units^2

Answer: area = 1/ 4 units^2

Answer 2

Pythagoras theorem

$a^2 + b^2 = c^2$

Given that it is a isosceles triangle, both legs are the equal

$a^2 + a^2 = c^2$

$2a^2 = c^2$

$c = \sqrt{2a^2}$

$c = \sqrt{2}a$

Solve a:

Perimeter = side 1 + side 2 + side 3

$a + a + \sqrt{2}a = \sqrt{2}+1$

$2a + \sqrt{2}a = \sqrt{2}+ 1$

$a(2 + \sqrt{2}) = \sqrt{2}+ 1$

$a =\dfrac{\sqrt{2}+ 1}{2 +\sqrt{2}}$

Find the area:

$\text {area = } \dfrac{1}{2} \times \text{base} \times \text{height}$

$\text {area = } \dfrac{1}{2} \times \dfrac{\sqrt{2}+ 1}{2 + \sqrt{2}} \times \dfrac{\sqrt{2}+ 1}{2 + \sqrt{2}}$

$\text {area = } \dfrac{(\sqrt{2}+ 1) (\sqrt{2} + 1)}{2(2 + \sqrt{2})(2 + \sqrt{2})}$

$\text {area = } \dfrac{2 + 2\sqrt{2} + 1 }{2(4 + 4\sqrt{2} + 2)}$

$\text {area = } \dfrac{3 + 2\sqrt{2} }{2(6 + 4\sqrt{2})}$

$\text {area = } \dfrac{3 + 2\sqrt{2} }{4(3 + 2\sqrt{2})}$

$\text {area = } \dfrac{1}{4} \text{ units}^2$