Question

Using pythagorean theorem find a, and b, where a is twice of b and the hypotonuse is 5√5

Answer 1

The pythagoras theorem is $a^{2} + b^{2} = c^{2}$ where C is the hypotenuse. 

Since a is twice of b, 
a = 2b 

The hypotenuse is C = 5√5. 

Let's put this into the pythagoras equation

$( 2b)^{2} + b^{2} = (5 \sqrt{5})^{2}$

$5b^{2} = (125)$

$b^{2} = 25$

$b=5$

since we know a=2b, then a = 2(5) 
a=10$since we know a=2b, then a = 2(5)$

Answer 2

Measure of Hypotaneous = 5√5

And it's already given that b = 2a

By using pythagoras theorem:

a²+ b² = (5√5)²

Let's substitute for b as 2a in above equation we'll get:

a² + (2a)² = (5√5)²

a² + 4a² = 125
5a² = 125
a² = 25
a = √25 = 5


Therefore,
a= 5
and
b = 2a = 2(5) = 10.