# help to determine sideshelp to determine sides of triangle hypotenuse=20cm sides: x and 2x

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Given the sides of a right angle triangle are"

x, 2x, and the hypotenuse (h) = 20.

But we know that: h^2 = side^2 + side^2

==> 20^2 = x^2 + (2x)^2

==> 400 = x^2 + 4x^2

==> 5x^2 = 400

Now we will divide by 5.

==> x^2 = 80.

==> x = sqrt80 = 4sqrt5.

==> 2x = 2*4sqrt5 = 8sqrt5.

**Then the sides are 4sqrt5 and 8sqrt5**

Another way would be to consider the angle opposite the side with length x.

The sine of this angle sin x = x / 20

The cosine of this angle is 2x/ 20

Now use the fundamental realtion (sin x)^2 + (cos x)^2 = 1

=> (x / 20)^2 + (2x/ 20)^2 = 1

=> x^2 / 20^2 + 4x^2 / 20^2 = 1

=> 5x^2 = 20^2

=> x^2 = 20^2 / 5

=> x = sqrt 80

Therefore the sides are sqrt 80 and 2* sqrt 80.

By the Pythagorean Theorem, we know that the sum of the squares of the legs equals the square of the hypotenuse. So we square all of the terms that you have given.

We get

x^2 + 4x^2 = 400

5x^2 = 400

x^2 = 80

And so x is the square root of 80 (that's one side) and the other side is twice that.

I agree to previous post that advice you to use Pythagorean theorem to determine the sides.

x^2 + 4x^2 = 400

5x^2 = 400

We'll divide by 5:

x^2 = 80

We'll calculate square root both sides:

x = sqrt 80

x = 4 sqrt 5

The negative value x = -4sqrt 5 will be rejected since a length of a side cannot be negative.

So, one leg is 4sqrt5 and the other leg is 8sqrt5.