math help please?

Question

The hypotenuse of a right triangle is twice as long as one of the legs and 7 inches longer than the other. What are the lengths of the sides of the triangle?

Answer 1

For the answer, I will convert square roots (SR) into decimals to do the computations as you didn't say if you needed to answer with SR or not....

You have 3 sides: a, b and c (hypotenuse)

and you have 3 equations:
1) c = a + 7
2) c = 2b
3) (a^2) + (b^2) = (c^2)

Combining eq's #1 & #2 we get 2b = a + 7 ----> a = 2b - 7

Plugging this equation and #2 into #3 we get:
(2b -7)^2 + b^2 = (2b)^2

Expanding this gives:
4(b^2) - 28b +49 + b^2 = 4(b^2)

Simplifying gives:
b^2 - 28b + 49 = 0

Using the quadratic equation gives two possible solutions for b.
b = 26.1245 or b = 1.8755

Using equation #2 from above means:
c = 52.249 or c = 3.751

Since equation #1 says that c = 7 + a, this means that "c" cannot equal 3.751 (since "a" would be negative),

so your 3 sides are:
a = 45.249, b = 26.1245 and c= 52.249

(Sorry this answer was lengthy)

Answer 2

The hypotenuse of a right triangle is twice as long as one of the legs → 30-60-90.

short leg = x
Hypotenuse = 2x
long leg = (sqrt3)x

So 2x = 7 + (sqrt3)x. Solve for x.

Answer 3

Call the length of the legs, A and B. Lets call the length of the hypotenuse, H.

Then we have:
A^2 + B^2 = H^2
H = 2A
H = B + 7

Therefore,
A = H /2
B = H -7

and then
(H/2)^2 + (H-7)^2 = H^2
which is
(0.25) (H^2) - 14H + 49 = 0

Solve using the quadratic formula
H = [14 +/- sqrt (14^2 - 4x0.25x49)] / 0.5

H = 52.24871131 or 3.751288694
Going back and solving for A and B, we see that one of these doesn't work, so it must be the other one.

Answer 4

Use the pythagorean theorem:

Let the hypotenuse length be 2x
Then the legs measure x and 2x-7

so x^2 + (2x-7)^2 = (2x)^2

x^2 + 4x^2 - 28x + 49 = 4x^2
x^2 - 28x + 49 = 0

Using the quadratic formula, x= 7rt(3) + 14

So the lengths are:

x = 7rt(3)+14
2x - 7 = 14rt(3)+21,
and 2x = 14rt(3)+28

Answer 5

Try this website,
http://www.webmath.com/
they give you the answer, but you might want to check with your teacher so you can figure these types of problems out for yourself, it is what they get paid for.