# Triangle problem and one more

• Sep 9th 2010, 02:30 PM
GrapeSoda
Triangle problem and one more
Attachment 18869Attachment 18867

I've been working on these for 2 hours, and still cant get the answer, thanks!!!

Nevermind the picture below, the problem is with the two problems in the link
• Sep 9th 2010, 04:59 PM
Wilmer
h(h+7) / 2 = 60
• Sep 9th 2010, 05:07 PM
GrapeSoda
Quote:

Originally Posted by Wilmer
h(h+7) / 2 = 60

which problem is that the answer to?

• Sep 9th 2010, 09:41 PM
Educated

We know that the area of a triangle is 1/2 * bh

Area = bh/2

Area = 20
h = h
b = h+6

20 = h*(h+6)/2
20 = (h^2 + 6h)/ 2

$\displaystyle 20 = \frac{1}{2}h^2 + 3h$

$\displaystyle 0 = \frac{1}{2}h^2 + 3h - 20$

$\displaystyle h = ...$

Use the same method to solve the other one that is similar.
If you know how to solve the third one, why not the first one? They are exactly the same. Unless you did trial and error to get the answer.
• Sep 9th 2010, 09:57 PM
Educated
For the second one, use the pythagorus theorum:

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

One leg is 2000ft long, b = 2000
The hypotenuse is 400ft longer than the other leg, a-400 = c

We substitute these known values into the equation and get:

$\displaystyle a^2 = 2000^2 + (a-400)^2$

$\displaystyle a^2 = 2000^2 + a^2 - 800a + 160000$

Now simplify and use the quadratic formula to solve for a, which is the length of the hypotenuse. (You might get 2 answers. In that case, use the one most appropriate)