hope you can help. My son has done a 50 question review for a test and cannot figure out one problem. I have tried, but keep coming up with the wrong asnwer as well.
here is the question; find the area of a right triangle, whose sides have lengths x cm, (3x+3) cm, and (3x+4) cm.
we started by drawing the traingle.
Here is what we have tried.
1/2 bh using all three of the sides above as base and height.
balancing individual equations so that one side ='s 0. (i.e 3x+3 = 0, 3x=-3, x=-1 etc.)
inserting a number for x (such as "2") and seeing if that works.
searched on line and through the text.
using the a squared + b squared = c squared.
The answer given is 84cm squared...
we have no idea what else to try.
can you get us started? Obviously there is something we are missing.
Thanks!

Yep, 84

It's really a few problems bundled as one. The first is solve for x, knowing you have a right triangle. Then find the sides and finally the area.

using a^2 + b^2 = c^2 and you know c will be (3x+4) because it's bigger than x or 3x+3

Substituting and solving x^2 + (3x+3)^2 = (3x+4)^2 you get

(X-7) (X+1) = 0

(X+1)=0 cannot be a solution, so (X-7)=0 means 7 is the value of x or the base.

Since 1/2 b * h is area and the base x and the height has to be (3x+3) or 24 or the height.

A = 1/2 * 7 * 24 or 7 * 12 or 84.

If the lenghts were all in cm, then 84 sqcm

Note: It's best to carry the units throughout all the calculations. Units should be correct too.