+118687 Hi Dragan 
Area of ABC = 56cm^2 Find the sum of the areas of the 2 side triangles.
\(0.5(t+r+q)*2h=56\qquad (1a)\\ h(t+r+q)=56\\ h(t+r)+hq=56\\ h(t+r)+hq=56\\ hq=56-h(t+r)\qquad (1b)\\~\\ \)
\(0.5(t+r)h+qh+0.5qh=56\\ 0.5(t+r)h+1.5qh=56\\ (t+r)h+3qh=112\\ 3qh=112-(t+r)h\\ \)
\(3[56-h(t+r)]=112-(t+r)h\\ 168-3h(t+r)=112-(t+r)h\\ (t+r)h-3h(t+r)=-56\\ -2h(t+r)=-56\\ h(t+r)=+28\\ \frac{1}{2}(t+r)h=14\\ \)
Sum of the areas of the 2 little side triangles is 14cm^2
Coding:
0.5(t+r+q)*2h=56\qquad (1a)\\
h(t+r+q)=56\\
h(t+r)+hq=56\\
h(t+r)+hq=56\\
hq=56-h(t+r)\qquad (1b)\\~\\
0.5(t+r)h+qh+0.5qh=56\\
0.5(t+r)h+1.5qh=56\\
(t+r)h+3qh=112\\
3qh=112-(t+r)h\\
3[56-h(t+r)]=112-(t+r)h\\
168-3h(t+r)=112-(t+r)h\\
(t+r)h-3h(t+r)=-56\\
-2h(t+r)=-56\\
h(t+r)=+28\\
\frac{1}{2}(t+r)h=14\\
+118687 Yes, that works too.
Did you actually want that question answered or was it just put there as a fun challenge question?
If it was just a fun challenge question that you already knew how to do then you should have said so from the very beginning.
+1490 No, I did not need the answer. I'm on this forum for fun, and to keep my mind occupied.
But, if there's a problem, I wont post the questions anymore. 
+118687 You can post questions, just spell out what they are for.
Challenge questions that you know the answer to are good, just don't be deceptive about them.
Spell out that it is a challenge and you already know the answer.
In this instance it might also have been nice if you had spelt out that the answer was simple, requiring no algebra.
I always tell the kids that they need to give us as much info as they have if they want us to help them.
If they say an answer is wrong they should always state why they believe this. That goes for you too.
I answered your question thinking I was helping you.
I would not have answered if I had understood it was a not a 'real' question.
However, it was nice to see your neat little answer. Using congruence and similar shapes is always interesting.
