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# Geometry

+2
346
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+1440

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Feb 15, 2020
edited by Dragan  Aug 14, 2020
edited by Dragan  Aug 14, 2020

#3
+1440
0

Can you solve it now?

Feb 19, 2020
#4
+112011
+2

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:

h(t+r+q)=56\\
h(t+r)+hq=56\\
h(t+r)+hq=56\\

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\\

Feb 19, 2020
edited by Melody  Feb 19, 2020
#5
+1440
0

What about this...?     56 / 4 = 14

Dragan  Feb 19, 2020
#6
+112011
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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.

Melody  Feb 19, 2020
#7
+1440
+1

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.

Dragan  Feb 19, 2020
#8
+112011
0

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.