Given the triangle is right-angled, find the value of a.

Question

I am doing my schoolwork, and I can't figure out how to do this problem. I can't find anything related to it by searching it up, and the lesson name is just called 'Problem Solving', so I have no title to search it up with. How do you do this?

The way I have been doing it is I apply the Pythagorean Theorem, and I get this: $$(a-2)^2 + (a+1)^2 = (a+4)^2$$ I then turn all of them into quadratic equations: $$(a^2 - 4a + 4) + (a^2 + 2a + 2) = (a^2 + 16a + 16)$$ This is where I have a problem. In the lesson, the first one is just $a^2$. Because of this, they don't show me how I should combine all of the equations together. In nearly every problem, I end up getting the same value on both sides of the equation, like this: $$a^2 - 30 + 30 = 0$$

Yes, but I am still very confused as to how to use it with this problem. — Wannabe Programmer, Sep 09 '21 at 16:04
I am still very confused. Can someone walk me through how to do this? — Wannabe Programmer, Sep 09 '21 at 16:30
@WannabeProgrammer Did you try to apply the Pythagorean theorem ? This is the first step... — TheSilverDoe, Sep 09 '21 at 16:31
I did, but I don't know how I should combine all the quadratic equations together when there are three of them. — Wannabe Programmer, Sep 09 '21 at 16:33
@WannabeProgrammer What equation did you obtain ? You should add your attempt to your question, otherwise we cannot help you. — TheSilverDoe, Sep 09 '21 at 16:34
In your equation, just put together the terms containing $a^2$, the terms with $a$, and the constant terms. It gives you a quadratic equation you should be able to solve. (and be careful when you develop $(a+1)^2$) — TheSilverDoe, Sep 09 '21 at 16:42
That is what I have been doing? I combine the ones on the left side, then I move over the ones of the right side and add them in, but with reversed signs. It gives me an unsolvable equation. — Wannabe Programmer, Sep 09 '21 at 16:44
@WannabeProgrammer Can you write what is the "unsolvable equation" ? — TheSilverDoe, Sep 09 '21 at 16:45
Just like the example I showed in my edit. a^2 - 30 + 30 = 0. How do I solve that? — Wannabe Programmer, Sep 09 '21 at 16:46

Vasili · Accepted Answer · 2021-09-09T16:56:53.727

0

You have some mistakes opening a binomial, you should get $$a^2-4a+4+a^2+2a+1=a^2+8a+16$$ or $a^2-10a-11=0 \implies a=11 \ (a=-1 \text{ does not make sense}) $
In general, $(x+y)^2=(x+y)\cdot(x+y)=x^2+2xy+y^2$

edited Sep 09 '21 at 16:56

answered Sep 09 '21 at 16:49

Vasili

10,690

score 0 · Answer 2 · edited Sep 10 '21 at 16:27

By applying pythagoras theorem in the given triangle, we get:

$(a+4)^2 = (a+1)^2 + (a-2)^2$.

Solving using the identity :

$(a+b)^2 = a^2 + b^2 + 2ab$

$\implies \not{a^2} + 16 + 8x = \not{a^2} + 1 + 2a + a^2 + 4 - 4a$.

$\implies 16 + 8a=a^2 - 2a + 5$.

$\implies 0 = a^2 - 2a + 5 - 16 - 8a$.

$\implies 0 = a^2 - 10a -11$.

$\implies 0 = a^2 - 11a + a -11$.

$\implies 0 = a(a - 11) + 1(a -11)$.

$\implies 0 = (a+1) ( a-11)$.

$\implies a = -1 , 11$.

Here, $a$ can not be equal to $-1$ because side cannot be equal to $0$ (if we substitute $a= -1$ in side).

Therefore,

$\boxed{\implies a = 11}$

Given the triangle is right-angled, find the value of a.

2 Answers2