The problem wants me to find x. I have attached a photo of the problem.

Question

asked Jul 17, 2023 209k views

1 Answer

Kernel · Answer 1 · 2023-07-22T04:13:01+0000

Solution

First we will label the triangle

In triangle ACD , we will use pythagoras theorem

$\begin{gathered} (hyp)^2=(opp)^2+(adj)^2 \\ x^2=4^2+y^2 \\ x^2=16+y^2 \end{gathered}$

In triangle ACB, Using pythagoras theorem

$\begin{gathered} (9+4)^2=x^2+z^2 \\ 13^2=x^2+z^2 \\ 169=x^2+z^2 \end{gathered}$

In triangle ABD , we will also use pythagoras theorem

$\begin{gathered} z^2=y^2+9^2 \\ z^2=y^2+81 \end{gathered}$

Now we have three equations

Let us find a replacement for y^2 in the first equation

From equ(3),

$\begin{gathered} x^2=16+y^2 \\ x^2=16+z^2-81 \\ x^2=z^2-65 \end{gathered}$

Now we will find a replacement for z^2 in the new equation

We will recall that

$\begin{gathered} 169=x^2+z^2 \\ 169-x^2=z^2 \\ z^2=169-x^2 \end{gathered}$

We will substitute for z^2 in x^2=z^2 - 65

$\begin{gathered} x^2=z^2-65 \\ x^2=169-x^2-65 \\ x^2+x^2=104 \\ 2x^2=104 \\ x^2=(104)/(2) \\ \\ x^2=52 \\ x=√(52) \\ \\ x=2√(13) \end{gathered}$