Finding x,y,z of a triangle

Question

My younger brother was asked to do this problem and was given the problem exactly as the problem was given with no additional information.

Basically, we are trying to find the values of x,y,and z but we are currently having a hard time. So the first thing we tried to do is set up proportions but realized that none of the three triangles can be proved similar by AA, SAS or SSS.

Then I thought about using the law of sines and here is what i did:

I am using triangles ACD and BCD

$$\frac{AC}{sin 90}=\frac{DC}{sin 90}$$

but AC=9 so, $$\frac{9}{sin 90}=\frac{DC}{sin 90}$$

which means $z=9$ and then here is where the problem occurs:

I decided to use Pythagorean theorem to find y but I get y=0. So now, I'm here to seek help. Any ideas?

Answer 1

You might start by recognizing that $ABD$ and $ADC$ are similar triangles, because they share the angles $\angle DAC$ and $90$ degrees. Therefore, $$\frac{y}{4} = \frac{z}{x} = \frac{9}{y}$$ We can rearrange $y/4 = 9/y$ to obtain $y^2 = 36$, so $y = 6$, which allows us to answer (b). Then by the Pythagorean theorem, $$4^2 + x^2 = y^2 = 36$$ so $x^2 = 20$, which means that $x = \sqrt{20}$, which allows us to answer (a). Finally, rearranging the first equation above gives us $$z = \frac{xy}{4} = \frac{6\sqrt{20}}{4}$$ which allows us to answer (c).

Answer 2

You can use just the Pythagorean theorem. There are three right triangles in your picture given by the three right angles (notice that $\angle ABD$ is right). With those three triangles we get the following equations. \begin{cases} z^2 = 25 + x^2 \\ y^2 = 16 + x^2 \\ 81 = z^2 + y^2 \end{cases} Substituting the first two into the third, you can solve for $x$ and get $x = \sqrt{20} = 2\sqrt{5}$. Then you can substitute this value of $x$ into the first two equations to solve for $y$ and $z$. I get that $y = \sqrt{36} = 6$ and that $z = \sqrt{45} = 3\sqrt{5}$.

Answer 3

$\Delta ADC, \Delta ABD, \Delta CBD$ are all right triangles clearly. Also, $x$ is the geometric mean of $AB$ and $BC$ because $DB$ is an altitude of $\Delta ADC$. That means that $x^2=4 \cdot 5$ so $x=\sqrt{20}=2\sqrt5$. A is then $4\sqrt5$. You can then calculate y by using Pythagorean Theorem. $y =\sqrt{4^2+(2\sqrt5)^2}=\sqrt{36}=6$ and $\frac{y}{2}=\frac{6}{2}=3$. Now for c we can use Pythagorean Theorem again. $z=\sqrt{5^2+(2\sqrt5)^2}=\sqrt{45}=3\sqrt5$ and $z+8=3\sqrt5+8$.