Similar Right Triangles

[vampirewitchreine]

Hi guys, I need a bit of help getting started on this.



Find the value of each variable.


Extra information: In the big triangle, the altitude does not evenly divide the right angle to a 45 degree on each side (Lets call the big triangle ABC starting from the bottom left and going clockwise. Let's call the altitude segment BD.)

 

[Mrspi]

Hi guys, I need a bit of help getting started on this.


View attachment 1562
Find the value of each variable.


Extra information: In the big triangle, the altitude does not evenly divide the right angle to a 45 degree on each side (Lets call the big triangle ABC starting from the bottom left and going clockwise. Let's call the altitude segment BD.)

When you draw an altitude to the hypotenuse in a right triangle, you create two smaller right triangles, which are similar to each other, and to the original BIG right triangle. The way you have described naming the triangles, we can say that

triangle ABC ~ triangle ADB ~ triangle BDC

You can verify this similarity....the three angles in each triangle are congruent to the three angles each of the other two triangles, and you can identify the pairs of congruent angles by looking at the NAMES of the triangles.

Looking at the two smaller triangles, we see that

triangle ADB ~ triangle BDC

If two triangles are similar, each pair of corresponding sides have the same ratio. So,

AD/BD = AB/BC = DB/DC

Substitute the numbers(or expressions) you are given, and we can fill in two fractions:

AD/BD = DB/DC
2/x = x/8

There! A proportion you can solve for x!

I'll BET you have several theorems which relate to this situation:

If an altitude is drawn to the hypotenuse in a right triangle, then
1) the altitude is the geometric mean between the two parts of the hypotenuse
2) each LEG of the original triangle is the geometric mean between the whole hypotenuse and the part of the hypotenuse which adjoins that leg.

(in your diagram, labeled as you've stipulated), we have AC/AB = AB/AD, and AC/BC = BC/DC)

 

[pka]

Hi guys, I need a bit of help getting started on this.


View attachment 1562
Find the value of each variable.


Extra information: In the big triangle, the altitude does not evenly divide the right angle to a 45 degree on each side (Lets call the big triangle ABC starting from the bottom left and going clockwise. Let's call the altitude segment BD.)

The value of \(\displaystyle x\) is called the mean proportional.
That is \(\displaystyle \frac{2}{x}=\frac{x}{8}\).
That is easily shown using similar triangles.

 

[vampirewitchreine]

When you draw an altitude to the hypotenuse in a right triangle, you create two smaller right triangles, which are similar to each other, and to the original BIG right triangle. The way you have described naming the triangles, we can say that

triangle ABC ~ triangle ADB ~ triangle BDC

You can verify this similarity....the three angles in each triangle are congruent to the three angles each of the other two triangles, and you can identify the pairs of congruent angles by looking at the NAMES of the triangles.

Looking at the two smaller triangles, we see that

triangle ADB ~ triangle BDC

If two triangles are similar, each pair of corresponding sides have the same ratio. So,

AD/BD = AB/BC = DB/DC

Substitute the numbers(or expressions) you are given, and we can fill in two fractions:

AD/BD = DB/DC
2/x = x/8

There! A proportion you can solve for x!

I'll BET you have several theorems which relate to this situation:

If an altitude is drawn to the hypotenuse in a right triangle, then
1) the altitude is the geometric mean between the two parts of the hypotenuse
2) each LEG of the original triangle is the geometric mean between the whole hypotenuse and the part of the hypotenuse which adjoins that leg.

(in your diagram, labeled as you've stipulated), we have AC/AB = AB/AD, and AC/BC = BC/DC)

The value of \(\displaystyle x\) is called the mean proportional.
That is \(\displaystyle \frac{2}{x}=\frac{x}{8}\).
That is easily shown using similar triangles.

Okay, since you've both given me the same proportion to use, I have
\(\displaystyle \frac{2}{x}=\frac{x}{8}\)
Which becomes \(\displaystyle x^2=16\)
Then I get \(\displaystyle \sqrt {x^2}= \sqrt {16}\)
Which is \(\displaystyle x=4\)




Also, I have a similar problem.... Should I solve this one like Pythagorean Theorem?

(Black was the info that the textbook provided, red was what I've already got worked out)

I got the altitude by working it out like the previous problem that you helped me with:
\(\displaystyle \frac{5}{a}=\frac{a}{20}\)
\(\displaystyle a^2=100\)
\(\displaystyle \sqrt {a^2}= \sqrt {100}\)
\(\displaystyle a=10\)

I got the hypotenuse of the larger triangle just by adding the measurements used for the other smaller triangles (20+5=25)

Now, how do I begin to solve for x and y?

 

[vampirewitchreine]

You must be tired Hazel; x and y are hypotenuses, and you have the other sides...
so phone up Pete Pythagoras

Yeah, I was really tired.....


Anyway, I managed to get x to be the \(\displaystyle \sqrt{125}\) \(\displaystyle or ~\approx 11.18\) and y to be \(\displaystyle \sqrt{500}\) \(\displaystyle or ~\approx 22.36\)



Staying on the same lines of similar triangles (I know that I keep posting a lot of different problems in this thread, but I've got, like, 20 of them and it's just a handful of them that have me stumped. I apologize if this is annoying or against the forum rules to keep doing it like this.)

Maybe it's the lack of other given answers that's stumping me here, but this one is throwing me through a bit of a loop.