# Math Help - A problem on triangles (area, ratios)

1. Hm.. I don't know about that, but I once had a drill exercise about those and the the class came to this conclusion when it concerns similar polygons.

That very principle also applies to volumes.

Say 2 circles of radius 2 and 5 respectively. The ratio of length is 2:5, that for the areas is 4:25 and that for the volumes is 8:125

2. That would help a lot in the coming exam. Thank you for imparting that knowledge!

3. Okay, to come to my solution about it, I used variables instead of giving specific values.

I let HM = x, this means that AH = 2x (due to similar triangles and I know that 2AM = AB) and using Pythagoras, I get AM = $x\sqrt{5}$

From this, we get AB = $2x\sqrt{5}$ and HB = 4x.

From there, I can get angles BAH and HAP which are respectively $\sin^{-1}\left(\frac{4x}{2x\sqrt{5}}\right)$ and $\sin^{-1}\left(\frac{x}{x\sqrt{5}}\right)$ respectively.

In degrees those give 63.43 and 26.57 degrees (I kept all the other numbers to retain the accuracy as much as possible).

And then I get angles ABP = 45, ACP = 45, APB = 180-(45+63.43) = 71.57, APC = 180-(45+26.57) = 108.43 degrees.

Since the sides don't matter now, I put AB = 2 cm and AH = 1 cm. And from those and using the angles I just found, using the sine rule, I get side BP = 2.12 cm and PC = 0.943 cm. The ratio then turns out to be 9:4

4. Originally Posted by Unknown008
Okay, to come to my solution about it, I used variables instead of giving specific values.

I let HM = x, this means that AH = 2x (due to similar triangles and I know that 2AM = AB) and using Pythagoras, I get AM = $x\sqrt{5}$

From this, we get AB = $2x\sqrt{5}$ and HB = 4x.

From there, I can get angles BAH and HAP which are respectively $\sin^{-1}\left(\frac{4x}{2x\sqrt{5}}\right)$ and $\sin^{-1}\left(\frac{x}{x\sqrt{5}}\right)$ respectively.

In degrees those give 63.43 and 26.57 degrees (I kept all the other numbers to retain the accuracy as much as possible).

And then I get angles ABP = 45, ACP = 45, APB = 180-(45+63.43) = 71.57, APC = 180-(45+26.57) = 108.43 degrees.

Since the sides don't matter now, I put AB = 2 cm and AH = 1 cm. And from those and using the angles I just found, using the sine rule, I get side BP = 2.12 cm and PC = 0.943 cm. The ratio then turns out to be 9:4
Thank you very much Unknown 008
But the answer sheet says otherwise. The ratio is actually BP:PC = 2:1. And I double checked this using some software too.

5. Oh, I did something wrong earlier. I get BP:PC = 2:1 now. I must have typed the wrong values in my calculator

6. I got your solution! Thanks a lot man! Unfortunately I'll have to find another way that uses less trigonometry since we aren't supposed to use calculators. But I think i know how now thanks to your approach!

7. I'd be glad to know how you did it. I must have overseen something, as always

8. ## a problem on triangles

Hello again gundanium,
Here is my geometry-algebra method

AB =6 ( any number could be used)
From P draw a perpendicular to AC meeting it @T
BC=6rad2 BP=y PC =6rad2-y PT=6rad2-y/2
TC =( 6rad2-y/2) AT =6-(6rad2-y/2)
AH =b BH =a both b and a are numbers found in my stage 1 post(after calcs)
PH=x

Write two equations in x and y and solve for x and y

bjh

9. Hm... yes, and from your method, I found yet a shorter method!

One slight thing I saw, is that PT = $\frac{6\sqrt2 - y}{\sqrt2}$

Anyway, triangles ABC and CPT are similar, meaning that the ratio BP:PC = ratio of AT:TC

Once you know TC (and you alreayd know AC), you can find AT and then the ratio.

10. Dear unknown008,
Thanks for the rad2 correction but I don't see any simple way to find PT which is equal TC.Can you show how?

bjh

11. You agree that angle PTC is 90 degrees and hence, triangle PTC is another isoseles right angled triangle, similar to triangle ABC.

From that, we know the length of PC, the length of PT = TC = PC/rad2 which is what you got and I corrected in my post.

And from there, yu use another ratio trick to get the ratio of BP:PC

12. I know everything you said,In addition triangle APT is similar to ABM ABH AHM In order to solve for PT we need one of the sides of APT to be defined. Do you agree? I don't see one

13. Originally Posted by bjhopper
Hello again gundanium,
Here is my geometry-algebra method

AB =6 ( any number could be used)
From P draw a perpendicular to AC meeting it @T
BC=6rad2 BP=y PC =6rad2-y PT=6rad2-y/2
TC =( 6rad2-y/2) AT =6-(6rad2-y/2)
AH =b BH =a both b and a are numbers found in my stage 1 post(after calcs)
PH=x

Write two equations in x and y and solve for x and y

bjh
Thanks but a few questions..

Why is PT = 6rad2-y/2?
Shouldn't it be 3rad(4-2y) by using the pythagorean theorem?
sin(45) = 1/2 = PT/6rad2-y = 3rad(4-2y)

IGNORE THIS sTUPID POST *Embarrassed* T.T

14. I was able to solve it by solving all the sides of triangle BHP by assuming values. But I needed a calculator to use sine law since it didn't give 'nice' angles (30, 45, 60). But I won't be able to do that in my exam since I can't use a calculator.

15. Originally Posted by bjhopper
I know everything you said,In addition triangle APT is similar to ABM ABH AHM In order to solve for PT we need one of the sides of APT to be defined. Do you agree? I don't see one
I think he used the PC = 6rad2-y and sin 45 to solve for PT. Though you need to leave y as a variable to make an equation to solve simultaneously to the equation with an x later on...

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