Triangle ratio problem

**Patrick_John** · June 30th 2007, 09:37 PM

Let $\bigtriangleup ABC$ be a right angled triangle such that $\angle A =90$ ° $, AB =AC$ and let $M$ be the mid point of the side AC. Take the point P on the side BC so that AP is vertical to BM. Let H be the intersection point of AP and BM.

(a) Find the ratio of the areas of the two triangles $\bigtriangleup ABH : \bigtriangleup AHM$ .

(b) Find the ratio BP : PC.

Now this is my triangle, to place P, I took "AP is vertical to BM" as "AP and BM are perpendicular".

Now the answer to (a) is :

$AB : AM = 2:1$

$\bigtriangleup ABH : \bigtriangleup AHM$

$= 2^2 : 1^2 = 4 : 1$

It is true that AB is twice large than AM, but why are they rising them to the 2th power?

To solve (b) I tried measuring PB and CP, and the comparing with CB, but I gave up soon, as I was getting something near 2.3 for CP. so I looked at the answer (2:1), and noticed the should be:

CB = 7
CP = 2.333333333
PB = 4.666666666

Though method used to calculate the ratio is very strange, I don't understand it:

M is in the middle of AC, $\bigtriangleup AHM = \bigtriangleup HCM$ --- (1)

on the other hand, $\frac{\bigtriangleup ABH}{\bigtriangleup HBP} = \frac{\bigtriangleup AHC}{\bigtriangleup HPC}$ --- (2)

From (1) and (2) $\frac{\bigtriangleup ABH}{\bigtriangleup HBP} = \frac{2\bigtriangleup AHM}{\bigtriangleup HPC}$ --- (3)

therefore from (3) and question (a)

$\frac{\bigtriangleup HPC}{\bigtriangleup HBP} = \frac{2\bigtriangleup AHM}{\bigtriangleup ABH} = \frac{(2)(1)}{4} = \frac{1}{2}$ --- (4)

$\frac{PC}{BP} = \frac{\bigtriangleup HPC}{\bigtriangleup HBP}$ because of (4)

The answer is BP : CP = 2 : 1

Can somebody explain me how to use this to get the ratio?

Thanks in advance.

**earboth** · June 30th 2007, 10:54 PM

Originally Posted by Patrick_John

...
Now the answer to (a) is :

$AB : AM = 2:1$

$\bigtriangleup ABH : \bigtriangleup AHM$

$= 2^2 : 1^2 = 4 : 1$

It is true that AB is twice large than AM, but why are they rising them to the 2th power?
...

Hello,

to calculate the area of a triangle you have to use two values: the length of the base and the length of the height.

In your case not only the base was lengthened by the factor 2 but the height also. If
$A_1 = \frac{1}{2} \cdot b_1 \cdot h_1$ is the area of the first triangle and you know that
$b_2 = 2 \cdot b_1 \text{ and } h_2 = 2 \cdot h_1$

then you get:

$A_2 = \frac{1}{2} \cdot b_2 \cdot h_2 = \frac{1}{2} \cdot 2 \cdot b_1 \cdot 2 \cdot h_1 = 2 \cdot 2 \cdot \frac{1}{2} \cdot b_1 \cdot h_1 = 2^2 \cdot A_1$

**earboth** · June 30th 2007, 11:11 PM

Originally Posted by Patrick_John

...
To solve (b) I tried measuring PB and CP, and the comparing with CB, but I gave up soon, as I was getting something near 2.3 for CP. so I looked at the answer (2:1), and noticed the should be:

CB = 7
CP = 2.333333333
PB = 4.666666666

Hello,

if $|AB| = |AC| = 5 ~cm$ then $|BC| = \sqrt{5^2+5^2} = 7 \cdot \sqrt{2}$

Though method used to calculate the ratio is very strange, I don't understand it:

M is in the middle of AC, $\bigtriangleup AHM = \bigtriangleup HCM$ --- (1)

Here you are refering to the value of the area of the triangle: In both triangles the bases and the heights have the same length therefore they must have the same area.

on the other hand, $\frac{\bigtriangleup ABH}{\bigtriangleup HBP} = \frac{\bigtriangleup AHC}{\bigtriangleup HPC}$ --- (2)

The same method is used here too.

Maybe you can complete the problem from here on ...

**red_dog** · June 30th 2007, 11:48 PM

I will note $S_{ABC}$ the area of the triangle $ABC$ .
Now:
For (1): $\triangle AHM$ and $\triangle HCM$ have the same height $h_1$ from $H$ .
$\displaystyle S_{AHM}=\frac{AM\cdot h_1}{2}=\frac{MC\cdot h_1}{2}=S_{HCM}$ .

For (2): $\triangle ABH$ and $\triangle HBP$ have the same height $h_2$ from $B$ .
$\displaystyle S_{ABH}=\frac{AH\cdot h_2}{2},S_{HBP}=\frac{PH\cdot h_2}{2}\Rightarrow\frac{S_{ABH}}{S_{HBP}}=\frac{AH }{PH}$ (i)
$\triangle AHC$ and $\triangle HPC$ have the same height $h_3$ from $C$ .
$\displaystyle S_{AHC}=\frac{AH\cdot h_3}{2},S_{HPC}=\frac{PH\cdot h_3}{2}\Rightarrow\frac{S_{AHC}}{S_{HPC}}=\frac{AH }{PH}$ (ii)
From (i) and (ii) yields $\displaystyle \frac{S_{ABH}}{S_{HBP}}=\frac{S_{AHC}}{S_{HPC}}$ (iii)

For (3) $\displaystyle S_{AHC}=\frac{AC\cdot h_1}{2}=\frac{2AM\cdot h_1}{2}=2S_{AHM}$ .
Replacing $S_{AHC}$ in (iii) $\Rightarrow$ (3).

For (4): $\displaystyle S_{HPC}=\frac{PC\cdot h_4}{2},S_{HBP}=\frac{BP\cdot h_4}{2}\Rightarrow$
$\displaystyle \frac{PC}{BP}=\frac{S_{HPC}}{S_{HBP}}=\frac{2S_{AH M}}{S_{ABH}}=\frac{2S_{AHM}}{4S_{AHM}}=\frac{1}{2}$

**Patrick_John** · July 1st 2007, 03:34 PM

This is still giving some problems,

Why in $\displaystyle \frac{PC}{BP}=\frac{S_{HPC}}{S_{HBP}}=\frac{2S_{AH M}}{S_{ABH}}=\frac{2S_{AHM}}{4S_{AHM}}=\frac{1}{2}$

$S_{ABH}$ is equal to $4S_{AHM}$ ?

**Quick** · July 3rd 2007, 06:43 PM

Originally Posted by Patrick_John

This is still giving some problems,

Why in $\displaystyle \frac{PC}{BP}=\frac{S_{HPC}}{S_{HBP}}=\frac{2S_{AH M}}{S_{ABH}}=\frac{2S_{AHM}}{4S_{AHM}}=\frac{1}{2}$

$S_{ABH}$ is equal to $4S_{AHM}$ ?

I'm going to say the same thing as earboth.

What's the area of a triangle? It's $\frac{1}{2}bh=A$

So what would happen if we doubled the base ( $2b$ ) and the height ( $2h$ )

Well let's substitute: $\frac{1}{2} (2b) (2h) \quad = \quad \frac{1}{2}(4bh) \quad = \quad 4\left(\frac{1}{2}bh\right)$

and we all know that: $4\left(\frac{1}{2}bh\right)=4A$

So when the legs of a triangle double, the area quadruples. That is the reason why
$S_{ABH}$ is equal to $4S_{AHM}$

**Patrick_John** · July 4th 2007, 07:03 AM

OK! Now I understand, thnks a lot everyone!

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