# Perimeter of triangle

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• April 4th 2012, 04:40 AM
Mhmh96
Perimeter of triangle
• April 4th 2012, 04:50 AM
Prove It
Re: Perimeter of triangle
I don't think you've been given enough information...
• April 4th 2012, 04:57 AM
Mhmh96
Re: Perimeter of triangle
How ?,i have the answer but i want to know the way to solve it .
• April 4th 2012, 04:58 AM
emakarov
Re: Perimeter of triangle
Can you find any of the shown segments? Also, a hint: use the right triangle altitude theorem.
• April 4th 2012, 05:03 AM
biffboy
Re: Perimeter of triangle
Get AM by Pythagoras (=8)
AngleACM=angleB (both=90-A)
sinB=6/CB But from triangle ACM sinACM=8/10 So sinB=8/10 So 8/10=6/CB So CB=15/2
In triangle BCM cosB=BM/CB So 6/10=BM/15/2 BM=9/2
Perimeter=AC+CB+BM+MA=10+7.5+4.5+8=30
• April 4th 2012, 05:04 AM
Mhmh96
Re: Perimeter of triangle
Really helpful ,thanks
• April 4th 2012, 05:13 AM
princeps
Re: Perimeter of triangle
Quote:

Originally Posted by Mhmh96

$|AM|=\sqrt{10^2-6^2}=8$

Since $\Delta AMC \sim \Delta BCM$ it follows :

$|BM| : 6 = 6 : 8$

$|BM|=\frac{9}{2} \Rightarrow |AB|=\frac{25}{2}$

Hence :

$|BC|=\sqrt{\left(\frac{25}{2}\right)^2-10^2}=\frac{15}{2}$

therefore perimeter is : $30$
• April 4th 2012, 05:41 AM
Mhmh96
Re: Perimeter of triangle
Thank you !
• April 5th 2012, 09:10 PM
Mhmh96
Re: Perimeter of triangle
Can you explain this part of your solution ?
http://store1.up-00.com/Mar12/x7688896.jpg
• April 6th 2012, 02:00 AM
earboth
Re: Perimeter of triangle
Quote:

Originally Posted by Mhmh96

Here is a slightly different approach:

1. Determine $|\overline{AM}| = 8$ by Pythagorean theorem.

2. Use Pythagorean theorem in triangle ABC:

$(8+\overline{BM})^2-10^2 = (\overline{BC})^2$ and in the small right triangle:
$6^2+ (\overline{BM})^2 = (\overline{BC})^2$

3. Subtract both equations columnwise and solve for $\overline{BM}$
• April 6th 2012, 08:25 AM
emakarov
Re: Perimeter of triangle
Quote:

Originally Posted by Mhmh96
Can you explain this part of your solution ?
http://store1.up-00.com/Mar12/x7688896.jpg

This is the right triangle altitude theorem. Its proof involves the fact that triangles AMC and CMB are similar.
• April 9th 2012, 07:31 AM
earboth
Re: Perimeter of triangle
Quote:

Originally Posted by emakarov
This is the right triangle altitude theorem. Its proof involves the fact that triangles AMC and CMB are similar.

Of course you are right but you also can use 2 right triangles to prove the altitude theorem:

(Using the labels of the attached diagram):

$CM^2+BM^2=BC^2$

$CM^2+AM^2=AC^2$

Add both equations columnwise:

$2CM^2 + AM^2 + BM^2 = BC^2+AC^2 = AB^2$

Since $AB = AM + BM$ the right side of the last equation becomes:

$2CM^2 + AM^2 + BM^2 = (AM + BM)^2 = AM^2 + 2AM \cdot BM + BM^2$

Collect like terms, divide through by 2 and you'll get the equation of the altitude theorem.