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- Apr 4th 2012, 04:40 AMMhmh96Perimeter of triangle
- Apr 4th 2012, 04:50 AMProve ItRe: Perimeter of triangle
I don't think you've been given enough information...

- Apr 4th 2012, 04:57 AMMhmh96Re: Perimeter of triangle
How ?,i have the answer but i want to know the way to solve it .

- Apr 4th 2012, 04:58 AMemakarovRe: Perimeter of triangle
Can you find any of the shown segments? Also, a hint: use the right triangle altitude theorem.

- Apr 4th 2012, 05:03 AMbiffboyRe: 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 - Apr 4th 2012, 05:04 AMMhmh96Re: Perimeter of triangle
Really helpful ,thanks

- Apr 4th 2012, 05:13 AMprincepsRe: Perimeter of triangle
$\displaystyle |AM|=\sqrt{10^2-6^2}=8$

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

$\displaystyle |BM| : 6 = 6 : 8$

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

Hence :

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

therefore perimeter is : $\displaystyle 30$ - Apr 4th 2012, 05:41 AMMhmh96Re: Perimeter of triangle
Thank you !

- Apr 5th 2012, 09:10 PMMhmh96Re: Perimeter of triangle
Can you explain this part of your solution ?

http://store1.up-00.com/Mar12/x7688896.jpg - Apr 6th 2012, 02:00 AMearbothRe: Perimeter of triangle
Here is a slightly different approach:

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

2. Use Pythagorean theorem in triangle ABC:

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

$\displaystyle 6^2+ (\overline{BM})^2 = (\overline{BC})^2$

3. Subtract both equations columnwise and solve for $\displaystyle \overline{BM}$ - Apr 6th 2012, 08:25 AMemakarovRe: Perimeter of triangle
- Apr 9th 2012, 07:31 AMearbothRe: Perimeter of triangle
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):

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

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

Add both equations columnwise:

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

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

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

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