Attachment 23521

Printable View

- April 4th 2012, 05:40 AMMhmh96Perimeter of triangle
- April 4th 2012, 05:50 AMProve ItRe: Perimeter of triangle
I don't think you've been given enough information...

- April 4th 2012, 05:57 AMMhmh96Re: Perimeter of triangle
How ?,i have the answer but i want to know the way to solve it .

- April 4th 2012, 05:58 AMemakarovRe: Perimeter of triangle
Can you find any of the shown segments? Also, a hint: use the right triangle altitude theorem.

- April 4th 2012, 06:03 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 - April 4th 2012, 06:04 AMMhmh96Re: Perimeter of triangle
Really helpful ,thanks

- April 4th 2012, 06:13 AMprincepsRe: Perimeter of triangle
- April 4th 2012, 06:41 AMMhmh96Re: Perimeter of triangle
Thank you !

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

http://store1.up-00.com/Mar12/x7688896.jpg - April 6th 2012, 03:00 AMearbothRe: Perimeter of triangle
- April 6th 2012, 09:25 AMemakarovRe: Perimeter of triangle
- April 9th 2012, 08:31 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):

Add both equations columnwise:

Since the right side of the last equation becomes:

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