Thread: Perimeter of triangle

I don't think you've been given enough information...

How ?,i have the answer but i want to know the way to solve it .

Can you find any of the shown segments? Also, a hint: use the right triangle altitude theorem.

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

Really helpful ,thanks

Originally Posted by Mhmh96

Since it follows :





Hence :



therefore perimeter is :

Thank you !

Can you explain this part of your solution ?

Originally Posted by Mhmh96

Here is a slightly different approach:

1. Determine by Pythagorean theorem.

2. Use Pythagorean theorem in triangle ABC:

and in the small right triangle:


3. Subtract both equations columnwise and solve for

Originally Posted by Mhmh96

Can you explain this part of your solution ?

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

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):





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.