# Math Help - Perimeter of triangle

2. ## Re: Perimeter of triangle

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

3. ## Re: Perimeter of triangle

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

4. ## Re: Perimeter of triangle

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

5. ## 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

7. ## Re: Perimeter of triangle

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 : $30$

Thank you !

9. ## Re: Perimeter of triangle

Can you explain this part of your solution ?

10. ## Re: Perimeter of triangle

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 $\overline{BM}$

11. ## Re: Perimeter of triangle

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.

12. ## Re: Perimeter of triangle

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):

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

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

$2CM^2 + AM^2 + BM^2 = BC^2+AC^2 = AB^2$
Since $AB = AM + BM$ the right side of the last equation becomes:
$2CM^2 + AM^2 + BM^2 = (AM + BM)^2 = AM^2 + 2AM \cdot BM + BM^2$