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# triangles

+1
1171
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In $\triangle RST$, $RS = 13$, $ST = 14$, and $RT = 15$. Let $M$ be the midpoint of $\overline{ST}$. Find $RM$.

Thanks

Aug 5, 2017

#1
+94544
+2

We  can use the Law of Cosines twice here

RT^2  =  ST^2 + RS^2  -  2(ST * RS) cos (RST)

15^2  =  14^2 + 13^2  - 2(14 * 13) cos (RST)   simpllfy

225 - 196 - 169  = -364 cos(RST)

-140 = -364cos(RST)

cos RST  = 140 / 364  =  5/13

And

RM^2  = SM^2 + RS^2  - 2(SM * RS) cos RST

RM^2  = 7^2 + 13^2 - 2 ( 7 * 13) (5/13)  simplify

RM^2 =  49 + 169 -  [ 2 *7 * 5 ]

RM^2  =  148

RM  = √148  =  √ [ 37 * 4 ]   =  2√37 ≈  12.166

Aug 5, 2017
#2
+94544
+2

Here's another approach using the Law of Cosines only once :

Refer to the following image :

RT^2  =  ST^2 + RS^2  -  2(ST * RS) cos (RST)

15^2  =  14^2 + 13^2  - 2(14 * 13) cos (RST)   simpllfy

225 - 196 - 169  = -364 cos(RST)

-140 = -364cos(RST)

cos RST  = 140 / 364  =  5/13

Now....in triangle RMS draw altitude RN and note that angle RSN  = angle RST........since triangle RNS is a right triangle with hypotenuse RS = 13......then.......if  cos RST = 5/13...it must be that  SN = 5.......so triangle RNS  has a hypotenuse of 13 and one leg   = 5.....so...the other leg (RN)  must = 12

And triangle RMN is another right triangle with RN  = 12.....and if M is the midpoint of ST, then SM  = 7.......but SN was shown to be = 5....so.....MN  = 2

And we have that  MN^2 + RN^2  = RM^2   ....so.....

2^2  + 12^2   = RM^2

148   = RM^2

√148  =  RM

Aug 5, 2017