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# Triangle help

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In right triangle ABC, $$\angle C = 90^\circ$$. Median $$\overline{AM}$$ has a length of 19, and median $$\overline{BN}$$ has a length of 13. What is the length of the hypotenuse of the triangle?

Apr 9, 2020

#1
+1

A median bisects the side in half(1/2).

Let the side AC=2a and CB=2B, thus, the medians divide the sides into a and b, respectively.

Systems of equations gives us...

a^2+(2b)^2=(13)^2

(2a)^2+b^2=(19)^2

a^2+4b^2=169

4a^2+b^2=361

Solve for 2a, and 2b 2\sqrt 85 and b=2\sqrt21

c^2=424 ...solve for c

probably made a mistake somewheree !

.

Apr 9, 2020
edited by tertre  Apr 9, 2020
edited by tertre  Apr 9, 2020
edited by tertre  Apr 9, 2020
#2
+1

Note  that   triangles  AMC  and NBC  are  both right

And note  that CM  =(1/2) BC    and   CN  =(1/2) AC

So.....we  have  this ststem of equationa

CM^2  + AC^2  = AM^2

CN^2  + (BC)^2  = BN^2        subbing  in, we have

[ (1/2) BC]^2  + AC^2  =  19^2          simplify

[ (1/2)AC]^2  + BC^2  = 13^2

BC^2/4  + AC^2  = 19^2

AC^2/4  + BC^2 = 13^2

BC^2/4 + AC^2  = 361    ⇒  BC^2  + 4AC^2  = 1441 ⇒  -BC^2  - 4AC^2 = -1444    (1)

AC^2/4  + BC^2  = 169    (2)

Add (1) and (2)   and  we  have  that

-4AC^2 + AC^2/4  = -1275

(-15/4)AC^2  = -1595        mutiply  both sides  by  -4/15

AC^2  = 340

And

340/4  + BC^2  =  169

340 + 4BC^2  =  676

4BC^2  = 336

BC^2 = 84

So the hypotenuse AB   can be  found as

AC^2  + BC^2  = AB^2

340  +  84  =  AB^2

424  =  AB^2

√424  =  AB

√ [ 4 * 106 ] =  AB

2√106  =  AB ≈  20.59

See  the image here :    Apr 9, 2020