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?
+4622 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 !
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+129899
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 :
