+4 In right triangle $ABC,$ $\angle C = 90^\circ.$ Median $\overline{AM}$ has a length of $7,$ and median $\overline{BN}$ has a length of $6.$ What is the length of the hypotenuse of the triangle?
+129850 
AC^2 + [(1/2)BC]^2 = AM^2
[(1/2)AC]^2 + BC^2 = BN^2
AC^2 + BC^2/4 = 49
AC^2/4 + BC^2 = 36
Multiply the first equation through by 4 and the second through by -1
4AC^2 + BC^2 = 196
-AC^2/4-BC^2 = - 36 add these
(15/4)AC^2 = 160
AC^2 = 4 * 160 / 15
AC^2 = 128/3
To find BC
4 (128/3) + BC^2 = 196
(512/3) + BC^2 = 196
BC^2 = 196 - 512/3
BC^2 = 76/3
AC^2 + BC^2 = AB^2
128/3 + 76/3 = AB^2
204/3 = AB^2
68 = AB^2
AB = sqrt (68) = 2sqrt (17) = hypotenuse length
+195 Since AM is a median, then M is the midpoint of BC. Therefore, BM=MC=6.
Similarly, since BN is a median, then N is the midpoint of AC. Therefore, AN=NC=7.
By the Pythagorean Theorem on right triangle ABM, AB2=AM2+BM2=72+62=85. Then, AB=85.
Similarly, by the Pythagorean Theorem on right triangle ACN, AC2=AN2+NC2=72+72=98. Then, AC=98.
Finally, by the Pythagorean Theorem on right triangle ABC, BC^2 = AB^2 + AC^2 = 85 + 98 = 183.
Therefore, the length of the hypotenuse is BC = sqrt(183).
+129850
AC^2 + [(1/2)BC]^2 = AM^2
[(1/2)AC]^2 + BC^2 = BN^2
AC^2 + BC^2/4 = 49
AC^2/4 + BC^2 = 36
Multiply the first equation through by 4 and the second through by -1
4AC^2 + BC^2 = 196
-AC^2/4-BC^2 = - 36 add these
(15/4)AC^2 = 160
AC^2 = 4 * 160 / 15
AC^2 = 128/3
To find BC
4 (128/3) + BC^2 = 196
(512/3) + BC^2 = 196
BC^2 = 196 - 512/3
BC^2 = 76/3
AC^2 + BC^2 = AB^2
128/3 + 76/3 = AB^2
204/3 = AB^2
68 = AB^2
AB = sqrt (68) = 2sqrt (17) = hypotenuse length
