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# triangle hypotenuse problem

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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?

Mar 5, 2024

#2
+129829
+1

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

Mar 6, 2024

#1
+193
0

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).

Mar 5, 2024
#2
+129829
+1

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

CPhill Mar 6, 2024