SOLUTION: In right triangle ABC with right angle at C
AB=17
BC=8
Find MN if M and N are midpoints of AB and BC respectively.
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-> SOLUTION: In right triangle ABC with right angle at C
AB=17
BC=8
Find MN if M and N are midpoints of AB and BC respectively.
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Question 1156764: In right triangle ABC with right angle at C
AB=17
BC=8
Find MN if M and N are midpoints of AB and BC respectively. Found 2 solutions by Cromlix, ikleyn:Answer by Cromlix(4381) (Show Source):
You can put this solution on YOUR website! Hi there,
By connecting M with N you form another right angled triangle MNB with the right angle at N.
Therefore, MN^2 + NB^2 = MB^2
MN^2 = MB^2 - NB^2
MN^2 = 8.5^2 - 4^2
MN^2 = 56.25
MN = 7.5.
The other way is:-
By trig ratio find what angle B is.
cos B = adjacent/hypotenuse
cos B = 4/8.5
B = 61.9 degrees
Now, using Cosine rule:-
b^2 = m^2 + n^2 - 2 x m x n x Cos (61.9)
b^2 = 16 + 72.25 - (2 x 4 x 8.5 x Cos (61.9))
b^2 = 56.2
b = 7.498 or 7.5 (1 decimal place)
Hope this helps. :-)
Triangle ABC is a right angled triangle with the hypotenuse AB = 17 and the leg BC = 8.
Hence, its other leg AC is = = = 15 units long.
The segment MN is the midline in this triangle, parallel to the leg AC.
Hence, the length of the segment MN is half the leng of the leg AC, i.e. 15/2 = 7.5 units. ANSWER
