MIT 1869

You know the length of hypotenuse parts 9 and 16, at which the hypotenuse of a right triangle is divided by a height. The task is to find the lengths of the sides of the triangle and the length of line x. This assignment was part of the Massachusetts Institute of Technology MIT entrance exams in 1869.

Correct answer:

c =  25
x =  12
a =  15
b =  20

Step-by-step explanation:

c1=9 c2=16  c=c1+c2=9+16=25
x2 = c1 c2  x=c1 c2=9 16=12
a2=x2+c12 a=x2+c12=122+92=15
b2=x2+c22 b=x2+c22=122+162=20   Verifying Solution:   c3=a2+b2=152+202=25

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Showing 1 comment:
Peter
Easier solution with almost no calculating. Because all 3 entered triangles are similar, the following holds: x / 9 = 16 / x => x = 12. And because every triangle that has an aspect ratio of 3: 4: 5 is right, it must hold: a = 3 * 5 = 15 ab = 4 * 5 = 20.
PS: Maybe at that time they wanted to know at MIT who was just calculate and who was even thinking.

3 years ago  1 Like




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