MIT 1869

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

Correct answer:

c =  25
x =  12
a =  15
b =  20

Step-by-step explanation:

$$\begin{gathered} c_1=9 \\ {c}_2=16 \\ c=c_1+c_2=9+16=25 \end{gathered}$$

$$\begin{gathered} x^2=c_1\cdot c_2 \\ x=\sqrt{c_1\cdot c_2}=\sqrt{9\cdot 16}=12 \end{gathered}$$

$$\begin{gathered} a^2=x^2+c_1^2 \\ a=\sqrt{x^2+c_1^2}=\sqrt{12^2+9^2}=15 \end{gathered}$$

$$\begin{gathered} b^2=x^2+c_2^2 \\ b=\sqrt{x^2+c_2^2}=\sqrt{12^2+16^2}=20 \\ \text{ Verifying Solution: } \\ {c}_3=\sqrt{a^2+b^2}=\sqrt{15^2+20^2}=25 \end{gathered}$$

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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.

9 days ago  1 Like Like

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