Question

In a right triangle with sides $a$ and $b,$ and hypotenuse $c,$ the altitude drawn on the hypotenuse is $x.$ Then:

$ab=x^2$

$\dfrac1a+\dfrac1b=\dfrac1x$

$a^2+b^2=2x^2$

$\dfrac1{x^2}=\dfrac1{a^2}+\dfrac1{b^2}$

$\dfrac1x=\dfrac ba$

Answer: D

Concepts:right triangle altitude to hypotenuse area reciprocal identity

Difficulty rating: 1810

Small Hint:

Compute the triangle’s area using either the legs or the hypotenuse and its altitude

Big Hint:

From $ab=cx,$ substitute $c^2=a^2+b^2$ and divide by $a^2b^2x^2$

Solution:

Equating two area formulas gives $\frac12ab=\frac12cx,$ so $ab=cx.$ Squaring and using $c^2=a^2+b^2$ yields $a^2b^2=x^2(a^2+b^2).$ Dividing by $a^2b^2x^2$ gives $\frac1{x^2}=\frac1{a^2}+\frac1{b^2}.$

Thus, the correct answer is D.

1956 AMC 12 Problem 38

Answer: D

Difficulty rating: 1810

Small Hint:

Big Hint:

Solution:

Problem 38 in Other Years