1952 AMC 12 Problem 22

Attempt Problem 22 of the 1952 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1952 AMC 12 solutions, or check the answer key.

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22.

On hypotenuse

AB

of a right triangle

ABC

a second right triangle

ABD

is constructed with hypotenuse

AB.

BC=1,

AC=b,

and

AD=2,

then

BD

equals:

\sqrt{b^2+1}

\sqrt{b^2-3}

\sqrt{b^2+1}+2

b^2+5

\sqrt{b^2+3}

Use triangle

ABC

first to express the common hypotenuse

AB

Then apply the Pythagorean theorem to triangle

ABD

, whose hypotenuse is

AB

In the first right triangle,

AB^2=AC^2+BC^2=b^2+1.

In the second right triangle,

AB

is the hypotenuse and

AD=2,

\begin{aligned} BD^2&=AB^2-AD^2\\ &=b^2+1-4=b^2-3. \end{aligned}

Therefore

BD=\sqrt{b^2-3}.

Thus, the correct answer is B.

Problem 22 in other years