1969 AMC 12 Problem 15

Attempt Problem 15 of the 1969 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 1969 AMC 12 solutions, or check the answer key.

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15.

In a circle with center at

O

and radius

r,

chord

AB

is drawn with length equal to

r

units. From

O

a perpendicular to

AB

meets

AB

at

M.

From

M

a perpendicular to

OA

meets

OA

at

D.

In terms of

r,

the area of triangle

MDA,

in appropriate square units, is:

\dfrac{3r^2}{16}

\dfrac{\pi r^2}{16}

\dfrac{\pi r^2\sqrt2}{8}

\dfrac{r^2\sqrt3}{32}

\dfrac{r^2\sqrt6}{48}

Answer: D

Concepts:chord equilateral triangle special right triangle triangle area

Difficulty rating: 1720

Small hint:

Because

OA=OB=AB=r,

triangle

OAB

is equilateral

Big hint:

Use the two

30

-

60

-

90

triangles to find

AD

and

MD

Solution:

Triangle

OAB

is equilateral. Since

OM\perp AB,

M

is the midpoint of

AB,

so

AM=\frac{r}{2}

and

\angle OAM=60^\circ.

In right triangle

AMD,

the hypotenuse is

AM=\frac{r}{2},

giving

AD=\frac r4,\qquad MD=\frac{r\sqrt3}{4}.

Hence

\begin{aligned} [MDA]&=\frac12(AD)(MD)\\ &=\frac{r^2\sqrt3}{32}. \end{aligned}

Therefore, the correct answer is D.

Problem 15 in other years