1958 AMC 12 Problem 21

Question

In the accompanying figure

\overline{CE}

and

\overline{DE}

are equal chords of a circle with center

O.

Arc

AB

is a quarter-circle. Then the ratio of the area of triangle

CED

to the area of triangle

AOB

is:

\sqrt2:1

\sqrt3:1

4:1

3:1

2:1

The equal chords from

E

make the pictured triangle symmetric about the perpendicular diameter

Compare each triangle’s base-height area in terms of the circle radius

Let the circle have radius

r.

In the pictured configuration,

\overline{CD}

is a diameter and

E

is the endpoint of the perpendicular radius, so

[CED]=\frac12(2r)(r)=r^2.

Since arc

AB

is a quarter-circle,

\angle AOB=90^\circ,

and

[AOB]=\frac12r^2.

The requested ratio is therefore

2:1.

Thus, the correct answer is E.

Problem 21 in other years