Point F is taken in side AD of square ABCD. At C a perpendicular is drawn to CF, meeting AB extended at E. The area of ABCD is 256 square inches and the area of triangle CEF is 200 square inches. Then the number of inches in BE is:

1963 AMC 12 Problem 25

Attempt Problem 25 of the 1963 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 1963 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

25.

Point

F

is taken in side

AD

of square

ABCD.

C

a perpendicular is drawn to

CF,

meeting

AB

extended at

E.

The area of

ABCD

256

square inches and the area of triangle

CEF

200

square inches. Then the number of inches in

BE

is:

12

14

15

16

20

Show that right triangles

CDF

and

CBE

are congruent

Then

CF=CE,

so use the area of right triangle

CEF

The complementary acute angles and the equal square sides give

\triangle CDF\cong\triangle CBE,

hence

CF=CE.

Since

CF\perp CE,

\begin{aligned} 200=[CEF] &=\frac12 CE\cdot CF\\ &=\frac12CE^2, \end{aligned}

CE=20.

The square side is

16,

and right triangle

CBE

gives

BE=\sqrt{20^2-16^2}=12.

Therefore, the correct answer is A.