1991 AMC 12 Problem 5

Attempt Problem 5 of the 1991 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 1991 AMC 12 solutions, or check the answer key.

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

In the arrow-shaped polygon shown, the angles at vertices

A,

C,

D,

E,

and

F

are right angles,

BC=FG=5,

CD=FE=20,

DE=10,

and

AB=AG.

The area of the polygon is closest to

288

291

294

297

300

Separate the arrowhead from the rectangular shaft

Use

AB=AG

and the two right angles at

C

and

F

to determine the arrowhead

The shaft

CDEF

is a

20

-by-

10

rectangle, so its area is

200.

Also

BG=BC+CF+FG

=5+10+5=20.

Because

\angle BAG=90^\circ

and

AB=AG,

triangle

ABG

is an isosceles right triangle with hypotenuse

20.

Its legs are

10\sqrt2,

so its area is

\frac12(10\sqrt2)^2=100.

The total area is

200+100=300.

Thus the correct answer is E.

Problem 5 in other years