1997 AMC 12 Problem 15

Attempt Problem 15 of the 1997 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 1997 AMC 12 solutions, or check the answer key.

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15.

Medians

\overline{BD}

and

\overline{CE}

of triangle

ABC

are perpendicular,

BD=8,

and

CE=12.

The area of triangle

ABC

24

32

48

64

96

View

BD

and

CE

as the diagonals of quadrilateral

BCDE

Triangle

ADE

is similar to

ABC

with scale factor

\frac{1}{2}

Quadrilateral

BCDE

has perpendicular diagonals

BD=8

and

CE=12,

so its area is

\frac12(8)(12)=48.

Since

D,E

are midpoints, triangle

ADE

has one fourth the area of

ABC,

making

BCDE

three fourths of it. Thus

[ABC]=\frac{48\cdot4}{3}=64,

so D is correct.

Problem 15 in other years