2008 AMC 10A Problem 20

Question

Trapezoid

ABCD

has bases

AB

and

CD

and diagonals intersecting at

K.

Suppose that

AB = 9,

DC = 12,

and the area of

\triangle AKD

is

24.

What is the area of trapezoid

ABCD?

92

94

96

98

100

Triangles

AKB

and

CKD

are similar with ratio

9 : 12

\triangle AKD

and

\triangle BKC

have equal areas; scale the others from

[AKD] = 24

Triangles

AKB

and

CKD

are similar with ratio

\dfrac{9}{12} = \dfrac{3}{4}.

Since

\triangle AKD

and

\triangle KCD

have bases

AK

and

KC

on the same line and share the same altitude from

D,

\dfrac{[KCD]}{[AKD]} = \dfrac{KC}{AK} = \dfrac{4}{3},

so

[KCD] = 32.

Similarly

[AKB] = 18.

Also

[BKC] = [AKD] = 24.

The total is

24 + 32 + 18 + 24 = 98.

Thus, the correct answer is D.

Problem 20 in other years