2023 AMC 8 Problem 24

Question

Isosceles triangle

ABC

has equal side lengths

AB

and

BC.

In the figures below, segments are drawn parallel to

\overline{AC}

so that the shaded portions of

\triangle ABC

have the same area. The heights of the two unshaded portions are

11

and

5

units, respectively. What is the height

h

of

\triangle ABC?

14.6

14.8

15

15.2

15.4

Equal shaded areas give an equation between the two unshaded triangle areas

Similar triangle areas scale as the square of their heights

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Let

a

be the area of

\triangle ABC.

The unshaded triangle in the left figure has height

11

and is similar to the full triangle, so its area is

a\left(\frac{11}{h}\right)^2.

Therefore, the shaded area there is

a\left(1-\left(\frac{11}{h}\right)^2\right).

In the right figure, the shaded triangle has height

h-5

and is similar to the full triangle, so its area is

a\left(\frac{h-5}{h}\right)^2.

Equating the shaded areas and canceling

a

gives

1-\left(\dfrac{11}{h}\right)^2 =\left(\dfrac{h-5}{h}\right)^2.

Simplifying yields

h^2 - 121 = h^2 - 10h + 25.

This simplifies to

10h = 146,

so

h=14.6.

Thus, A is the correct answer.

Problem 24 in other years