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 $\triangle ABC?$

$14.6$

$14.8$

$15$

$15.2$

$15.4$

Answer: A

Concepts:similarity area ratio

Difficulty rating: 1930

Video solution:

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Written solution:

Let $a$ be the area of $\triangle ABC.$ The unshaded triangle in the left figure has height $11,$ so the shaded area there is $a\left(1-\left(\frac{11}{h}\right)^2\right).$

In the right figure, the shaded triangle has height $h-5,$ 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.

2023 AMC 8 Problem 24

Answer: A

Difficulty rating: 1930

Video solution:

Written solution:

Problem 24 in Other Years