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 $\triangle ABC?$

$14.6$

$14.8$

$15$

$15.2$

$15.4$

Video solution:

Click to load, then click again to play

Written solution:

Let $a$ be the area of $\triangle ABC.$

Note that the smaller triangles are similar to $\triangle ABC.$ This means that the ratio of their areas is the ratio of their side lengths squared.

Then we get that $a - a \cdot \left(\dfrac{11}{h}\right)^2$ $= a \cdot \left(\dfrac{h - 5}{h}\right)^2.$ The left side is the area of the whole triangle minus the area of the unshaded region of the left triangle.

The right hand side is the area of the shaded triangle. We get this by finding the ratio of their side lengths (which is the same as the ratio of their heights) and squaring it.

Simplifying yields $h^2 - 121 = h^2 - 10h + 25.$ This simplifies to $10h = 146$ $h = 14.6.$

Thus, A is the correct answer.

2023 AMC 8 Problem 24

Video solution:

Written solution:

Problem 24 in Other Years