2025 AMC 10A Problem 15

Attempt Problem 15 of the 2025 AMC 10A 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 2025 AMC 10A solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

15.

In the figure below, ABEFABEF is a rectangle, ADDE,AD \perp DE, AF=7,AF = 7, AB=1,AB = 1, and AD=5.AD = 5. What is the area of ABC?\triangle ABC?

38\dfrac{3}{8}

49\dfrac{4}{9}

1813\dfrac{1}{8}\sqrt{13}

715\dfrac{7}{15}

1815\dfrac{1}{8}\sqrt{15}

Answer: A
Concepts:similarityPythagorean Theoremquadratic
Difficulty rating: 1730
Solution:

Let x=BC.x = BC. Since ABEFABEF is a rectangle with AB=1AB = 1 and AF=7,AF = 7, and AD=5,AD = 5, we get AC=1+x2,AC = \sqrt{1 + x^2}, CE=7x,CE = 7 - x, and CD=51+x2.CD = 5 - \sqrt{1 + x^2}. The triangles ABC\triangle ABC and EDC\triangle EDC are similar, so 7x1+x2=51+x2x.\frac{7 - x}{\sqrt{1 + x^2}} = \frac{5 - \sqrt{1 + x^2}}{x}. Clear denominators and square to get 24x2+14x24=0,24x^2 + 14x - 24 = 0, which factors as (4x3)(3x+4)=0.(4x - 3)(3x + 4) = 0. The positive root is x=34.x = \tfrac{3}{4}. So the area is 12341=38.\tfrac12 \cdot \tfrac34 \cdot 1 = \tfrac{3}{8}. Thus, A is the correct answer.

← Problem 14#14
Full Exam

Problem 15 in Other Years