Triangle problem

Geometry Level 2

Rectangle ABCD has AB = 8 and BC = 6. Point M is the midpoint of diagonal AC, and E is on AB with ME perpendicular to AC. What is the area of triangle AME?

Source: 2009 AMC10 B

9 75/8 65/8 85/8

Well Max by well max , 31, USA

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1 solution

By pythagorean theorem, A C = 8 2 + 6 2 = 64 + 36 = 100 = 10 AC=\sqrt{8^2+6^2}=\sqrt{64+36}=\sqrt{100}=10 . It follows that A M = M C = 5 AM=MC=5 .

Since △ A M E ∼ △ A B C \triangle AME \sim \triangle ABC , we have

M E A M = B C A B \dfrac{ME}{AM}=\dfrac{BC}{AB}

M E 5 = 6 8 \dfrac{ME}{5}=\dfrac{6}{8}

M E = 5 ( 6 ) 8 = 30 8 ME = \dfrac{5(6)}{8}=\dfrac{30}{8}

The area of △ A M E \triangle AME is

A = 1 2 × 5 × 30 8 = 150 16 = 75 8 A=\dfrac{1}{2} \times 5 \times \dfrac{30}{8} = \dfrac{150}{16} = \boxed{\dfrac{75}{8}}

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