2017 AMC 10 B

Complete problem set with solutions and individual problem pages

Problem 15 Medium

Rectangle ABCD has AB=3 and BC=4. Point E is the foot of the perpendicular from B to diagonal \overline{AC}. What is the area of \triangle AED? (2017 AMC 10B Problem, Question#15)

 

  • A.

    1

  • B.

    \frac {42}{25}

  • C.

    \frac {28}{15}

  • D.

    2

  • E.

    \frac {54}{25}

Answer:E

First, note that AC=5 because ABC is a right triangle. In addition, we have AB\cdot BC=2\left[ ABC\right]=AC\cdot BE, so BE=\frac {12}5. Using similar triangles within ABC, we get that AE=\frac 95 and CE=\frac {16}5. Let F be the foot of the perpendicular from E to AB. Since EF and BC are parallel,\triangle AFE is similar to \triangle ABC. Therefore, we have \dfrac{AF}{AB}=\dfrac{AE}{AC}=\dfrac{9}{25}. Since AB=3,AF=\frac {27}{25} . Note that AF is an altitude of \triangle AED from AD, which has length 4. Therefore, the area of \triangle AED is \dfrac{1}{2}\cdot\dfrac{27}{25}\cdot4=\dfrac{54}{25}.

Alternatively, we can use coordinates. Denote D as the origin. We find the equation for AC as y=-\frac 43x+4, and BE as y=\frac 34x+\frac 74. Solving for x yields \frac {27}{25}. Our final answer then becomes \frac 12\cdot \frac {27}{25}\cdot 4=\frac {54}{25}.

We note that the area of ABE must equal area of AED because they share the base and the height of both is the altitude of congruent triangles. Therefore, we find the area of ABE to be \dfrac{1}{2}*\dfrac{9}{5}*\dfrac{12}{5}=\dfrac{54}{25}.

We know all right triangles are 5-4-3, so the areas are proportional to the square of like sides. Area of ABE is \left( \dfrac{3}{5}\right)^{2} of ABC=\frac {54}{25}. Using similar logic in Solution 3, Area of AED is the same as ABE.

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