2007 AMC 10A Problems/Problem 18

From AoPSWiki

Problem

Consider the $12$ -sided polygon $ABCDEFGHIJKL$ , as shown. Each of its sides has length $4$ , and each two consecutive sides form a right angle. Suppose that $\overline{AG}$ and $\overline{CH}$ meet at $M$ . What is the area of quadrilateral $ABCM$ ?

$\text{(A)}\ \frac {44}{3}\qquad \text{(B)}\ 16 \qquad \text{(C)}\ \frac {88}{5}\qquad \text{(D)}\ 20 \qquad \text{(E)}\ \frac {62}{3}$

Solution

We can obtain the solution by calculating the area of rectangle $ABGH$ minus the combined area of triangles $\triangle AHG$ and $\triangle CGM$ .

We know that triangles $\triangle AMH$ and $\triangle CGM$ are similar because $\overline{AH} \parallel \overline{CG}$ . Also, since $\frac{AH}{CG} = \frac{3}{2}$ , the ratio of the distance from $M$ to $\overline{AH}$ to the distance from $M$ to $\overline{CG}$ is also $\frac{3}{2}$ . Solving with the fact that the distance from $\overline{AH}$ to $\overline{CG}$ is 4, we see that the distance from $M$ to $\overline{CG}$ is $\frac{8}{5}$ .

The area of $\triangle AHG$ is simply $\frac{1}{2} \cdot 4 \cdot 12 = 24$ , the area of $\triangle CGM$ is $\frac{1}{2} \cdot \frac{8}{5} \cdot 8 = \frac{32}{5}$ , and the area of rectangle $ABGH$ is $4 \cdot 12 = 48$ .

Taking the area of rectangle $ABGH$ and subtracting the combined area of $\triangle AHG$ and $\triangle CGM$ yields $48 - (24 + \frac{32}{5}) = \boxed{\frac{88}{5}}\ \text{(C)}$ .

2007 AMC 10A (Problems)
Preceded by Problem 17	Followed by Problem 19
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25

Personal tools

Navigation

Search

Toolbox

2007 AMC 10A Problems/Problem 18

From AoPSWiki

Problem

Solution

See also