2006 AMC 10B Problem 23

Below is the professionally curated solution for Problem 23 of the 2006 AMC 10B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2006 AMC 10B solutions, or check the answer key.

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Concepts:area ratio system of equations

Difficulty rating: 1950

23.

A triangle is partitioned into three triangles and a quadrilateral by drawing two lines from vertices to their opposite sides. The areas of the three triangles are $3,$ $7,$ and $7,$ as shown. What is the area of the shaded quadrilateral?

$15$

$17$

$\dfrac{35}{2}$

$18$

$\dfrac{55}{3}$

Solution:

Split the quadrilateral into two triangles of areas $R$ and $S,$ so the shaded area is $T=R+S.$

Comparing triangles that share an altitude, base ratios give $\tfrac{R}{3}=\tfrac{T+7}{10}$ and $\tfrac{S}{7}=\tfrac{T+3}{14}.$

Then $T=R+S=3\cdot\tfrac{T+7}{10}+7\cdot\tfrac{T+3}{14},$ so $10T=3(T+7)+5(T+3)=8T+36,$ giving $T=18.$

Thus, the correct answer is D.

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