2014 CEMC Gauss (Grade 8) Problems/Problem 16

Problem

In the diagram, $ABCD$ is a rectangle.

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If the area of triangle $ABP$ is $40$ , then the area of the shaded region is

$\text{ (A) }\ 20 \qquad\text{ (B) }\ 40 \qquad\text{ (C) }\ 60 \qquad\text{ (D) }\ 50 \qquad\text{ (E) }\ 80$

Solution 1

Let $DP = x$ and $AD = BC = h$ . We then have $CD = 10$ , so $CP = 10 - x$ .

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Using our variables, we have:

$[ADP] = \frac{xh}{2}$

$[BCP] = \frac{(10 - x)h}{2}$

$[ABP] = \frac{10h}{2}$

The shaded area is just the sum of the areas of triangles $ADP$ and $BCP$ :

$[ADP] + [BCP] = \frac{xh}{2} + \frac{(10 - x)h}{2}$

$=\frac{xh + (10 - x)h}{2}$

$=\frac{(10 - x + x)h}{2}$

$=\frac{10h}{2}$

We can notice that this is equal to $[ADP]$ , so the shaded area also has an area of $\boxed {\textbf {(B) } 40}$ .

~anabel.disher

Solution 2

We can notice that the triangle shares the same base as the rectangle, and has the same height. Thus, the unshaded region must be half the area of the full rectangle.

This means that the shaded region's area is the same as the area of the unshaded region in the rectangle. Thus, the shaded area also has an area of $\boxed {\textbf {(B) } 40}$ .

~anabel.disher

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2014 CEMC Gauss (Grade 8) Problems/Problem 16

Problem

Solution 1

Solution 2