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Difference between revisions of "2014 CEMC Gauss (Grade 8) Problems/Problem 16"

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{{CEMC box|year=2014|competition=Gauss (Grade 8)|num-b=15|num-a=17}}

Latest revision as of 11:37, 18 October 2025

Problem

In the diagram, $ABCD$ is a rectangle.

An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.

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 $CP = 10 - x$.

An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.

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

2014 CEMC Gauss (Grade 8) (Problems • Answer Key • Resources)
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Problem 15 		Followed by
Problem 17
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CEMC Gauss (Grade 8)
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