2012 CEMC Gauss (Grade 8) Problems/Problem 10

Problem

The rectangle shown has side lengths of 8 and 4.

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The area of the shaded region is

$\text{ (A) }\ 32 \qquad\text{ (B) }\ 16 \qquad\text{ (C) }\ 64 \qquad\text{ (D) }\ 12 \qquad\text{ (E) }\ 4$

Solution 1

Let $x$ be the base of one of the shaded triangles that isn't equal to 4 necessarily, as shown here:

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We know that the shaded triangles are two right triangles because the interior angles of a rectangle are right angles.

Since the full side lengths is $8$ , the length of the base of the other shaded triangle is $8 - x$ . This means that the total area of both of the triangles is:

$\frac{4 \times x}{2} + \frac{4 \times (8 - x)}{2} = \frac{4 \times x + 4 \times (8 - x)}{2} = \frac{4 \times x + 4 \times 8 - 4 \times x}{2} = \frac{32}{2} = \boxed {\textbf {(C) } 16}$ .