2001 CEMC Gauss (Grade 8) Problems/Problem 6

The following problem is from both the 2001 CEMC Gauss (Grade 8) #6 and 2001 CEMC Gauss (Grade 7) #8, so both problems redirect to this page.

Problem

The area of the entire figure shown is

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$\text{ (A) }\ 16 \qquad\text{ (B) }\ 32 \qquad\text{ (C) }\ 20 \qquad\text{ (D) }\ 24 \qquad\text{ (E) }\ 64$

Solution

We can see all of the right triangles are equivalent because their legs are all equal to $4$ . Thus, we only need to find the area of one triangle, and then multiply that by $3$ .

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From triangle $ABC$ , its area is:

$\frac{1}{2} \times 4 \times 4 = 2 \times 4 = 8$

Thus, the total area would be:

$8 \times 3 = \boxed {\textbf {(D) } 24}$

~anabel.disher

2001 CEMC Gauss (Grade 8) (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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CEMC Gauss (Grade 8)

2001 CEMC Gauss (Grade 7) (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
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CEMC Gauss (Grade 7)

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2001 CEMC Gauss (Grade 8) Problems/Problem 6

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