Difference between revisions of "2012 CEMC Gauss (Grade 8) Problems/Problem 18"

Latest revision as of 21:02, 18 October 2025

The following problem is from both the 2012 CEMC Gauss (Grade 8) #18 and 2012 CEMC Gauss (Grade 7) #21, so both problems redirect to this page.

A triangular prism has a volume of $120 cm^{3}$ . Two edges of the triangular prism measure 3 cm and 4 cm, as shown.

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The height of the prism, in cm, is

$\text{(A)}\ 12 \qquad \text{(B)}\ 20 \qquad \text{(C)}\ 10 \qquad \text{(D)}\ 16 \qquad \text{(E)}\ 8$

The volume of the triangular prism will be the area of the base multiplied by its height.

Let $A$ and $h$ be the area of the base and the height, respectively. We then have:

$A = \frac{3 cm * 4 cm}{2} = 6 cm^2$

$A * h = V$

$6 cm^2 * h = 120 cm^3$

$h = \frac{120 cm^3}{6 cm^2} = 20 cm$

Thus, the answer is $\boxed {\textbf {(B) } 20}$ .

2012 CEMC Gauss (Grade 8) (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
CEMC Gauss (Grade 8)

2012 CEMC Gauss (Grade 7) (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
CEMC Gauss (Grade 7)

@@ Line 1: / Line 1: @@
+{{Duplicate|[[2012 CEMC Gauss (Grade 8) Problems|2012 CEMC Gauss (Grade 8) #18]] and [[2012 CEMC Gauss (Grade 7) Problems|2012 CEMC Gauss (Grade 7) #21]]}}
 ==Problem==
 A triangular prism has a volume of <math>120 cm^{3}</math>. Two edges of the triangular prism measure 3 cm and 4 cm, as shown.
@@ Line 19: / Line 20: @@
 Thus, the answer is <math>\boxed {\textbf {(B) } 20}</math>.
+{{CEMC box|year=2012|competition=Gauss (Grade 8)|num-b=17|num-a=19}}
+{{CEMC box|year=2012|competition=Gauss (Grade 7)|num-b=20|num-a=22}}