Difference between revisions of "2012 CEMC Gauss (Grade 7) Problems/Problem 21"

Revision as of 16:51, 22 April 2025

This page has been proposed for deletion. Reason: this is also Problem 18 on the Grade 8 test, so please move (and make sure to make a redirect) Note to sysops: Before deleting, please review: • What links here • Discussion page • Edit history

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

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

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 $\boxed {\textbf {(B) } 20}$ .

@@ Line 1: / Line 1: @@
 ==Problem==
-A triangular prism has a volume of <math>120 cm^{3}. Two edges of the triangular prism measure 3 cm and 4 cm, as shown.
+{{Delete|this is also Problem 18 on the Grade 8 test, so please move (and make sure to make a redirect)}}
+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.
 {{Template:Image needed}}
 The height of the prism, in cm, is
-</math>\text{(A)}\ 12 \qquad \text{(B)}\ 20 \qquad \text{(C)}\ 10 \qquad \text{(D)}\ 16 \qquad \text{(E)}\ 8 <math>
+<math>\text{(A)}\ 12 \qquad \text{(B)}\ 20 \qquad \text{(C)}\ 10 \qquad \text{(D)}\ 16 \qquad \text{(E)}\ 8 </math>
 ==Solution==
 The volume of the triangular prism will be the area of the base multiplied by its height.
-Let </math>A<math> and </math>h<math> be the area of the base and the height, respectively. We then have:
+Let <math>A</math> and <math>h</math> be the area of the base and the height, respectively. We then have:
-</math>A = \frac{3 cm * 4 cm}{2} = 6 cm^2<math>
+<math>A = \frac{3 cm * 4 cm}{2} = 6 cm^2</math>
-</math>A * h = V<math>
+<math>A * h = V</math>
-</math>6 cm^2 * h = 120 cm^3<math>
+<math>6 cm^2 * h = 120 cm^3</math>
-</math>h = \frac{120 cm^3}{6 cm^2} = 20 cm<math>
+<math>h = \frac{120 cm^3}{6 cm^2} = 20 cm</math>
-Thus, the answer is </math>\boxed {\textbf {(B) } 20}$.
+Thus, the answer is <math>\boxed {\textbf {(B) } 20}</math>.