Question

4a: what error was made in student as work? (worth 2 points) *
student a:
$v=\frac{1}{3}bh$
$v = \frac{1}{3}(12)(8)$
$v=(4)(8)$
$v = 32in^{3}$

Explanation:

Step1: Identify base - area formula

The base of the pyramid is a square with side length 12 in. The area of the base $B$ for a square is $B = s^2$, where $s = 12$ in, so $B=12\times12 = 144$ in².

Step2: Analyze student's work

Student A used the wrong value for the base - area. They used 12 instead of calculating the area of the square base ($12\times12$). Also, the height $h = 8$ in is used correctly in the volume formula $V=\frac{1}{3}Bh$, but the incorrect base - area value led to an incorrect result.

Answer:

Student A used the side - length of the base as the base - area instead of calculating the area of the square base. The correct base - area $B = 12\times12=144$ in², and the volume should be $V=\frac{1}{3}(144)(8)=384$ in³. The error was using 12 as the base - area instead of 144.