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Charles begins finding the volume of a trapezoidal prism using the formula [tex]A=\frac{1}{2}\left(b_1+b_2\right) h[/tex] to find the prism's base area.

[tex]\[
\begin{array}{l}
A=\frac{1}{2}((x+4)+(x+2)) x \\
A=\frac{1}{2}(2x+6) x \\
A=(x+3)x \\
A=x^2+3x
\end{array}
\][/tex]

Which expression can be used to represent the volume of the trapezoidal prism?

A. [tex]2x^3 + 6x^2[/tex]

B. [tex]x^3 + 6x^2[/tex]

C. [tex]x^3 + 3x^2[/tex]

D. [tex]2x^3 + 3x^2[/tex]

Sagot :

Sure, let's solve this step-by-step.

To find the volume of the trapezoidal prism, we use the fact that the volume [tex]\( V \)[/tex] is given by

[tex]\[ V = \text{Base Area} \times \text{Height} \][/tex]

From the problem, we have already derived the expression for the area [tex]\( A \)[/tex] of the trapezoidal base as follows:

[tex]\[ A = \frac{1}{2}((x+4)+(x+2))x \][/tex]

[tex]\[ A = \frac{1}{2}(2x + 6)x \][/tex]

[tex]\[ A = (x + 3)x \][/tex]

[tex]\[ A = x^2 + 3x \][/tex]

Let's assume the height (which I'll denote as [tex]\( H \)[/tex]) of the prism is [tex]\( x \)[/tex]. Thus, the volume [tex]\( V \)[/tex] can be computed as:

[tex]\[ V = A \times H \][/tex]

Substitute [tex]\( A = x^2 + 3x \)[/tex]:

[tex]\[ V = (x^2 + 3x) \times x \][/tex]

Simplify this expression:

[tex]\[ V = x^3 + 3x^2 \][/tex]

Therefore, the expression that represents the volume of the trapezoidal prism is:

[tex]\[ x^3 + 3x^2 \][/tex]

Now, let's match this resulting volume with one of the answer choices:

1. [tex]\( 2x^3 + 6x^2 \)[/tex]
2. [tex]\( x^3 + 6x^2 \)[/tex]
3. [tex]\( x^3 + 3x^2 \)[/tex]
4. [tex]\( 2x^3 + 3x^2 \)[/tex]

The correct answer is:

[tex]\[ x^3 + 3x^2 \][/tex]

So, the correct choice is:
[tex]\( x^3 + 3x^2 \)[/tex]
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