anoroclupita2008
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Given a prism with a right triangle base and the dimensions [tex]\( h = x + 1 \)[/tex], [tex]\( b = x \)[/tex], and [tex]\( l = x + 7 \)[/tex], what is a correct expression for the volume of the prism?

A. [tex]\( \quad V = \frac{1}{3}\left(x^3 + 8x^2 + 7x\right) \)[/tex]

B. [tex]\( \quad V = \frac{1}{2}\left(x^3 + 8x^2 + 7x\right) \)[/tex]

C. [tex]\( V = x^2 + 8x + 7 \)[/tex]

D. [tex]\( \quad V = x^3 + 8x^2 + 7x \)[/tex]



Answer :

GinnyAnswer
To find the volume of a prism with a right triangle base, we need to follow these steps:

1. Identify the formula for the volume of a prism: The volume [tex]\( V \)[/tex] of a prism is given by the product of the area of the base [tex]\( A_{\text{base}} \)[/tex] and the length (or height) of the prism [tex]\( l \)[/tex]:
[tex]\[ V = A_{\text{base}} \times l \][/tex]

2. Find the area of the right triangle base: For a right triangle, the area [tex]\( A_{\text{base}} \)[/tex] is calculated as:
[tex]\[ A_{\text{base}} = \frac{1}{2} \times \text{base} \times \text{height} \][/tex]

Given:
- The base of the triangle [tex]\( b = x \)[/tex]
- The height of the triangle [tex]\( h = x + 1 \)[/tex]

Therefore:
[tex]\[ A_{\text{base}} = \frac{1}{2} \times x \times (x + 1) = \frac{1}{2} \times x \times x + \frac{1}{2} \times x = \frac{1}{2} (x^2 + x) \][/tex]

3. Substitute the area of the base and the length into the volume formula:

Given:
- The length of the prism [tex]\( l = x + 7 \)[/tex]

Therefore:
[tex]\[ V = A_{\text{base}} \times l = \left( \frac{1}{2} (x^2 + x) \right) \times (x + 7) \][/tex]

4. Simplify the expression:
[tex]\[ V = \frac{1}{2} (x^2 + x) (x + 7) \][/tex]

Distribute:
[tex]\[ (x^2 + x)(x + 7) = x^2(x + 7) + x(x + 7) = x^3 + 7x^2 + x^2 + 7x = x^3 + 8x^2 + 7x \][/tex]

Therefore:
[tex]\[ V = \frac{1}{2} (x^3 + 8x^2 + 7x) \][/tex]

5. Check against the answer choices:
- [tex]\( \frac{1}{3}(x^3+8x^2+7x) \)[/tex] does not match our expression.
- [tex]\( \frac{1}{2}(x^3+8x^2+7x) \)[/tex] matches our derived expression [tex]\(\boxed{B}\)[/tex].
- [tex]\( x^2 + 8x + 7 \)[/tex] does not match our volume expression.
- [tex]\( x^3 + 8x^2 + 7x \)[/tex] does not match our volume expression after including the [tex]\( \frac{1}{2} \)[/tex].

Thus, the correct answer is:

[tex]\[ \boxed{B} \][/tex]

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