MHB Therefore, the volume of the rectangular prism is 84 units cubed.

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The discussion centers on calculating the volume of a rectangular prism with a surface area of 136 square units and edge measurements of 6 and 2. The volume is confirmed to be 84 cubic units. The dimensions are defined as length, width, and height, with the height calculated to be 7 units using the surface area formula. The final volume formula applied is V = length × width × height, resulting in 6 × 2 × 7 = 84. The problem-solving process highlights the importance of understanding the relationships between surface area and volume in geometry.
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What is the answer to this question. The surface area of a rectangular prism is 136 square units. Some edge measurements are 6 and 2.What is the volume?
 
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Re: Math homework

682357 said:
What is the answer to this question. The surface area of a rectangular prism is 136 square units. Some edge measurements are 6 and 2.What is the volume?
We are not here to do your homework. We can, however, guide you through.

So what have you been able to do with this problem so far?

-Dan
 
Re: Math homework

I got the answer 84units3 (or cubed)
 
Re: Math homework

682357 said:
What is the answer to this question. The surface area of a rectangular prism is 136 square units. Some edge measurements are 6 and 2.What is the volume?

682357 said:
I got the answer 84units3 (or cubed)

The dimensions are length $\ell$, width $w$ and height $h$. Suppose we know $\ell$ and $w$ to be 6 and 2 respectively. A rectangular prism (cuboid) has 3 pairs of matching faces, and it surface area $A$ is given by:

$$A=2(\ell w+\ell h+wh)$$

Using the given surface area and $\ell$ and $w$, we may state:

$$136=2(6\cdot2+6h+2h)$$

$$68=12+8h$$

$$h=7$$

And so the volume $V$ (in units cubed) is given by:

$$V=\ell wh=6\cdot2\cdot7=84\quad\checkmark$$
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...

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