# The rectangular solid below has surface area of 270 in². What is the value of x? ?

### 8 Answers

- Engr. RonaldLv 71 month ago
SA = 2bh + 2bw + 2hw

270 = 2(11)x + 2(11)x + 2x^2

270 = 22x + 22x + 2x^2

270 = 44x + 2x^2

2x^2 + 44x - 270 = 0

x^2 + 22x - 135 = 0

(x - 5)(x + 27) = 0

x = 5 , x = - 27

The length of x is x = 5 in

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- PhilipLv 61 month ago
Notation: rs = rectangular solid.

Surface area of rs = [4(11x) + 2x^2] [in]^2 = 270 [in]^2.

2x^2 +44x -270 = 0.

x^2 +22x -135 = 0.

(x-5)(x+27) = 0.

Clearly, x = 5 [in] since x must be > 0.

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- KrishnamurthyLv 71 month ago
The surface area of the rectangular solid = 270 in^2

2x^2 + 4*11x = 270

2x^2 + 44x - 270 = 0

2(x^2 + 22x - 135) = 0

2(x - 5)(x + 27) = 0

x = 5

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- Ian HLv 71 month ago
2x^2 + 44x - 270 = 0 div by 2 and factor

(x + 27)(X - 5) = 0

Check using x = 5, A = 50 + 44*5 = 270

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- Anonymous1 month ago
Sweet mother of god. Hint: there are six sides.

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- llafferLv 71 month ago
If you look at this as a rectangular solid with a square base, the surface area equation becomes:

SA = 2s² + 4sh

Using SA = 270, s = x and h = 11, we get:

270 = 2x² + 4x(11)

270 = 2x² + 44x

0 = 2x² + 44x - 270

0 = x² + 22x - 135

This factors:

0 = (x + 27)(x - 5)

x = -27 and 5

We can't have a negative side length, so throwing that out:

x = 5

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- fcas80Lv 71 month ago
The 6 sides have four areas of 11x, and two areas of x^2.

4(11x) + 2x^2 = 270

2x^2 + 44x - 270 = 0

x^2 + 22x - 135 = 0

x = - 27 (reject; can't be negative), x = 5

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