Surface Area 135.5: Solve for X in Cuboid Dimensions (X+5, X+2, X+3)

To solve this problem, let's determine the value of $X$ using the given dimensions of the cuboid and its surface area:

• The dimensions of the cuboid are $X+5$, $X+2$, and $X+3$.
• Surface area formula: $\text{SA} = 2(lw + lh + wh)$.
• Substitute $l = X+5$, $w = X+2$, and $h = X+3$ into the equation to get:

Surface Area, $\text{SA} = 2((X+5)(X+2) + (X+5)(X+3) + (X+2)(X+3)) = 135.5$.

First, simplify each term separately:

• $(X+5)(X+2) = X^2 + 2X + 5X + 10 = X^2 + 7X + 10$.
• $(X+5)(X+3) = X^2 + 3X + 5X + 15 = X^2 + 8X + 15$.
• $(X+2)(X+3) = X^2 + 3X + 2X + 6 = X^2 + 5X + 6$.

Next, substitute these into the surface area formula:

$2\left((X^2 + 7X + 10) + (X^2 + 8X + 15) + (X^2 + 5X + 6)\right) = 135.5$

Combine like terms:

$2(3X^2 + 20X + 31) = 135.5$

Distribute the 2:

$6X^2 + 40X + 62 = 135.5$

Subtract 135.5 from both sides to set the equation to zero:

$6X^2 + 40X + 62 - 135.5 = 0$

Simplify to:

$6X^2 + 40X - 73.5 = 0$

Now, solve this quadratic equation using the quadratic formula: $X = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a}$.

Here, $a = 6$, $b = 40$, $c = -73.5$.

Calculate the discriminant:

$b^2 - 4ac = 40^2 - 4 \cdot 6 \cdot (-73.5)$

$= 1600 + 1764 = 3364$

Taking the square root of the discriminant:

$\sqrt{3364} = 58$

Now solve for $X$:

$X = \frac{{-40 \pm 58}}{12}$

Calculate the two possible values:

$X_1 = \frac{{-40 + 58}}{12} = \frac{18}{12} = 1.5$

$X_2 = \frac{{-40 - 58}}{12}$ (which results in a negative and thus non-viable solution given the dimensions context).

Only the positive value $X = 1.5$ makes sense in the context of cuboid dimensions.

Therefore, the solution to the problem is $X = 1.5$.