The length of a rectangular solid is three times the width. Find the volume and the length of its diagonal if the total surface area is 198 square inch
L = 3W
L * W = 3W * W = 198
Solve for L and W, then use Pythagorean Theorem.
L^2 + W^2 = Diagonal^2
The length of a rectangular solid is three times the width and the height is twice the width. Find the volume and the length of its
The length of a rectangular solid is three times the width. Find the volume and the length of its diagonal if the total surface area is 198 square inch
To find the volume and length of the diagonal, we first need to find the dimensions of the rectangular solid. Let's assign a variable to the width, say "w" in inches.
Given that the length is three times the width, we can write the length as "3w" inches.
Now, let's calculate the total surface area. The formula for the surface area of a rectangular solid is 2lw + 2lh + 2wh, where l is the length, w is the width, and h is the height.
In this case, we don't have the height, but since we only need the surface area, we can assume the height is 1 inch. So, the surface area formula simplifies to 2lw + 2lh + 2wh = 198 square inches.
Substituting the given values, we get:
2(3w)(w) + 2(3w)(1) + 2(w)(1) = 198
Simplifying further:
6w^2 + 6w + 2w = 198
6w^2 + 8w = 198
3w^2 + 4w - 99 = 0
To solve this quadratic equation, we can use the quadratic formula:
w = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 3, b = 4, and c = -99. Substituting these values, we get:
w = (-4 ± √(4^2 - 4(3)(-99))) / (2(3))
w = (-4 ± √(16 + 1188)) / 6
w = (-4 ± √1204) / 6
Now, we have two possible values for the width: w₁ = (-4 + √1204) / 6 and w₂ = (-4 - √1204) / 6. Since we cannot have a negative width, we discard w₂.
Using w = (-4 + √1204) / 6, we can find the length and calculate the volume.
Length:
l = 3w = 3((-4 + √1204) / 6) = (-12 + 3√301) / 6 = (-2 + √301) / 2
Volume:
V = lwh = ((-2 + √301) / 2) * ((-4 + √1204) / 6) * 1
= ((-2 + √301)(-4 + √1204)) / 12
To find the length of the diagonal, d, we can use the Pythagorean theorem:
d^2 = l^2 + w^2 + h^2
Since h = 1, we have:
d^2 = ((-2 + √301) / 2)^2 + ((-4 + √1204) / 6)^2 + 1
= ((-2 + √301)^2 + (-4 + √1204)^2 + 6) / 36
Finally, we take the square root to find the length of the diagonal:
d = √((((-2 + √301)^2 + (-4 + √1204)^2 + 6) / 36)