The base of a cube is horizontal. A plane cuts through the cube, creating a pentagon-shaped cross section. The plane is _1_ and
may pass through a pair of adjacent _2_ of the base. Options for 1: a.vertical b.slanting Options for 2: a.vertices b.sides
1 answer:
For this problem, imagine a cube laying flat on the floor. If you zoom in on this, it would show a full square. The cutting plane should be able to form a cross-sectional shape of a pentagon. This is a closed figure comprising of 5 sides. The cutting plane should not be vertical. If it was, then it would only for a linear cross section. If it is horizontal, it would superimpose on the square base. So, if the cutting plane is slanting, it would form a pentagon cross section as shown in the right side of the picture attached. Then, it would pass between two adjacent vertices, and three adjacent sides. So, the complete statement would be:
<span>The base of a cube is horizontal. A plane cuts through the cube, creating a pentagon-shaped cross section. The plane is slanting and may pass through a pair of adjacent vertices of the base.</span>
<span>The base of a cube is horizontal. A plane cuts through the cube, creating a pentagon-shaped cross section. The plane is slanting and may pass through a pair of adjacent vertices of the base.</span>
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400 meters
Answer:
-2, 8/3
Step-by-step explanation:
You can consider the area to be that of a trapezoid with parallel bases f(a) and f(4), and width (4-a). The area of that trapezoid is ...
A = (1/2)(f(a) +f(4))(4 -a)
= (1/2)((3a -1) +(3·4 -1))(4 -a)
= (1/2)(3a +10)(4 -a)
We want this area to be 12, so we can substitute that value for A and solve for "a".
12 = (1/2)(3a +10)(4 -a)
24 = (3a +10)(4 -a) = -3a² +2a +40
3a² -2a -16 = 0 . . . . . . subtract the right side
(3a -8)(a +2) = 0 . . . . . factor
Values of "a" that make these factors zero are ...
a = 8/3, a = -2
The values of "a" that make the area under the curve equal to 12 are -2 and 8/3.
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<em>Alternate solution</em>
The attachment shows a solution using the numerical integration function of a graphing calculator. The area under the curve of function f(x) on the interval [a, 4] is the integral of f(x) on that interval. Perhaps confusingly, we have called that area f(a). As we have seen above, the area is a quadratic function of "a". I find it convenient to use a calculator's functions to solve problems like this where possible.
