Which of the following completes the proof? (6 points)
By the midpoint formula
By definition of congruence
Given
By construction
Answer:
1192 ft²
Step-by-step explanation:
Figure 3 is a trapezoidal prism.
The total surface area of a trapezoidal prism is made up of 2 congruent trapezoid bases and 4 rectangular faces connecting the bases.
The formula for the area of a trapezoid is:
where a and b are the bases, and h is the height.
From observation of the given diagram, the bases are 16 ft and 19 ft, and the height is 12 ft. Therefore, the area of each trapezoid base is:
To calculate the areas of all the rectangular faces, we first need to calculate the slant (s) of the trapezoid base by using the Pythagoras Theorem:
The area of a rectangle is the product of its width and length.
Therefore, the sum of the areas of the rectangular faces is:
To find the total surface area of the given trapezoidal prism, sum the area of the two trapezoid bases and the area of the rectangular faces:
Therefore, the total surface area of the given trapezoidal prism is 1192 ft², rounded to the nearest foot.
The width of the garden is 6 feet. We found this by expressing the length in terms of width, substituting into the perimeter equation, simplifying to find the value of width.
In solving this problem, we will follow a few simple steps. First, we know that the perimeter of a rectangle is given by the formula: Perimeter = 2*length + 2*width. We know from the problem that the perimeter is 48 feet and the length of the garden is 6 feet longer than 2 times its width.
Let's denote the width as 'w'. Then, the length would be 2w + 6. Substituting these into the perimeter equation, we have: 48 = 2*(2w+6) + 2*w. Simplifying this equation gives 48 = 4w + 12 + 2w, which further simplifies to 48 = 6w + 12. If we now deduct 12 from both sides, we have: 36 = 6w. Finally, dividing by 6 gives us the width: w = 6 feet.
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14. (2x – 1)(x + 7) = 0
Answer: The required intersection is the line 'VS'.
Step-by-step explanation: We are given to find the intersection of the planes WZVS and plane STUV.
We know that,
If two planes intersect each other, then their intersection is a straight line.
As shown in the given figure, the point 'V' and 'S' lie on both the planes WZVS and STUV.
So, the line joining these two points, i.e., the line VS also lie on both the planes.
Therefore, the intersection of both the planes is the line 'VS'.
Thus, the required intersection is the line 'VS'.