SOLUTION: Find the area of the region in the xy - plane under the line y = x/2 above the x-axis, and between the lines x=8 and x=24 . find area?

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Question 763107: Find the area of the region in the xy - plane under the line y = x/2 above the x-axis, and between the lines x=8 and x=24 . find area?
Found 3 solutions by Alan3354, josgarithmetic, MathLover1:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
Find the area of the region in the xy - plane under the line y = x/2 above the x-axis, and between the lines x=8 and x=24
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That makes a rectangle and a right triangle.
The rectangle is 16 by 4.
The triangle's base is 16 and its height is 8.

Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
TRAPEZOID! Base is 24-8=16 and average height is %288%2F2%2B24%2F2%29%281%2F2%29=%284%2B12%29%281%2F2%29=8

Area is {base*AverageHeight}
16%2A8 square units.

Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
Find the area of the region in the xy - plane under the line y+=+x%2F2 above the x-axis, and between the lines x=8 and x=24


the region in the xy - plane is right-angular trapezoid and its area is:
A=%281%2F2%29%28h%5B1%5D%2Bh%5B2%5D%29%2Aa
in this case a=24-8=16, h%5B1%5D=4, and h%5B2%5D=12 because if x=24 then y+=+24%2F2=12
A=%281%2F2%29%284%2B12%29%2A16
A=%281%2Fcross%282%29%29%2816%29%2Across%2816%298
A=16%2A8
A=128