Please help. Find the area of the following shape
1 answer:
There are several ways to answer this. All involve finding a way to calculate the area of shapes we're familiar with and using those areas to find the area of this unusual shape. I've included three different ways, all of which yield the same total area.
In the first case, you cut the shape into two shapes by drawing a perpendicular line from point C to segment AE. That will give you a square and a trapezoid. The area of the square is (2 m)(5 m) = 10 m², and the area of the trapezoid is (0.5)(9 m - 5 m)(4 m + 4 m - 2 m) = 12 m². So the area of the entire shape is 10 m² + 12 m² = 22 m².
In the second case, you cut the shape into two shapes by drawing a perpendicular line from point C to segment AB. That will give you a rectangle and a triangle. The area of the rectangle is (2 m)(9 m) = 18 m². The area of the triangle is (0.5)(4 m - 2 m)(9 m - 5 m) = 4 m². So the area of the entire shape is 18 m² + 4 m² = 22 m².
In the third case, you can imagine that this shape is a piece of a larger rectangle with sides 4 m and 9 m with an area of 36 m². The area of this shape would be the difference between 36 m² and the area of the imaginary trapezoid that fills in rest of the rectangle. That trapezoid would have an area of (0.5)(4 m - 2 m)(9 m + 5 m) = 14 m². So the area of the shape given would be 36 m² - 14 m² = 22 m².
In any case, the area of the shape is 22 m².
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Answer:
34.43
Step-by-step explanation:
A differential of length in terms of t will be ...
dL(t) = √(x'(t)^2 +y'(t)^2)
where ...
x'(t) = 4cos(4t)
y'(t) = 7cos(7t)
The length of c(t) will be the integral of this differential on the interval [0, 2π].
Dividing that interval into 10 equal pieces means each one has a width of (2π)/10 = π/5. The midpoint of pieces numbered 1 to 10 will be ...
(π/5)(n -1/2), so the area of the piece will be ...
sub-interval area ≈ (π/5)·dL((π/5)(n -1/2))
It is convenient to let a spreadsheet or graphing calculator do the function evaluation and summing of areas.
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The attachment shows the curve c(t) whose length we are estimating (red), and the differential length function (blue) we are integrating. We use the function p(n) to compute the midpoint of the sub-interval. The sum of sub-interval areas is shown as 34.43.
The length of the curve is estimated to be 34.43.
