#Math problem for AREA of unknown space

33 messages · Page 1 of 1 (latest)

shadow ingot
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Someone solve the area for the red colored spot. I replace 8 with 4 because that was the incorrect measurement. Plus, (4,0) is at the bottom indicating both sides are the same in length

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primal jayBOT
flint radish
shadow ingot
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got it but im not exactly sure what the square root expression is for

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and the x/2

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i originally assumed those are the areas of whats inside the circle

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are those the left/right sides of the circle you are referring to?

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or is the square root expression just for the downward curve on the left side of the circle and x/2 is referring to the slope of the diagonal line?

shadow ingot
flint radish
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Are you familiar with integration?

shadow ingot
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i assume its calculus

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but i never took it

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i took like trig

flint radish
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Yeah it is. Finding the are will be tricky without it

shadow ingot
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but what are the steps to summary if u are able to answer that?

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im assuming u use integration to find the area of the other shapes

flint radish
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Split the red shape in two at the intersection of the line and semi circle.
Integrate the right half, you can just find the area normally for triangle on the left half.
The sum will be the red area

shadow ingot
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thanks

shadow ingot
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i think

flint radish
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,w x/2 = 4 - sqrt(16-(x-4)^2)

keen wedgeBOT
flint radish
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,w int 4-sqrt(16-(x-4)^2), [x, 8/5,4]

keen wedgeBOT
flint radish
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,w 0.6120 + (4/5*8/5)/2

keen wedgeBOT
flint radish
shadow ingot
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ur right i got 1.252, i literally had to redo it lol

blazing blaze
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You don't need integration to do this. Mark the centre of the circle as C and the point where the line and circle meet (at the peak of the red section) as P. Mark the point on the x-axis where the circle touches ie. (4, 0) as X. The section below the circle and right of X is a square minus a quarter circle. The minor segment between the line and the circle can be found once you join CP and find the angle and the centre. The red section is then the triangle minus the other two.

blazing blaze
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The Red Area is:

$A = \frac{8\times 4}{2} - \left[4^2-\frac{\pi(4)^2}{4}\right] - \frac{4^2}{2}\left{\left[\pi-2\tan^{-1}{\left(\frac{1}{2}\right)}\right]-\sin{\left[\pi-2\tan^{-1}{\left(\frac{1}{2}\right)}\right]}\right}$

$\quad =16- 4^2 + 4\pi - 8\pi + 16\tan^{-1}{\left(\frac{1}{2}\right)} + 16\sin{\left[\tan^{-1}{\left(\frac{1}{2}\right)\right]}}\cos{\left[\tan^{-1}{\left(\frac{1}{2}\right)\right]}}$

$\quad = 16\tan^{-1}{\left(\frac{1}{2}\right)} - 4\pi + \frac{16\times 2}{\left(\sqrt{5}\right)^2}$

$\quad = 16\tan^{-1}{\left(\frac{1}{2}\right)} - 4\pi + \frac{32}{5}$

$\quad \approx 1.251991...$