How shall I find the area of shaded region?

It is given that the figure is a square with each side as 14 cm. and taking each side of the square as radius 4 quadrants are drawn find the area of the shaded region.

Note by Avn Bha
6 years, 4 months ago

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Here's my solution: soln soln

Guiseppi Butel - 6 years, 4 months ago

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it is much simpler and sauber.thanks

Nick Bonje - 6 years, 4 months ago

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I suggest to find the area A=4A1+A2A = 4 A_1 + A_2 , where:

  • A1A_1 is the area of the green chord
  • A2A_2 is the area of yellow square

Since ADQ\triangle ADQ and PDC\triangle PDC are equilateral triangles, PDQ\angle PDQ is  30\space 30^\circ. Let R=14R= 14 cm, then we have:

A1=Area of segment DPQArea of DPQ=30360πR212(2Rcos75)(Rsin75)A_1 = \text {Area of segment DPQ} - \text {Area of } \triangle DPQ = \dfrac {30^\circ}{360^\circ} \pi R^2 - \frac {1}{2} (2R \cos{75^\circ} )(R \sin {75^\circ} )

 =πR212R2sin75cos75=(π1212sin150)R2=(π1214)R2\quad \space = \dfrac {\pi R^2}{12} - R^2 \sin {75^\circ} \cos {75^\circ} = \left( \dfrac {\pi}{12} - \dfrac {1}{2} \sin {150^\circ} \right) R^2 = \left( \dfrac {\pi}{12} - \dfrac {1}{4} \right) R^2

A2=(2Rcos75)2=4R2cos275=(2cos150+2)R2=(23)R2A_2 = (2R\cos {75^\circ})^2 = 4R^2 \cos^2 {75^\circ} = (2\cos {150^\circ} + 2)R^2 = (2-\sqrt{3})R^2

Therefore, A=4A1+A2=4(π1214)R2+(23)R2=(π3+13)R2=61.76876175A = 4A_1+A_2 = 4\left( \dfrac {\pi}{12} - \dfrac {1}{4} \right) R^2 + (2-\sqrt{3})R^2 = \left( \dfrac {\pi}{3} + 1 -\sqrt{3} \right) R^2 = \boxed {61.76876175}

Chew-Seong Cheong - 6 years, 4 months ago

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Novel approach!

Guiseppi Butel - 6 years, 4 months ago

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Whose diagram are you referencing?

Guiseppi Butel - 6 years, 4 months ago

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Sorry, the diagram did not turn out well earlier.

Chew-Seong Cheong - 6 years, 4 months ago

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could we use coordinate geometry to find the points of intersection and then use integration?

Des O Carroll - 6 years, 4 months ago

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My first thought was to do just that, but then I came up with a purely geometric approach, which is probably what Avn is after.

Brian Charlesworth - 6 years, 4 months ago

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I would divide the shaded area up into four subregions of equal area, the dividing lines being the line segments joining the midpoints of directly opposite sides of the square. Let the origin be the lower left corner of the square, and focus on the subregion furthest from the origin. Now form a sector formed by the origin and the arc of this subregion. This sector has a central angle of 3030^{\circ} and sides length 1414. Now form two triangles joining the origin to the lower left corner of the subregion and to one of the endpoints of the arc described above. The area of the subregion will then be the area of the sector minus the areas of these two triangles.

Each of these triangles has side lengths 1414 and 727\sqrt{2} with an angle of 1515^{\circ} between them. The area of each is then (12)(14)(72)sin(15)=(492)(31)(\frac{1}{2})(14)(7\sqrt{2})\sin(15^{\circ}) = (\frac{49}{2})(\sqrt{3} - 1). The area of the sector is (493)π(\frac{49}{3})\pi, and thus the area of the subregion is

(493)π49(31)=(493)(π+333)(\frac{49}{3})\pi - 49(\sqrt{3} - 1) = (\frac{49}{3})(\pi + 3 - 3\sqrt{3}).

The area of the entire shaded region is then just four times the area of the subregion, giving an answer of

1963(π+333)=61.769\dfrac{196}{3} * (\pi + 3 - 3\sqrt{3}) = 61.769 to 33 decimal places.

This represents an area of about 31.531.5% of that of the square. When the quarter-circles are drawn more accurately, this appears to the eye as a reasonable percentage coverage.

Brian Charlesworth - 6 years, 4 months ago

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I often prefer the geometric approach, but that one was already done. So here is the calculus approach for your viewing pleasure.

First, given the nature of this figure, it's symmetrical and can be split into 4 equal parts by connecting the intersections with a vertical and horizontal line (symmetric figures are nice like that). The calculus portion focuses on the upper-right portion.

To write three of the curves as functions:

Upper-right circle: f(x)=142x2f(x)=\sqrt{14^{2}-x^{2}}

Upper-left circle: g(x)=142(x14)2g(x)=\sqrt{14^{2}-(x-14)^{2}}

Lower-right circle: h(x)=14142x2h(x)=14-\sqrt{14^{2}-x^{2}}

Staying in the domain and range of [0,14][0,14] , if you set f(x)=g(x)f(x)=g(x) you'll get (7,73)(7,7\sqrt{3}) , and from f(x)=h(x)f(x)=h(x) you'll get (73,7)(7\sqrt{3},7).

Four times the area of the upper-right portion represents the area of the shaded regions, and is represented with:

4773(142x27)dx=4773(142x2)dx28773dx4\int_{7}^{7\sqrt{3}}( \sqrt{14^{2}-x^{2}}-7)dx = 4\int_{7}^{7\sqrt{3}}( \sqrt{14^{2}-x^{2}})dx-28\int_{7}^{7\sqrt{3}}dx

The second half of that is 28(737)=196(31)28(7\sqrt{3}-7) = 196(\sqrt{3}-1).

For the first half, use the following trig substitutions:

x=14sinux = 14\sin u

dx=14cosududx = 14\cos udu

x=7u=sin112=π6x=7 \Rightarrow u=\sin^{-1}\frac{1}{2} = \frac{\pi}{6}

x=73u=sin132=π3x=7\sqrt{3} \Rightarrow u=\sin^{-1}\frac{\sqrt{3}}{2} = \frac{\pi}{3}

Which turns the first half of the integral into:

4π6π3(142142sin2u)(14cosu)du=4π6π3(142(1sin2u))(14cosu)du\Rightarrow 4\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}( \sqrt{14^{2}-14^{2}\sin^{2}u})(14\cos u)du=4\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}( \sqrt{14^{2}(1-\sin^{2}u}))(14\cos u)du

=4π6π3(142cos2u)(14cosu)du=784π6π3(cos2u)du=392π6π3(1+cos2u)du=4\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}( \sqrt{14^{2}\cos^{2}u})(14\cos u)du=784\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}( \cos^{2}u)du=392\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}(1+\cos 2u)du

=392(u+12sin2u)π6π3=196(π3)=392(u+\frac{1}{2}\sin 2u)|_\frac{\pi}{6}^\frac{\pi}{3}=196(\frac{\pi}{3})

The first part minus the second part gives 196(π3)196(31)=196(π33+1)196(\frac{\pi}{3}) - 196(\sqrt{3}-1) = \boxed{196(\frac{\pi}{3}-\sqrt{3}+1)}.

Louis W - 6 years, 4 months ago

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You can make an algorithm count the number of black pixels within the drawing. Then calculate the percentage of pixels within the shape out of the pixels that are within the square and multiply by 14^2.

Lucian Dobre - 6 years, 4 months ago

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