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# Tricky!

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
2 years, 5 months ago

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Here's my solution:

soln

· 2 years, 5 months ago

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it is much simpler and sauber.thanks · 2 years, 5 months ago

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

• $$A_1$$ is the area of the green chord
• $$A_2$$ is the area of yellow square

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

$$A_1 = \text {Area of segment DPQ} - \text {Area of } \triangle DPQ = \dfrac {30^\circ}{360^\circ} \pi R^2 - \frac {1}{2} (2R \cos{75^\circ} )(R \sin {75^\circ} )$$

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

$$A_2 = (2R\cos {75^\circ})^2 = 4R^2 \cos^2 {75^\circ} = (2\cos {150^\circ} + 2)R^2 = (2-\sqrt{3})R^2$$

Therefore, $$A = 4A_1+A_2 = 4\left( \dfrac {\pi}{12} - \dfrac {1}{4} \right) R^2 + (2-\sqrt{3})R^2 = \left( \dfrac {\pi}{3} + 1 -\sqrt{3} \right) R^2 = \boxed {61.76876175}$$ · 2 years, 5 months ago

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Novel approach! · 2 years, 5 months ago

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Whose diagram are you referencing? · 2 years, 5 months ago

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Sorry, the diagram did not turn out well earlier. · 2 years, 5 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: $$f(x)=\sqrt{14^{2}-x^{2}}$$

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

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

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

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

$4\int_{7}^{7\sqrt{3}}( \sqrt{14^{2}-x^{2}}-7)dx = 4\int_{7}^{7\sqrt{3}}( \sqrt{14^{2}-x^{2}})dx-28\int_{7}^{7\sqrt{3}}dx$

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

For the first half, use the following trig substitutions:

$$x = 14\sin u$$

$$dx = 14\cos udu$$

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

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

Which turns the first half of the integral into:

$\Rightarrow 4\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}( \sqrt{14^{2}-14^{2}\sin^{2}u})(14\cos u)du=4\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}( \sqrt{14^{2}(1-\sin^{2}u}))(14\cos u)du$

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

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

The first part minus the second part gives $$196(\frac{\pi}{3}) - 196(\sqrt{3}-1) = \boxed{196(\frac{\pi}{3}-\sqrt{3}+1)}$$. · 2 years, 5 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 $$30^{\circ}$$ and sides length $$14$$. 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 $$14$$ and $$7\sqrt{2}$$ with an angle of $$15^{\circ}$$ between them. The area of each is then $$(\frac{1}{2})(14)(7\sqrt{2})\sin(15^{\circ}) = (\frac{49}{2})(\sqrt{3} - 1)$$. The area of the sector is $$(\frac{49}{3})\pi$$, and thus the area of the subregion is

$$(\frac{49}{3})\pi - 49(\sqrt{3} - 1) = (\frac{49}{3})(\pi + 3 - 3\sqrt{3})$$.

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

$$\dfrac{196}{3} * (\pi + 3 - 3\sqrt{3}) = 61.769$$ to $$3$$ decimal places.

This represents an area of about $$31.5$$% of that of the square. When the quarter-circles are drawn more accurately, this appears to the eye as a reasonable percentage coverage. · 2 years, 5 months ago

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could we use coordinate geometry to find the points of intersection and then use integration? · 2 years, 5 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. · 2 years, 5 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. · 2 years, 5 months ago

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