Find the area of the shaded region

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When we first learn to find areas by integration, we take representative rectangles vertically. The rectangles have base #dx# (a small change in #x#) and heights equal to the greater #y# (the one on upper curve) minus the lesser #y# value (the one on the lower curve). We then integrate from the smallest #x# value to the greatest #x# value. For this new problem, we could use two such intergrals (See the answer by Jim S), but it is very valuable to learn to turn our thinking #90^@#. We will take representative rectangles horiontally. The rectangles have height #dy# (a small change in #y#) and bases equal to the greater #x# (the one on rightmost curve) minus the lesser #x# value (the one on the leftmost curve). We then integrate from the smallest #y# value to the greatest #y# value. Notice the duality ## The phrase "from the smallest #x# value to the greatest #x# value." indicates that we integrate left to right. (In the direction of increasing #x# values.) The phrase "from the smallest #y# value to the greatest #y# value." indicates that we integrate bottom to top. (In the direction of increasing #y# values.) Here is a picture of the region with a small rectangle indicated: The area is #int_1^2 (y-1/y^2) dy = 1# #x=1/y^2# #y^2=1/x# #y=sqrtx/x# (we can see from the graph) #sqrtx/x=x# ## #x^2=sqrtx# ## #x^4-x=0# ## #x(x^3-1)=0# ## #x=1# (we can also see from the graph) One of many ways the area of the shaded region can be expressed could be as the area of the triangle #AhatOB=...

Area of given figure = Area of ABGE + Area of GCFD Area of ABGE : Area of rectangle = Length ⋅ Width length BE = 6 m, width GE = 2 m Area of ABGE = 6(2) = 12 m 2 Area of GCFD : Area of rectangle = Length⋅ Width length CD = 6 m, width FD = 2 m Area of GCFD = 6(2) = 12 m 2 Area of shaded portion = 12 + 12 = 24 m 2 Example 2 : Find the area of the shaded portion Area of shaded portion = Area of rectangle ABCD - Area of square GEHF Area of ABCD : Area of rectangle = Length ⋅ Width Length AB = 20 cm Width AC = 16 cm Area of ABCD = 20 (16) = 320 cm 2 Area of GEHF : Area of square (GEHF) = side ⋅ side Length GE = 6 cm Area of square GEHF = 6 ⋅ 6 Area of square (GEHF) = 36 cm 2 Area of shaded region = 320 - 36 = 284 cm 2 Example 3 : Find the area of the shaded portion

This compilation of meticulously crafted printable area of compound shapes worksheets for students of 6th grade, 7th grade, and 8th grade extends two levels of composite figures to prep up finding the area. Simple plane shapes like triangles, rectangles, squares, parallelograms, rhombus, trapezoids, circles, semicircles, and quadrants compose the figures in our pdf area of composite shapes worksheets. The combinations include two or more overlapping and non-overlapping shapes with whole-number and decimal dimensions. Free worksheets are also included.

Use the formula for area of a triangle: #a = (b xx h)/2# #a = (25 xx 25)/2# #a = 312.5# Therefore, the area of the shaded half square is 312.5 square centimetres. I could be wrong about it being impossible to find the area of the complete shaded area. I have flagged my answer so that other contributors can think about it. Hopefully this helps. The area of the green space can be calculated as the difference between the area of the square and the area of the semicircle. The area of the square is easy to calculate as we are given the side length, and we know the area of the semicircle is half the area of the circle with the same radius. Then, to solve the problem, the main task is to find #r#. The picture seems to indicate that the diagonal of the square is tangent to the semicircle. We will operate based on that assumption. We can label the picture as follows: Imposing the image on a coordinate plane with the lower left corner of the square at #(0,0)#, let the coordinate for #A# be #(x_0,y_0)#. Noting that the diagonal of the square is a segment of the line #y=x# we can rewrite #A=(x_0,x_0)#. Because the diagonal is tangent to the circle and #bar(AC)# is a radius of the circle, their intersection forms a #90^@# angle, and thus we have #angleOAC = 90^@# Then, by symmetry, #triangleABC# is a #45-45-90# right triangle, meaning #bar(AB) = bar(BC)#. But we know #bar(AB)=x_0# from the coordinates of #A#, and so, applying symmetry again, we have #bar(BC) = bar(OB) = x_0#. As the su...

The basic idea is right. First calculate the area of the sector by observing that $\dfrac$. Then find the area of the triangle by noting that the triangle is equilateral, and subtract it off. This will give you the white region. Now subtract the white region from the area of the circle. Area of white strip=Area of sector subtending$\frac$(12)$^2$=36$\sqrt3=$62.35 sq units Area of white strip=75.36 -62.35=13.01 sq units Area of circle=$\pi r^2=3.14\times12\times12=452.16$sq units Shaded area=Area of circle-Area of white strip=452.16-13.01=439.15 sq units Notice, the area of the equilateral triangle (indicated by red) $$=\frac$$

This is a partial answer using geometry. Draw a chord $GH$. The area in question is the area of circular segment $GHB$ minus circular segment $GH$ of the larger circle. The area of circular segment is the area of sector minus area of isosceles triangle. All we need to find is the angles $GEH$ and $GAH$. Note that we know all sides in $\triangle GAE$ therefore we can use law of cosine to find $\cos \angle GEA=\frac$. I'll leave the remaining calculations to you. I have thought of another way to calulate the shaded area, without using any calculus. First let's draw lines from the bottom right corner to the intersection points, lines from the center of the square to the intersection points, and finally a line connecting the center of the square to the bottom right corner. We obtained two similar triangles, and we actually know all of their sidelengths, since the greater circle has radius $10$ cm, the smaller circle has radius $5$ cm, and the center of the two circles are $5 \sqrt \approx \\ \approx 30.24 + 2\times 21.28 - 48.67 = 24.13.$$ So the area of the shaded region is approximately $24.13$ $cm^2$. Let me know if I made a mistake anywhere. You could probably generalize this to a square with a different sidelength as well. I'm not sure if you are familiar with definite integrals, but this sort of problem can easily be solved with them. This is how I would go about it: First place the construction into the $(x,y)$ plane, maybe like this (you can place the origin elsewhere,...

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