Guide :

draw a circle and cut it to form a rectangle by taking equal angles so that their areas will equal

convert circle into rectangle.

Research, Knowledge and Information :

Area of a Circle by Cutting into Sectors - Math is Fun

Area of a Circle by Cutting into Sectors . Here is a way to find the formula for the area of a circle: Cut a circle into equal sectors ... Which resembles a rectangle:
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Basic Geometric Shapes: Square, Circle, Rectangle, and ...

... square, circle, rectangle, ... and they come together to form four right angles. A circle, ... take their name from the type of angles which can be found ...
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Geometry Flashcards | Quizlet

Triangles are similar if all their corresponding angles are equal and their ... The angles formed by parallel lines cut ... form a circle, so the sum of these angles ...
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The areas of a rectangle, a triangle and a circle are equal ...

They created hexagons and pentagons with these areas using their ... Draw the three shapes with equal areas ... a rectangle, a triangle and a circle are equal, ...
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Area of diagonal-generated triangles (video) | Khan Academy

... by the diagonals of a rectangle are equal. ... Area of diagonal-generated triangles. ... the areas of the different triangles. So let's do the orange ...
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Perimeter & area (video) | Area | Khan Academy

And a rectangle is a figure that has 4 sides and 4 right angles. So this is a rectangle ... equal. So let me draw a ... So the area of rectangle ABCD is equal to the ...
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Shape: Quadrilateral | Think Math!

When we talk about "dissecting" a parallelogram and rearranging the parts to form a rectangle ... their sides, angles, ... (draw a circle within a quadrilateral so ...
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Paralleograms_and_rectangles - Home - AMSI

First property of a parallelogram − The opposite angles are equal. ... the endpoints of any two diameters of a circle form a rectangle, ... so their discussion ...
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How to: Draw Shapes with the OvalShape and RectangleShape ...

How to: Draw Shapes with the OvalShape and RectangleShape Controls ... (100, 100) ' To draw a rounded rectangle, ... To draw a circle that has a custom border.
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Suggested Questions And Answer :

draw a circle and cut it to form a rectangle by taking equal angles so that their areas will equal

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area under a curve at the right of a circle

Remember that the area under a curve has to be bounded, so limits need to be applied to define the boundaries. Usually it's the area between the curve and one axis or both. Occasionally it's the area between two curves or the area produced by the intersection of two curves. Sometimes, as in the case of a circle or ellipse which already encloses an area, it's the whole or part of an area inside the curve. The "formula" is based on the area of very thin rectangles, infinitely thin, in fact, which are laid side by side to fill the area. An integral is applied which sums the areas of the rectangles over the whole region specified to get the area of that region. The most common formula is integral(ydx) where y=f(x) defines the curve. Limits of x are then given to enclose the area, so that definite integration is applied. The best way to approach any problem involving finding areas related to a curve is to sketch the curve and draw the area that needs to be found. Imagine a large curve had been drawn on the ground and you had a roll of sticky tape. You work out where the area is and cut strips of tape and stick them down so that you fill the required area. You can only lay the strips side by side, no criss-crossing and no laying the tape in a different direction. You can only cut the strips into rectangles using a cut straight across, not at an angle. You end up with not quite filling, or slightly over-filling the space because the tape will overlap the curve slightly. But when you step back it will look like the area has been properly covered by tape. If you used narrower tape the area would be even better filled. That's the principle on which the integration is based, because the area of each rectangle is the length of the strip y times the width we call dx (the width of the strips of tape). The area outside a circle must be bounded. You need the equation of the circle so that you can relate x and y. For example, the general equation of a circle is (x-h)^2/r^2 + (y-k)^2/r^2=1, where r is the radius and (h,k) is the centre. So y=k+sqrt(r^2-(x-h)^2) is half the circle, because the other half is k-sqrt(r^2-(x-h)^2).  If you need the area between the circle and the x axis between a and b, you need the lower part of the curve given by the second expression; if you need the upper part of the curve you need the first expression. If the circle is coloured red and the outside of the circle is blue, the area between the lower circle and the x axis will be entirely blue, whereas the area between the upper circle and the x axis will be red and blue. Get the picture? Once you've decided what you want, you compose the integral: integral(ydx)=integral((k-sqrt(r^2-(x-h)^2))dx) for b Read More: ...

How do you find the area of a sector?

I assume by sector you mean the equivalent of a pie or pizza slice you cut out of a whole pie or pizza. You need to know the angle of the sector and the radius (or diameter) of the circle to which it belongs. The angle at the centre of a circle is 360 degrees and the area of a circle is (pi)r^2, where (pi) is a number a little larger than 3. The area of the sector is in the same proportion to the area of the circle as the sector angle is to 360. So (area of sector)/((pi)r^2)=(angle of the sector)/360. For example, if you divide a circle into 4 equal segments, the area of each segment is (pi)r^2/4 because the angle of each segment is 90 degrees, or a right-angle. If by sector you mean the piece cut off by a chord, which is just the result of joining two points on the circle's circumference, to find the area you need to know the length of the chord and the diameter of the circle. There's no simple proportion formula for this. The formula is something like r^2(sin^-1(a/r)-(a/r)sqrt(1-(a/r)^2)), where a is half the length of the chord. If, for example, a=r/2, the sector area is r^2((pi)/6-(1/4)sqrt(3)).  
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A sphere with a diameter of 10 cm is cut by a plane 3 cm above its center. what is the area of the circle formed? (Leave answer in units of (pi))

The radius of the sphere is 10/2=5cm. The way to picture the problem is to represent the plane as a line parallel to the horizontal diameter of the sphere, represented by a circle, cutting through the circle. If you draw a perpendicular radius at right angles to this line from the centre of the circle, and another radius from where the line meets the circumference, you will have a right-angled triangle where the sloping radius, length 5, is the hypotenuse, the vertical side, length 3, and a horizontal segment of the line, length sqrt(5^-3^)=sqrt(16)=4. This line segment forms the radius of the circular cross-section cut by the plane, so the area of the cross-section is (pi)*4^2=16(pi).
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find the center of mass of the solid of uniform density biunded by the graphs of the equations

x^2+y^2=a^2 is a cylinder with its circular cross-section in the x-y plane; the cylinder is sliced at an angle by the plane z=cy, forming a wedge. y=0 is the x-z plane and z=0 is the x-y plane; these planes and the conditions that y and z are positive impose further constraints on the shape. The cylinder is sliced in half by the x-z plane so the cross-section will be a semicircle and only the upper half is required. The semicircular end rests on the z=0 plane, and doesn't extend beyond the vertex of the wedge. The length of the wedge is ac, because the slope of the angle of the wedge is tan^-1(c) to the y axis and the radius of the wedge is a. Viewed from the side, along the x axis, the wedge is a right-angled triangle with sides a (height), ac (length) and hypotenuse a√(1+c^2). In everyday terms, the wedge looks like the end of a lipstick that has been sliced across. Going back to the side view of the wedge, we can see that the flat end has a height a. As we move along the wedge towards the point the height changes and shrinks to zero when we reach the vertex. The length at a point P(y,z) is a-y. This height is the distance of a chord to the circumference, starting at height a. The first task is to work out the area of a segment. We do this by subtracting the area of a triangle from the area of a sector, given the distance of the chord from the circumference. This distance is a-y, so the distance from the centre of the circle is, conveniently, y. y/a=cosø where ø is half the angle subtended by the arc of the sector. So ø=cos^-1(y/a) and the area of the sector is given by 2ø/2π=ø/π=A/πa^2, or A=a^2ø (ø is measured in radians). The area of the isosceles triangle formed by joining the radial ends of the arc of the sector is B=aysinø=y√(a^2-y^2), and the area of the segment is A-B=a^2cos^-1(y/a)-y√(a^2-y^2). (Note that when y=0, this expression becomes πa^2/2, the area of the semicircle.) The volume of the segment is (a^2cos^-1(y/a)-y√(a^2-y^2))dz where infinitesimal dz is the thickness of the segment. The mass of the segment is directly proportional to its volume because the density is uniform, so the volume is sufficient to represent the mass. The centre of gravity (COG) of the wedge can be found by summing moments about a point G(x,y,z) which is the COG. This sum has to be zero. Since a semicircle is symmetrical in the x-y plane about the y axis, we know the x coord of G must be zero. We need G(0,g,h) where g is the height along the z axis of the COG and h the z position. We need to find the COG of a segment first. The length of the chord of a segment is 2√(a^2-y^2). Earlier we discovered this when we were finding out the areas of the sector and isosceles triangle. Consider just two dimensions. The length of the chord is one dimension and if we create a rectangle from this length and an infinitesimal width dy, the area will be 2√(a^2-y^2)dy. If we think of the rectangle as having mass, its mass is proportional to its area, so area is a sufficient measure for mass. Let's call the COG of the segment point C(x,y)=(0,q), then the moment of the rectangle about the COG=C(0,q) is mass times distance from C=2(q-y)√(a^2-y^2)dy. If we sum the rectangles over the interval y to a we have 2∫((q-y)√(a^2-y^2)dy)=0 for y≤y≤a. The interval looks strange, but it means that y is just a general value, which is determined when we return to the wedge. What we are trying to do is to find q relative to y. We need to evaluate the definite integral, so we split it: 2q∫(√(a^2-y^2)dy)-2∫(y√(a^2-y^2)dy. Call these (a) and (b). To evaluate (a), let y=asinu, then dy=acosudu; √(a^2-y^2)=acosu and the integral becomes 2a^2q∫(cos^2(u)du). Since cos(2u)=2cos^2(u)-1, cos^2(u)=½(cos(2u)+1), so we have a^2q∫((cos(2u)+1)du)=a^2q(sin(2u)/2+u). Since sin(2u)=2sinucosu, this becomes a^2q(sinucosu+u)=a^2q(y/a * √(1-(y/a)^2)+sin^-1(y/a))=qy√(a^2-y^2)+a^2qsin^-1(y/a). Now we apply the limits for y: πa^2q/2-qy√(a^2-y^2)+a^2qsin^-1(y/a). To evaluate (b), let u=a^2-y^2, then du=-2ydy and the integral is +∫√udu = (2/3)u^(3/2)=(2/3)(a^2-y^2)^(3/2). Applying the limits we have: -(2/3)(a^2-y^2)^(3/2). (a)+(b)=πa^2q/2-qy√(a^2-y^2)+a^2qsin^-1(y/a)-(2/3)(a^2-y^2)^(3/2)=0. Multiply through by 6: 3πa^2q-6qy√(a^2-y^2)+6a^2qsin^-1(y/a)-4(a^2-y^2)^(3/2)=0. From this q=4(a^2-y^2)^(3/2)/(3πa^2-6y√(a^2-y^2)+6a^2sin^-1(y/a)). Fearsome though this looks, when y=0 and z=0, q=4a^3/(3πa^2)=4a/(3π), which is in fact the COG of a semicircle. When y=a and z=ac, q=0. Strictly speaking, we should write q(y) instead of just q, showing q to be a dependent variable rather than a constant. To find the COG of the wedge, we need to find the moments of the COGs of the segments. For a segment distance y above and cy along the z axis, we have the above expression for q. So q(y) is the y coord of the COG of the segment and cy is the z coord. The x coord is 0. The mass of the segment is (a^2cos^-1(y/a)-y√(a^2-y^2))dz, as we saw earlier, and this mass can be considered to be concentrated or focussed at the COG of the segment (0,q(y),cy). The COG of the wedge is at G(0,g,h) so the distance of the mass of the segment from G is g-q(y) and h-cy in the y and z directions. The y-moment is (g-4(a^2-y^2)^(3/2)/(3πa^2-6y√(a^2-y^2)+6a^2sin^-1(y/a))(a^2cos^-1(y/a)-y√(a^2-y^2))dz and the z-moment is (h-cy)(a^2cos^-1(y/a)-y√(a^2-y^2))dz. The sums of these individual components are zero. These lead to rather complicated integrals, requiring considerably more space than is available to work out, even if I knew how to do so!
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how many 22" x 16" rectangles will I need to fill a 12' x 8' area

The area in square inches is 144*96=13824. The area of a rectangle is 22*16=352. The maximum number of rectangles is 13824/352=39.273. But if we cannot use part of a rectangle the maximum is 39. 40 rectangles would allow the area to be filled completely but it would be necessary to cut up some rectangles to fit. Since 96 is exactly divisible by 16, the 8' side of the area will allow 6 rectangles to be placed with their 16" edges forming an 8' edge. The 22" edge will not cover 12', but 6 rectangles will cover 6*22=132 out of 144 inches, leaving 12 inches short. The size of this remaining area is 12" by 96". So far we've used 6*6=36 rectangles, and the maximum is 40. If 3 more rectangles are split into two, we have 6 rectangles of size 11" by 16" which will partly fill the 12" gap along the 8' edge of the area, leaving only 1" by 8' to fill. Now we have used 39 rectangles, 3 of which were divided in half. We would need to cut 6 1" by 16" strips from another 22" by 16" rectangle to fill the remaining area. So we used 36 complete rectangles, plus another 3 cut in half, plus part of another cut into 6 1" strips, making 40 rectangles in all completely covering the area, and part of a rectangle unused.
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equilateral math equation help

The side of the square is the square root of its area=sqrt(16)=4. The area of the triangle is half the base times the height. So we need the height of the equilateral triangle. The perpendicular from vertex E bisects AD and the angle AED. The equilateral triangle now consists of two back to back right-angled congruent triangles. Angle EAD=60 (same as  the other two angles of triangle AED) so half the angle is 30 degrees. The common side of the two right-angled triangles is the height of the equilateral triangle and has length=sqrt(4^2-2^2), by Pythagoras, because the hypotenuse is length 4 (same as the side of the square) and the shortest side is 2, half of AD. The height is sqrt(12)=sqrt(4*3)=2sqrt(3). The area of the triangle is 1/2*4*2sqrt(3)=4sqrt(3)=6.93 sq units approx. Another way of finding the area is to take the two right-angled triangles and join them together along their common hypotenuse forming a rectangle of sides 2sqrt(3) and 2. The hypotenuse is a diagonal of the rectangle. The area of the rectangle (2sqrt(3)*2=4sqrt(3)=6.93) must be the same size as the area of the equilateral triangle, because the rectangle and the equilateral triangle are made up of the same two right-angled triangles.
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what is the distance of 800 sqft, 1800sqft and 1250 sqft

Distance is measured in feet (or other linear measurement) and area in square feet (or other measure of area. Linear measurements have one dimension and area measurements have two, so you can't compare them or convert one into the other. A line doesn't enclose anything, because it's just a line. An area is an enclosure like a circle, rectangle, triangle, square, and so on. To make an area out of a line you would have to bend it and join the ends together. The best we can do with the figures in the question is to bend a line to make a shape. Let's go for a 45 degree right-angled triangle. The area of this type of triangle is half the area of a square split into half along a diagonal, and the area of a square is the square of the length of the side, so the area of the triangle is half this value. Now look at the figures representing the area of the triangle. To find the area of the square we simply double the values: 1600, 3600 and 2500. The sides of the corresponding squares would be 40, 60 and 50 (40*40=1600, 60*60=3600 and 50*50=2500). So we know that the sides of each triangle forming the right angle must be 40, 60 or 50. The length of the third side of the triangle is the length of the diagonal of the square and would have respective lengths 56.57ft, 84.85ft and 70.71ft. The length of the lines forming these triangles would therefore be 136.57ft, 204.85ft and 170.71ft. These are the perimeter lengths of the triangles. So now I hope you can see the difference between feet and square feet. We could have bent our line into a circle and the line would be the circumference of the circle and the answers would be quite different. Or we could have made another shape, all giving different lengths for the lines making the shape.
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find the dimensions of the rectangle with the largest area that can be inscribed in a semi-circle with a radius of 4

A parallelogram inscribed in a circle is a rectangle or a square.   The perpendicular bisector of two opposite sides of the rectangle corresponds to a diameter of the circle. So, a rectangle inscribed in a semicircle rests on the diameter. ( Proving things mentioned above is skipped in here.) Label each vertex of the rectangle A, B, C and D counterclockwise as placing base AB on the diameter and vertices C and D on the circumference. The midpoint of AB corresponds to the circle's center O. Let angle∠AOD=x (0 Read More: ...

area of major segment of a circle

The chord and the two radii form an isosceles triangle with equal sides equal to 5cm. If the chord bisector is drawn in we get two back to back congruent right-angled triangles, and we can find the angle at the centre, because it's half the bisected angle. The bisected chord makes a side of the right-angled triangle (sqrt(50))/2=5sqrt(2)/2. The sine of the angle is 5sqrt(2)/(2*5)=sqrt(2)/2, making the angle 45 degrees. Therefore the angle at the centre is 90 degrees making the area of the minor sector (1/4)(pi)(5^2). Therefore the area of the major sector is (3/4)(25(pi))=(75/4)(22/7)=58.93 approx.
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