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what is area?

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Definition of Area

Math explained in easy language, plus puzzles, games, quizzes, worksheets and a forum. For K-12 kids, teachers and parents.
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Area | Define Area at

Area definition, any particular extent of space or surface; part: the dark areas in the painting; the dusty area of the room. See more.
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Area | Definition of Area by Merriam-Webster

Define area: a part or section within a larger place — area in a sentence
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Area - definition of area by The Free Dictionary

area To calculate the area of a rectangle, multiply the length by the width. The area of this rectangle is 50 square feet. ar·e·a (âr′ē-ə) n. 1. A roughly ...
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What is Area in Math? - Definition & Formula - Video & Lesson ...

Area is the size of a two-dimensional surface. This lesson will define area, give some of the most common formulas, and give examples of those...
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Area - Wikipedia

Area is the quantity that expresses the extent of a two-dimensional figure or shape, or planar lamina, in the plane. Surface area is its analog on the two-dimensional ...
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Area - Definition in Relation to Math - ThoughtCo

Here are the definition and examples of formulas used to determine area along with authentic and real life reasons why this math concept is important.
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Area - math word definition - Math Open Reference

Definition and meaning of the math word area. Includes a list of other pages that refer to this word
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Definition and examples area | define area - geometry - Free ...

Area is defined as the number of square units that covers.....complete information about the area, definition of an area, examples of an area, step by step solution ...
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What is Area? - Math is Fun Area - Maths Resources

What is Area? Area is the size of a surface! Example: These shapes all have the same area of 9: It helps to imagine how much paint would cover the shape.
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Suggested Questions And Answer :

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: ...

my fence or rectangular area will be 20x6

Define the shape of the rectangular area by establishing a relationship between the length and width of the rectangle. For example, L = 2W + 5, or W = 3L – 4. Be sure to include the appropriate units (inches, feet, yards, miles, or meters). Using the fact that A = LW, together with the relationship defined in step 2, eliminate one of the variables to set up a quadratic equation. Solve the quadratic equation using any of the techniques learned in this unit. The solution(s) will be one of the dimensions; use step 2 to find the other. Now determine the perimeter so that you will know how much fencing to buy
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area of a region between two curves?

The picture below shows the required area between the two curves: y=e^(-x^2) (red) and y=1-cos(x) (blue). The vertical blue line shows the limit of the first quadrant (x=(pi)/2). Each grid square in the picture has an area of 0.04 and the area is roughly 14 complete squares: 14*0.04=0.56 is a rough estimate of the area between the curves. The points where the curves cross is the upper limit of the enclosed area which includes the y axis. I assume that the area beyond that up to x=(pi)/2 is not required because it is not enclosed by the y axis. The intersection is when x=0.941944 approx., that is, when e^(-x^2)=1-cos(x) (x in radians). At this point y=0.411783 approx. These define the upper limit for definite integration. Using S[low,high](...) to denote integration, we integrate to get the area between the two curves: S[0,0.941944]((e^(-x^2)-1+cos(x))dx). As far as I know, there's no solution to the indefinite integral of e^(-x^2), but the above definite integral can be evaluated by approximating dx. If dx=0.01, for example, then we can sum the areas of thin rectangles of width 0.01 over the range. This is tedious but it will yield a reasonable approximation. Let f(x)=e^(-x^2)-1+cos(x). Consider a starting point represented by a rectangle of width 0.01 and height f(0). Its area is 1*0.01=0.01. Now consider a different rectangle with the same width but height f(0.01) and area=0.0099985. The two rectangles "trap" the curve between them, so the true area under the curve between the limits 0 and 0.01 is somewhere between the two rectangular areas. If we take the average of these two areas we get 0.00999925. Then we move to f(0.01) and f(0.02). The two areas this time are 0.01f(0.01) and 0.01f(0.02) and the average area is 0.005(f(0.01)+f(0.02)). A third pair of rectangles will be averaged at 0.005(f(0.02)+f(0.03)). The last pair of rectangles will be narrower than 0.01: the first has a height of f(0.94) and a width of 0.001944, the second has a height of f(0.941944) which approximates to zero (to the nearest 10^-7). So the last average area is 0.000972f(0.94). The total area between the curves and the y axis can be expressed as a series: 0.005(f(0)+f(0.01) + f(0.01)+f(0.02) + ... + f(0.92)+f(0.93) + f(0.93)+f(0.94)) + 0.000972f(0.94). As you can see, f(0.01) to f(0.93) are all repeated, so we have 0.005(f(0)+f(0.94)) + 0.000972f(0.94) + 0.01(f(0.01)+...+f(0.93)). Since f(0)=1 and f(0.94)=).00308040 approx. So the first expression is 0.005*1.00308040=0.0050154020. 0.000972f(0.94)=0.000972*0.00308040=0.000003 approx.  To make matters more manageable, I'll divide the range of values into summed groups: 0.01-0.10, 0.11-0.20, 0.21-0.30, etc., up to 0.81-0.90, then 0.91-0.94. Here are the group results (these figures have a superfluous accuracy and will be rounded off later): 0.01-0.10:  9.946463769  0.11-0.20:  9.630991473 0.21-0.30:  9.036361176 0.31-0.40:  8.183088444 0.41-0.50:  7.104969671 0.51-0.60:  5.842140126 0.61-0.70:  4.437880465 0.71-0.80:  2.935475192 0.81-0.90:  1.375462980 0.91-0.93:  0.104328514 TOTAL: 58.59716181 AREA SUBTOTAL: 0.5859716181 TOTAL AREA: 0.585972+0.005018=0.591 approx. Control summation: 5.415713508. Control area subtotal: 0.5415713508 (based on increments of 0.1 across the range 0.1 to 0.9. We would expect this figure to be approximately the same as the more accurate summation.) It appears that the area between the curves and the y axis is of the order 0.591; this figure is close to the rough estimate given near the beginning of this answer.  
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how many square feet in an area 149.3' x 60.8' x 140' x 60'

Area has only two dimensions, whereas you have multiplied four dimensions! You say the area is irregular but you haven't defined the shape, so I'm going to assume you have a 4-sided area (quadrilateral) and you have provided the perimeter measurements. If we split the irregular quadrilateral into two triangles by joining the opposite corners, we may be able to use Heron's formula to find the area of the quadrilateral by adding the areas of the two triangles together. First we need to find the semiperimeter of the two triangles. We'll call the diagonal length x. Triangle 1 has a semiperimeter, S1, of 1/2(149.3+60.8+x) and triangle 2 a semiperimeter, S2, of 1/2(140+60+x). The area of triangle 1, A1=sqrt(S1(S1-149.3)(S1-60.8)(S1-x)) and A2=sqrt(S2(S2-140)(S2-60)(S2-x)). The area of the quadrilateral is A1+A2. S1=1/2(210.1+x)=105.05+x/2 and S2=1/2(200+x)=100+x/2. A1=sqrt((105.05+x/2)(x/2-44.25)(x/2+44.25)(105.05-x/2)) A2=sqrt((100+x/2)(x/2-40)(x/2+40)(100-x/2)) ​We now have to make another assumption, because we don't know what x is and the areas depend on it. The area is almost a rectangle or parallelogram, because 210 and 200 are similar measures, as are 60 and 60.8. Let's assume that triangle 2 is a right-angled triangle, then x=sqrt(140^2+60^2) (Pythagoras), so x=152.315 and x/2=76.158 (approx values). We can now calculate A1 and A2. A1=4484.92 and A2=4200, therefore the total area is 8684.92 sq ft. Note that A2 is half the area of a rectangle with sides 140 and 60=1/2(60*140)=4200. If we assume that triangle 1 is right-angled instead, what difference does it make? Its hypotenuse, x=sqrt(149.3^2+60.8^2)=161.21 and A1=4538.72 sq ft. If we calculate A2 using Heron's formula we get A2=4141.80, so total area is 8680.52 sq ft. So the difference between the two areas is less than 4.5 sq ft. We could average the two estimates of area: 8682.7 sq ft approx.
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Find the area in sq units bounded by |3x| + |y| = 36

Find the area in sq units bounded by |3x| + |y| = 36. The modulus function takes the positive value of whatever is inside the brackets. Whatever is inside the brackets can be either positive or negative. Since we have two modulus functions, this means that we actually have 4 equations, not just one. These equations are, 3x + y = 36 3x - y = 36 -3x + y = 36 -3x - y = 36 The intersection of these four lines can be found by pairing off any two equations (provided they don't have the same gradient) and solving for x and y. The intersection points are: A(0, -36), B(12, 0), C(0, 36), D(-12, 0) These four points define a parallelogram shape which is the graph of |3x| + |y| = 36. The area enclosed by this graph, |3x| + |y| = 36, is the area of the parallelogram ABCD. This area is 4 times the area of triangle  ABO, where O is the origin (0, 0). area = (1/2)*base*height = (1/2)*12*36 = 216 Answer: enclosed area is 216 sq units
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integral from 0 to infinity of (cos x * cos(x^2)) dx

The behaviour of this function f(x)=cos(x)cos(x^2) is interesting. The integral is the area between the curve and the x axis. If the functions cos(x) and -cos(x) are plotted on the same graph, the latter form an envelope for f(x). Between x=(pi)/2 and 3(pi)/2, the curve has 4 maxima and 3 minima; between x=3(pi)/2 and 5(pi)/2 there are 7 maxima and 6 minima; between x=(2n-1)(pi)/2 and (2n+1)(pi)/2 there are 3n+1 maxima and 3n minima (integer n>0), a total of 6n+1. (These are based purely on observation, and need to be supported by sound mathematical deduction.) As n becomes large the envelope appears to fill as the extrema become closer together. As x tends to infinity n also tends to infinity. The envelope apparently has as much area above (positive) as below (negative) the x axis so the total area will be zero as the positive and negative areas cancel out. The question is: do the areas cancel out exactly? As x gets larger, the curve starts to develop irregularities and patterns, but it stays within the envelope, and positive irregularities appear to be balanced by negative irregularities, so the overall symmetry appears to be preserved. f(x)=0 when cos(x)=0 or cos(x^2)=0, which means that x=(2n-1)(pi)/2 or sqrt((2n-1)(pi)/2), where n>0. Between 3(pi)/2 and 5(pi)/2, for example, we have sqrt(3(pi)/2), sqrt(5(pi)/2), ..., sqrt(13(pi)/2), because sqrt(13(pi)/2)<3(pi)/20, and this lies between (2n-1) and (2n+1); so n is defined by 2n-1<(2m+1)^2(pi)/2<2n+1.  For m-1 we have 2(n-z)-1<(2m-1)^2(pi)/2<2(n-z)+1, where z is related to the number of zeroes in the current "batch". For example, take m=3: 2n-1<49(pi)/2<2n+1; 49(pi)/2=76.97 approx., so 2n-1=75, and n=38. Also 2(n-z)-1<25(pi)/2<2(n-z)+1 so, because 25(pi)/2=39.27 approx., 2(n-z)+1=41, n-z=20, and z=18. When m=2, 2n-1=39, n=20; 2(20-z)-1<9(pi)/2<2(20-z)+1; 2(20-z)-1=13, 20-z=7, z=13. The actual number of zeroes, Z, including the end points is 2 more than this: Z=z+2. Now we have an exact way to calculate the number of zeroes in each batch. So Z and n are both related to m. The number of extrema, E=Z-1=z+1. In fact, E=int(2(pi)(m-1)+1), where int(a) means the integer part of a, so as m increases, there are proportionately more extrema over the range (2m-1)(pi)/2 to (2m+1)(pi)/2. The figure of 6n+1 deduced earlier by observation approximates to the mathematical findings, because 2(pi) is approximately equal to 6. But we still need to show, or disprove, that the areas above and below the x axis are equal and therefore cancel out. Unfortunately, if we consider the area under the first maximum (between x=(pi)/2 and sqrt(3(pi)/2)), and the area above the first minimum (between x=sqrt(3(pi)/2) and sqrt(5(pi)/2)), they are not the same, so do not cancel out. More...
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Prove that the medians of a triangle partition the triangle into six triangles of equal area.

In a triangle ABC the points D, E and F are the midpoints of AB, BC and CA respectively. The medians intersect at the point O inside the triangle. Drawing a picture will help you to follow the argument that follows. Let Z(XYZ) mean the area of triangle XYZ, where X, Y and Z are the representative vertices of a triangle XYZ. Z(COE)=Z(EOB)=x because the triangles have the same height and the same length base because CE=EB. Similarly Z(AOF)=Z(FOC)=z and Z(BOD)=Z(DOA)=y. The values of x, y and z are defined here as a shorthand convenience. Write x, y and z in the relevant triangles and you will see that the following equations are true. Z(CAE)=2z+x=Z(EAB)=2y+x, Z(ABF)=2y+z=Z(FBC)=2x+z and Z(BCD)=2x+y=Z(DCA)=2z+y. The area of ABC is 2(x+y+z). From these equalities, 2z+x=2y+x, therefore z=y; 2y+z=2x+z, so x=y. Therefore x=y=z and the area of ABC=2(3x)=6x. In other words, the area of ABC is 6 times the area of one of the equal-area triangular partitions.
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integral sin(1/x) from 0 to 1

Integral in addition to its meaning as "antiderivative" is the area beneath a curve. The curve sin(1/x) is not undefined when x=0 because sin(infinity) is undefinable, even though the sine function ranges between -1 and 1. When x=2/(pi), sin(1/x)=1; when x=1/(pi) sin(1/x)=0. At x=1/2(pi) the function is zero again and goes to zero for every x=1/n(pi), where n is an integer. If the curve is undefined in the given range it follows that the area under the curve is not defined either. As x approaches 0, the sine wave "concertinas" with increased frequency, but with amplitude 2 (swinging between 1 and -1) bunching up towards the origin. The area fluctuates between positive and negative with increased intensity and it could be argued that the positive areas above the x axis and the negative areas below the axis cancel each other out, so the net area is zero near the origin.
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how do I calculate a perimeter

Two-sided object? A 4-sided object would be more sensible and you can find the area of such an object, but a 2-sided object would have an undefinable area. If you mean we have the length of two sides, then it does make sense and the answer would be that the object has two sides of the same length and another two sides of the same length, like a football field. We also have to assume that the object is a rectangle otherwise we can't define its area unless we know the height. If it's not a rectangle but a parallelogram (a skewed rectangle) you could find the area but not the perimeter. So it's a rectangle. Two sides of length 32 metres gives us 64m plus two of length 66m gives 64 + 132 metres which is 196 metres. The area is just the product of the two side lengths = 32*66 = 2,112 square metres.
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Guyssss please help me "Create a problem situation in which only the positive solution is meaningful" what am I going to do? I can't understand it

First, you need to understand the question, right? You need to know what gives you more than one solution. The answer is an equation which contains a polynomial of at least degree 2. That means a quadratic equation or a cubic equation, or more, that has more than one solution. A typical example is a quadratic that has a positive root and a negative root: (x-a)(x+b)=0 or x^2+x(b-a)-ab=0, where a and -b are the roots. Next we need a situation which implies that we need a positive solution only. An example is a geometrical figure like a triangle or rectangle which must, of course, have sides of positive length. Now we have to think of a situation which introduces the square of a number. In the case of geometrical figures this is likely to be the area. Now we have to invent a problem. We start with a solution. Here's an example: A rectangle has dimensions 9 by 4 so its area is 36 (never mind the dimensions at the moment). Then we create an unknown, x. Now we use x to define the dimensions of the rectangle. If we say that the length is 5 more than the width and the area is 36, then the solution is: let x=width then length=x+5, then the area is x(x+5)=36. So x^2+5x=36, or the quadratic equation: x^2+5x-36=0=(x-4)(x+9). The roots are 4 and -9, but we can reject -9. Now we have the basics of the problem, let's make it more interesting. First add dimensions. Let's use feet. Replace the rectangle with a familiar rectangular object: a box, swimming pool, a garden, etc. OK here's the question. A swimming pool's length is 5 feet more than its width. What is its perimeter if the area of the base is 36 square feet? We know how to find the width (4 feet) from the solution above, so we find the length (4+5=9 feet). The perimeter adds an extra to the problem, because it uses the length and the width. Perimeter=2(length+width)=26 feet.
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