5.2e Area, Volume, and Temperature – Composition of Shapes


5.2e Area, Volume, and Temperature – Area of a Circle. If we find that we do not have a formula for a particular shape we encounter, we can divide it up into shapes we do know. Shapes that are attached to each other and are all shaded are things that we are adding together. Shapes that are cut out of other shapes are things that are subtracted. We will see this in example one and two. Example one, shows a shape that has an irregular form. We need to divide the shape up into shapes we can recognize. The first of which is a triangle. We can see that a triangle forms at the top of the shape. Secondly, we see that there is a rectangle that forms below it. And lastly, we see that there is half of a circle. Let’s start with the triangle. The area of a triangle is one-half base times height, which means we need to find the base and the height. It seems as if we have neither of these values but, then you need to assess what values are given and see what you can determine from the shape. As you can see, the bottom of this shape says that it is nine inches. This means that the opposite side of a rectangle would also be nine inches. Therefore, the base of this triangle is nine inches. Similarly, we can see that this point that is a dotted line is six inches. We can also see that a similar vertical line is four inches on the other end. If you consider that it would be four inches for this portion of the shape that would mean that the second portion of the shape would have to be two inches, in order to add to the 6-inch total. This, therefore, gives us our height. We can now fill in the base of nine inches, and the height of two inches. This will give us one-half times nine times two, which will result in nine inches squared. Next, let us look at the rectangle. The rectangle’s formula is length times width. We must determine where the length and the width are. We see that the length is nine inches. We can also see that the width is four inches. We can now plug these into our formula in which we do length or nine inches times width or four inches, which will give us thirty-six inches squared. We now have found two of the shapes. The last shape is half of a circle. Since we only have the formula for a circle, we then look at it. We know that the formula of a circle is pi r squared. If we would only like half of this, we can therefore divide it by two and then we know half of a circle would be one-half of the pi r squared. We now can use this formula to determine the area of the half circle. We have one-half times pi times the radius squared. We must determine the radius of this half circle. If we look at the shape, we see that the whole diameter is four inches. Remembering that a radius is the diameter divided by two will help us determine the radius of the circle, which is four inches divided two or two inches. Therefore, the radius equals two inches. We plus this in and we can now finish solving for the half circle. We find that one-half times pi and two inches give us one-half pi and four inches squared. Half of four inches squared is two inches squared that we now need to multiply by pi, which is three point one four. When we multiply these together, we find the answer to be 6.28 inches squared. We have now found the area of the half circle. Since we have the area of the triangle, the rectangle, and the half circle and they are all filled in, we want to keep all of these areas in our total. Therefore, our total area is going to be the triangle, or nine inches squared plus the rectangle, which is thirty-six inches squared plus the half circle which is 6.28 inches squared. When we add all of these together, we find that the total area is 51.28 inches squared. In example two, we have a similar problem and that there are multiple shapes that we must determine the area that is shaded. As you can see, there is one large circle that I am labeling in red. There are then two smaller circles that are labeled in blue. These circles are the same size. We therefore must first find the bigger circle or the one in red. The area for the circle, remember, is pi r squared which means we need to find the radius. Remembering that the radius is the distance from the center to the edge, we see that we already have a value that measures the large circle from the center to the edge, and that it is two millimeters. We can then see that we will have pi times four millimeters squared. We will then replace pi with its value of 3.14 and multiply it by the four millimeters squared. When we multiply the two of these together, we find that the big circle equals 12.56 millimeters squared. This would be the area of the shape if the entire shape was shaded in. Since there is white portions or portions that are un-shaded, we’re going to need to find their size to be able to subtract them out. We therefore, need to find the small circle’s area. Once again, we will have pi r squared. We then need to determine the r, if we look at the small circle, we see that we only have a distance that touches edge to edge or goes all the way across. This means that this is the diameter. So the diameter, remember, is from side to side and so it is two millimeters. We would like to find the radius which is the diameter divided by two. This means we have to do two millimeters and divide it by two to find that the radius is one millimeters of the small circle. We can place this into our formula, and see that we have pi times one millimeters squared. When we replace pi with three point fourteen and do the multiplication times the one millimeters squared, we find that we get three point one four millimeters squared for each small circle. Since there are two, we will need to add them together. We find that there is 6.28 millimeters squared that is taken up by the white circles. We must therefore subtract this to get our total area, taking the large circle and subtracting out the smaller circles. This will give us the area that is shaded, which is 6.28 millimeters squared. Remembering shapes that have pieces cut out of them, you must use subtraction. If there are no pieces cut out, then all of the shapes are added together.

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