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Perimeter and Area Page 1 of 57
PERIMETER AND AREA
Objectives: After completing this section, you should be able to do the following:
- Calculate the area of given geometric figures.
- Calculate the perimeter of given geometric figures.
- Use the Pythagorean Theorem to find the lengths of a side of a right triangle.
- Solve word problems involving perimeter, area, and/or right triangles.
Vocabulary: As you read, you should be looking for the following vocabulary words and their definitions:
- polygon
- perimeter
- area
- trapezoid
- parallelogram
- triangle
- rectangle
- circle
- circumference
- radius
- diameter
- legs (of a right triangle)
- hypotenuse
Formulas: You should be looking for the following formulas as you read:
- area of a rectangle
- area of a parallelogram
- area of a trapezoid
- area of a triangle
- Heron’s Formula (for area of a triangle)
- circumference of a circle
- area of a circle
- Pythagorean Theorem
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We are going to start our study of geometry with two-dimensional figures. We will look at the one-dimensional distance around the figure and the two- dimensional space covered by the figure.
The perimeter of a shape is defined as the distance around the shape. Since we usually discuss the perimeter of polygons (closed plane figures whose sides are straight line segment), we are able to calculate perimeter by just adding up the lengths of each of the sides. When we talk about the perimeter of a circle, we call it by the special name of circumference. Since we don’t have straight sides to add up for the circumference (perimeter) of a circle, we have a formula for calculating this.
Example 1: Find the perimeter of the figure below
Solution: It is tempting to just start adding of the numbers given together, but that will not give us the perimeter. The reason that it will not is that this figure has SIX sides and we are only given four numbers. We must first determine the lengths of the two sides that are not labeled before we can find the perimeter. Let’s look at the figure again to find the lengths of the other sides.
Circumference (Perimeter) of a Circle
C = 2 π r
r = radius of the circle π = the number that is approximated by 3.
perimeter
circumference
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The area of a shape is defined as the number of square units that cover a closed figure. For most of the shape that we will be dealing with there is a formula for calculating the area. In some cases, our shapes will be made up of more than a single shape. In calculating the area of such shapes, we can just add the area of each of the single shapes together.
We will start with the formula for the area of a rectangle. Recall that a rectangle is a quadrilateral with opposite sides parallel and right interior angles.
Example 2: Find the area of the figure below
Solution: This figure is not a single rectangle. It can, however, be broken up into two rectangles. We then will need to find the area of each of the rectangles and add them together to calculate the area of the whole figure.
There is more than one way to break this figure into rectangles. We will only illustrate one below.
Area of a Rectangle
A = bh
b = the base of the rectangle h = the height of the rectangle
rectangle
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We have shown above that we can break the shape up into a red rectangle (figure on left) and a green rectangle (figure on right). We have the lengths of both sides of the red rectangle. It does not matter which one we call the base and which we call the height. The area of the red rectangle is A= bh= 4 × 14 = 56
We have to do a little more work to find the area of the green rectangle. We know that the length of one of the sides is 8 units. We had to find the length of the other side of the green rectangle when we calculated the perimeter in Example 1 above. Its length was 7 units.
Thus the area of the green rectangle is A = bh= 8 × 7 = 56. Thus the area of the whole figure is area ofredrectangle+ areaofgreenrectangle= 56 + 56 = 112. In
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way to figure out the lengths of the other two sides that are not given.
Our next formula will be for the area of a trapezoid.
A trapezoid is a quadrilateral that has one pair of sides which are parallel. These two sides are called the bases of the trapezoid. The height of a trapezoid is a segment that connects the one base of the trapezoid and the other base of the trapezoid and is perpendicular to both of the bases.
Example 4: Find the area of the figure
Solution: For this trapezoid, the bases are shown as the top and the bottom of the figure. The lengths of these sides are 45 and 121 units. It does not matter which of these we say is b 1 and which is b 2. The height of the trapezoid is 20 units. When we plug all this into the formula, we get ( ) ( 121 45 ) 20 1660 2
A = b 1 +b 2 h= + = square
units.
Area of a Trapezoid
A (b 1 b 2 ) h 2
=^1 +
b 1 = the one base of the trapezoid b 2 = the other base of the trapezoid h = the height of the trapezoid
trapezoid
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Our next formulas will be for finding the area of a triangle (a three-sided polygon). We will have more than one formula for this since there are different situations that can come up which will require different formulas
The height of a triangle is the perpendicular distance from any vertex of a triangle to the side opposite that vertex. In other words the height of triangle is a segment that goes from the vertex of the triangle opposite the base to the base (or an extension of the base) that is perpendicular to the base (or an extension of the base). Notice that in this description of the height of a triangle, we had to include the words “or an extension of the base”. This is required because the height of a triangle does not always fall within the sides of the triangle. Another thing to note is that any side of the triangle can be a base. You want to pick the base so that you will have the length of the base and also the length of the height to that base. The base does not need to be the bottom of the triangle.
You will notice that we can still find the area of a triangle if we don’t have its height. This can be done in the case where we have the lengths of all the sides of the triangle. In this case, we would use Heron’s formula.
Area of a Triangle
For a triangle with a base and height
A bh 2
=^1
b = the base of the triangle h = the height of the triangle
Heron’s Formula for a triangle with only sides A = s(s−a)(s−b)(s− c ) a = one side of the triangle b = another side of the triangle c = the third side of the triangle s = ½ ( a + b + c)
triangle
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Solution: In this figure there are two dashed lines. One of them is extended from the side of the triangle that has a length of 1.7. That dashed line is show to be perpendicular to the dashed line that has a length of 2.6. This is how we indicate that the dashed line that has a length of 2.6 is the height of the triangle and the base of the triangle is the side of length 1.7. The height here is outside of the triangle. Also, the dashed line that is extended from the base is not part of the triangle and its length is not relevant to finding the area of the triagnle. Since we have both the height and the base of this triangle, we can just plug these numbers into the formula to get ( 1. 7 )( 2. 6 ) 2. 21 2
A=^1 bh= = square units.
Example 7: Find the area of the figure.
Solution: You should notice that we do not have a height for this triangle. This means that we cannot use the formula that we have been using to find the area of this triangle. We do have the length of all three sides of the triangle. This means that we can use Heron’s Formula to find the area of this triangle.
For this formula, it does not matter which side we label a, b, or c. For our purposes, we will let a be 6, b be 7, and c be
- Now that we have a, b, and c, we need to calculate s so that we can plug a, b, c , and s into the formula. We get
Heron’s Formula
A = s(s−a)(s−b)(s− c ) a = one side of the triangle b = another side of the triangle c = the third side of the triangle s = ½ ( a + b + c)
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s = 1
(^6 +^7 +^8 )=^1
(21) = 10.5. Now we can plug everything in to
Heron’s formula to find the area of this triangle to be A= s( s−a)(s−b)(s− c)= 10. 5 ( 10. 5 − 6 )( 10. 5 − 7 )( 10. 5 − 8 )= 20. 33315257 square units.
Another formula that we are interested in is the Pythagorean Theorem. This applies to only right triangles. The Pythagorean Theorem relates the lengths of the sides of a right triangle.
When using the Pythagorean Theorem, it is important to make sure that we always use the legs of the triangle for a and b and the hypotenuse for c.
Example 8: Find the length of the third side of the triangle below.
Solution: The figure is a right triangle (as indicated by the box in one of the angles of the triangle). We need to decide what the side we are looking for is in terms of a leg or the hypotenuse of the triangle. The hypotenuse is the side of the triangle opposite the right angle. That would be the side that has length 26 in our picture. Thus c will be 26 in our formula. This means that the other two sides of
Pythagorean Theorem
a^2 + b^2 = c^2
a = leg (one side of the triangle that makes up the right angle) b = leg (another side of the triangle that makes up the right angle) c = hypotenuse (side opposite the right angle)
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Circumference (Perimeter) of a Circle
C = 2 πr
r = radius of the circle π = the number that is approximated by 3.
Example 9: Find the perimeter and area of the circle below.
Solution: The number 11 in the figure above is the length of the diameter of this circle. We need the radius to be able to use the formulas. The
radius is half of the diameter. Thus r = 1
(diameter) = 1 2
We are now ready to find both the perimeter (circumference) and
area by plugging 5.5 into each formula for r.
Perimeter:
C = 2 πr = 2 ( π)( 5. 5 ) = 11 π ≈ 34. 557519
The answer 11π units is an exact answer for the perimeter. The answer 34.557519 units is an approximation.
Area:
A =πr^2 =π( 5. 5 )^2 = 30. 25 π≈ 95. 0331778 The answer 30.25π
square units is an exact answer for the area. The answer 95.0331778 square units is an approximation.
Example 10: Find the perimeter and area of the semicircle below.
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Solution: In this problem, we have a semicircle (a half of a circle). The diameter of this half circle is 120. Thus the radius is ( 120 ) 60 2
r =^1 =.
Perimeter: We now can find the perimeter. We will find the circumference of the whole circle and then divide it by 2 since we only have
half of a circle. This will give us C = 2 π ( 60 )= 120 π as the
circumference for the whole circle and 60π units as the circumference of half of the circle.
Unfortunately this is not the answer for the perimeter of the figure given. The circumference of a circle is the curved part. The straight line segment in our figure would not have been part of the circumference of the whole circle, and thus it is not included as part of half of the circumference. We only have the red part shown in the picture below.
We still need to include the straight segment that is 120 units long. Thus the whole perimeter of the figure is
curved part+straightpart= 60 π + 120 ≈ 308. 495559 units.
Area:
We can also find the area. Here we will plug 60 in for r in the
area formula for a whole circle and then divide by 2 for the half
circle. This will give us A =π ( 60 )^2 = 3600 π square units for
the area of the whole circle and 1800π square units for the area of the half circle. Now are we done with finding the area or is there more that we need to do like we did in finding the perimeter? If we think back to the definition of the area, (it is the number of square units needed to cover the figure) we should see that there is nothing further to do. The inside of
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start
10 yards
3 yards
12 yards
end
We need to figure out how long this line is. We might be tempted to say that we have two right triangles and that the line we are looking for is just the sum of each triangle’s hypotenuse. The only problem with that is that we do not know the lengths of the legs of each of the right triangles. Thus we are not able to apply the Pythagorean Theorem in this way.
We are on the right track though. This is a Pythagorean Theorem problem. If we add a couple of line segments, we can create a right triangle whose hypotenuse is the line we are looking for.
As you can see above, the orange segments along with the red segment that was 10 yards and the dashed segment make up a right triangle whose hypotenuse is the dashed segment. We can figure out how long each of the legs of this triangle are fairly easily. The orange segment that is an extension of the red segment is another 3 yards. Thus the vertical leg of this triangle is 10 + 3 = 13 yards long. The orange segment which is horizontal is the same length as the green segment. This means that the horizontal leg of this triangle is 12 yards long. We can now use the Pythagorean Theorem to calculate how far away from the original starting point we are.
Pythagorean Theorem
a^2 + b^2 = c^2
a = leg b = leg c = hypotenuse
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2
2
2 2 2
c
c
c
c
Here an exact answer does not really make sense. It is enough to approximate that we end up 17.6918 yards from where we started.
Volume and Surface Area Page 19 of 57
area of the paper that it would take to cover the outside of an object without any overlap. In most of our examples, the exposed sides of our objects will polygons whose areas we learned how to find in the previous section. When we talk about the surface area of a sphere, we will need a completely new formula.
The volume of an object is the amount of three-dimensional space an object
takes up. It can be thought of as the number of cubes that are one unit by one unit by one unit that it takes to fill up an object. Hopefully this idea of cubes will help you remember that the units for volume are cubic units.
Surface Area of a Rectangular Solid (Box)
SA = 2 ( lw+ lh+ wh )
l = length of the base of the solid w = width of the base of the solid h = height of the solid
volume
Volume of a Solid with a Matching Base and Top
V = Ah
A= area of the base of the solid h = height of the solid
Volume of a Rectangular Solid (specific type of solid with matching base and top)
V = lwh
l = length of the base of the solid w = width of the base of the solid h = height of the solid
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Example 1: Find the volume and the surface area of the figure below
4 .2 m 3.8^ m
2.7 m
Solution: This figure is a box (officially called a rectangular prism). We are given the lengths of each of the length, width, and height of the box, thus we only need to plug into the formula. Based on the way our box is sitting, we can say that the length of the base is 4.2 m; the width of the base is 3.8 m; and the height of the solid is 2.7 m. Thus we can quickly find the volume of the box to be
V = lwh =( 4. 2 )( 3. 8 )( 2. 7 )= 43. 092 cubic meters.
Although there is a formula that we can use to find the surface area of this box, you should notice that each of the six faces (outside surfaces) of the box is a rectangle. Thus, the surface area is the sum of the areas of each of these surfaces, and each of these areas is fairly straight-forward to calculate. We will use the formula in the problem. It will give us
SA= 2 ( lw+ lh+ wh)= 2 ( 4. 2 * 3. 8 + 4. 2 * 2. 7 + 3. 8 * 2. 7 )= 75. 12
square meters.
A cylinder is an object with straight sides and circular ends of the same
size. The volume of a cylinder can be found in the same way you find the volume of a solid with a matching base and top. The surface area of a cylinder can be easily found when you realize that you have to find the area of the circular base and top and add that to the area of the sides. If you slice the side of the cylinder in a straight line from top to bottom and open it up, you will see that it makes a rectangle. The base of the rectangle is the circumference of the circular base, and the height of the rectangle is the height of the cylinder.
cylinder
