In This Chapter
^ Computing the perimeter and area of polygons
^ Coming full circle with the circumference and area of a circle
^ Putting shapes together to create interesting structures
^ Using the distance formula in Cartesian coordinates to measure sides
Reas and perimeters of Polygons (geometric figures with segments for sides) and circles have many practical applications. You need the
Perimeter of your yard before you order new fencing. You want the total area of your living room before purchasing carpeting. You need to know the area of a walkway around your circular pool when you’re choosing between cement or gravel. The challenges to doing problems involving perimeter and area are in determining what type of figure you have and then in finding and applying the correct formula.
Keeping the Cows in the Pasture
Cows and sheep are standard fare when it comes to problems involving fencing in a pasture. You may have a set amount of fencing available, or you may need a particular minimum area for your herd. The pasture doesn’t have to be rectangular, but rectangles are nice figures to work with in these problems.
Working With a set amount of fencing
You have several rolls of fencing and you need to create a rectangular area for your herd of cattle. How can you best make use of the fencing? In this case, the better or best use is when you can create the largest possible area.
In Figure 19-1, you see how 60 feet of fencing is used to create several different rectangular regions, each with a different area.
10
Figure 19-1:
Sixty feet of fencing creates several
Areas. 1 C
20
15
15
25
29
5
The areas of the figures created by the same 60 feet of fencing are 200 square units, 125 square units, 29 square units, and 225 square units. The way a certain amount of fencing is used makes a huge difference in the area.
The Problem: You have 680 feet of fencing and want to create a rectangular pasture whose length is 10 feet more than twice its width. What dimensions do you make your pasture?
^.VLA/V The perimeter of a rectangular figure is determined with the formula P = 2/ + 2w Where / And W Represent the length and width of the rectangle. If your length is to be 10 more than twice the width, Then let W Represent the width and 10 + 2w represent the length. Replace the perimeter, P, With 680 and solve the equation for W.
680 = 2 (10 + 2w) + 2w 680 = 20 + 4w + 2w 680 = 20 + 6w 660 = 6W
W=
660 6
110
The width of the pasture is 110 feet. Replacing the W With 110 in 10 + 2W, You get that the length is 10 + 2(110) = 10 + 220 = 230 feet. Checking the perimeter, you get that P = 2(230) + 2(110) = 460 + 220 = 680 feet.
Having a nice, rectangular pasture or yard is fine and dandy, but what if you need to keep the little boy sheep separated from the little girl sheep — or some other such arrangement?
The Problem: You have 1,200 yards of fencing and want to create a rectangular pasture that has two dividing fences running down the middle. If the width is to be 30 feet less than the length, then what is the area of the pasture?
Figure 19-2 shows you the layout of the fencing. Let the length be represented by X And the width be represented by X - 30. Solve for the length; then determine the width. And, finally, compute the area by multiplying the length times the width.
Figure 19-2:
The dividers keep the sheep separated.
X – 30
X
The total perimeter plus the dividers is equal to 1,200 yards. So you add up all the fencing to get 1,200 = X + X + X - 30 + X - 30 + X - 30 + X - 30 = 6x – 120. The equation is now 1,200 = 6x – 120. Solving for X By adding 120 to each side, 6x = 1,320. Now, dividing each side by 6, you get that X = 220. The length is 220 yards. The width is 30 yards less than that or 190 yards. The area is 220 x 190 = 41,800 square yards.
Aiming for a needed area
You’re told that llamas need a certain amount of grazing area in order to thrive. Armed with the area constraints, you can go about figuring out how to create the needed room for a hungry llama.
The Problem: You need to create a rectangular area encompassing 6,400 square feet, in which the length is 40 feet less than three times the width. One side of the rectangular area will be along a river, so you don’t need to put a fence there. What is the least amount of fencing needed to partition off that rectangular area?
268 Part IV: Taking the Shape of Geometric Word Problems _
S, VLA*
The two different scenarios are:
Have the length of the area be along the river. Let the width be along the river.
Look at Figure 19-3 for the two different layouts. The W Represents the width, and the 3W - 40 represents the length.
Figure 19-3:
Use the river instead of fencing.
3w – 40
W
3w – 40
W
In both cases, the area is length times width or (3W - 40)w. So solve the equation A = /w, Replacing the A With 6,400.
6400 = (3w – 40) W 6400 = 3w2 – 40w
0 =3W2-40W-6400
0 = (3w — 160)( W + 40)
160
3 ,
W=
40
The two solutions of the quadratic equation are 53Ki and -40. The negative solution makes no sense, of course. Letting the width be 533*3, then the length, which is 40 less than three times that is 120 feet. The dimensions of the area needed are 53K by 120.
Now determine which layout will use the least amount of fencing. If you let the width be along the river, then you need two lengths and one width of the fencing or 293K feet. If you let the length run along the river, then you need one length and two widths or 226J3 feet — the better choice for economy’s sake.
Getting the Most Out of Your Resources
You can’t always create a rectangular yard or area, but the rectangular shape does seem to be the most popular with most builders and their projects. If you could use any shape you wanted, you could do a lot better with your resources because you could economize on fencing and maximize the area enclosed.
Triangulating the area
A triangular area requires just three sides of fencing. The area of a triangle is found using one of two formulas, depending on whether you have a perpendicular measure from one of the sides to the opposite vertex or just the measures of the sides.
The area of the triangle shown in Figure 19-4 is computed:

A = 2 bh, or
A = JS (s — A)(s — B)(s — C), where S Is half the total perimeter. (This is called Heron’s formu/a.)
Figure 19-4:
A triangle’s height is perpendicular to the base.
B
The Problem: You have 72 feet of fencing and need to enclose an area that’s triangular. Which triangle will give you more area: a right triangle that has sides measuring 18, 24, and 30 feet, or an equilateral triangle that’s 24 feet on each side?
^VLA* Find the area of the right triangle using the two perpendicular sides as the base
And height. The area is A = 2(18)(24) = 216 square feet. To find the area of the
Equilateral triangle, use Heron’s formula. The perimeter of that triangle is 72 feet (the amount of fencing you have), so half the perimeter, S, Is 36 feet. Computing
The area of the equilateral triangle, A = ^36 (36 — 24)(36 — 24)(36 — 24) =
36 (12) . 249.42 square feet. So the equilateral triangle has the greater area, even though the perimeter of each triangle is the same.
Squaring off with area
In the "Working with a set amount of fencing" section, earlier in this chapter, you see how the same amount of fencing creates several different areas. What size rectangle creates the greatest area? Does the shape of the rectangle or proportion of the lengths of the sides of a rectangle depend on the perimeter? Or is there some optimum shape?
The Problem: A rectangle is to have a perimeter of 36 feet and the greatest area possible. What are the dimensions of the rectangle that has the greatest area?
V£?LA/v Let the width of the rectangle be 1 foot, 2 feet, 3 feet, and so on. For each
Width, determine the length of the rectangle by subtracting the width from 18. You subtract from 18 because the perimeter of a rectangle is P = 2(/ + w). When the perimeter is 36 feet, you get 36 = 2(/ + w). Dividing each side of the equation by 2, 18 = / + W, And / = 18 – W. When you get the width and length, compute the area of the resulting rectangle. Table 19-1 lays it all out.
|
Table 19-1
|
Rectangles with a Perimeter of 36 Feet
|
|
Width
|
Length (18 – w)
|
Area (l X W)
|
|
1 foot
|
17 feet
|
17 square feet
|
|
2 feet
|
16 feet
|
32 square feet
|
|
3 feet
|
15 feet
|
45 square feet
|
|
4 feet
|
14 feet
|
56 square feet
|
|
5 feet
|
13 feet
|
65 square feet
|
|
6 feet
|
12 feet
|
72 square feet
|
|
7 feet
|
11 feet
|
77 square feet
|
8 feet
10 feet
80 square feet
|
Width
|
Length (18 – w)
|
Area (l X W)
|
|
9 feet
|
9 feet
|
81 square feet
|
|
10 feet
|
8 feet
|

80 square feet
|
You see that the area values start decreasing when you pass the point where the rectangle is a square. A rectangle that’s actually a square has the greatest possible area for a given perimeter. This statement is most easily proved using calculus. For now, the demonstration with the table should suffice.
Taking the hex out with a hexagon
A hexagon is a six-sided polygon. A Regu/ar Hexagon is special, because all the sides are the same measure, and all the interior angles are 120 degrees. An even more special feature of the regular hexagon is the fact that it’s made up of six equilateral triangles, all nestled together. In Figure 19-5, you find two hexagons, one made up of a rectangle topped by a trapezoid, and the other a regular hexagon. You get to find out which has the greater area for a set amount of perimeter.
Figure 19-5:
Hexagons have six sides.
5 in
4 in
6 in
12 in
5 in
6 in
4 in
6 in
6 in
6 in
6 in \ 6 in 6 in A in
The Problem: Each of the hexagons shown in Figure 19-5 has a perimeter of 36 inches. Which has the greater area?
The house-shaped hexagon on the left is a rectangle topped by a trapezoid. Add the two areas together to get the total area. The area of the rectangle is
12 X 4 = 48 square inches. The area of a trapezoid is A = 2H _b 1 + B2), which
Is half the height of the trapezoid times the sum of the two parallel bases.
In the case of the trapezoid in the figure, the area is A = (4)(6 + 12) = 36
Square inches. Add the area of the rectangle and the area of the trapezoid together to get 48 + 36 = 84 square inches.
The regular hexagon is made up of six equilateral triangles; the sides of each of the triangles is 6 inches. Use Heron’s formula to find the area of one of the triangles, and then just multiply by 6. The area of one of the triangles
/9 (9 — 6)( 9 — 6)( 9 — 6) = / 9 (3)3
Is A = J9(9 — 6)(9 — 6)(9 — 6) = J9(3) = /243 . 15.59 square inches. Multiply that area by 6 to get 93.54 square inches. The regular hexagon has
The greater area.
Coming full circle with area
Each of the problems in this section deals with making the most of perimeter to get the biggest possible area. A common theme that you find is that a Regu/ar Polygon has the greatest area of any other polygon of its type. Also, the more sides you add to a regular polygon, the more the polygon seems to resemble a circle. In this section, you compare the area of a hexagon that has a perimeter of 36 inches with the area of a circle that has a circumference of 36 inches.
The Problem: Which has the greater area: a regular hexagon with a perimeter of 36 inches or a circle with a circumference of 36 inches?
First, refer to the problem in the preceding section, "Taking the hex out with a hexagon," and you find that a regular hexagon with a perimeter of 36 inches has an area of about 93.54 square inches. To find the area of a circle that has a Circumference Of 36 inches, you need to find the radius of the circle, first.
The circumference of a circle is found with: C = 2nr, where R Is the radius. The area of a circle is found with: A = Nr2, Where R Is the radius.
If the circumference of a particular circle is 36 inches, then 2nr = 36. Dividing each side of the equation by 2n, you get that R Is about 5.73 inches. Use 5.73 as the radius in the formula for the area of the circle, and you get that the area is about 103.15 square inches. The area of the circle is almost 10 square inches larger than that of the hexagon. The area of a circle will always be greater than a polygon with the same perimeter.
Putting in a Walk-Around
You’ve finally hit the big time and decide you can afford to put in a pool and party area in the backyard. A pool has water in it — well, let’s hope so. With water comes mud and a mess, so you need to put a nice cement walk around the perimeter of the pool.
Determining the area around the outside
When you have an existing pool or other area that needs to be surrounded, then you take measurements of the structure in the middle and determine what you want around the outside — how wide and how deep.
The Problem: You have a rectangular pool that’s 40 feet long and 30 feet wide. You want to put in a cement walkway that’s 6 feet wide on all four sides, and the corners will be 6-foot squares. How many square feet of cement will you need? (Refer to Figure 19-6 which shows both the square corners for this problem and rounded corners for the next problem.)
Figure 19-6:
The pools have cement walkways.
To figure the total area of the cement walkway, divide the walkway into four rectangles. Include the square corners in rectangles that go across the top and bottom. The rectangles across the top and bottom now have dimensions 42 feet by 6 feet (the 42 comes from 30 + 6 + 6 for the two ends). Multiplying 42 x 6, you get 252 square feet for each of the two sections. The side sections are 40 feet by 6 feet, so each of their areas are 40 x 6 = 240 square feet. Double each section type and add the areas together: 2(252) + 2(240) = 504 + 480 = 984 square feet.
But what if you want curved corners? It may be that you want to make mowing easier by making the corners curves. Or, perhaps, you think that the curved corners are more aesthetically pleasing. Also, there’s always the chance that you’re just very frugal and want to save some money.
The Problem: You want a 6-foot cement walkway around the outside of your 40 foot by 30 foot rectangular pool. The walkway is to be 6-feet wide at the corners, too, so they’ll be pieces of circles with a radius of 6 feet. How many square feet, total, will your walkway contain? (Refer to Figure 19-6 to see what this type walkway would look like.)
^VLA* The four corners of the walkway are each one-fourth of the same 6-foot -
Radius circle. Just find the area of a circle with a radius of 6 feet, and add it onto the four rectangular sections. The four rectangular sections have dimensions matching the sides of the pool. Two sections are 40 x 6, and the other two are 30 x 6.
The total area of the walkway is the area of the circle plus twice the area of each rectangular section.
A = n (6)2 + 2 (40 x 6) + 2 (30 x 6) . 113.10 + 480 + 360 = 953.10
The total area is about 953 square feet. That’s about 31 square feet less than the walkway with square corners.
Adding up for the entire area
You’re contemplating putting in a circular above-ground swimming pool, but you’re not sure whether you have enough room. There’s not only the pool itself, but also the 3-foot-wide deck to consider.
The Problem: Your backyard is 120 feet long and 40 feet wide. The neighborhood ordnance dictates that a pool and its surround can’t take up more than 25 percent of your yard. How big a pool can you get if you’re going to put a 3-foot-wide deck around a circular pool? (Figure 19-7 shows you a possible scenario for the pool and the yard.)
120
Figure 19-7: 40
The pool fits in the yard.
First, determine the total square footage allowed by the ordnance. Then try to maximize the size of the pool. Your main constraint will be the width of the yard — you may have to settle for a smaller pool than allowed or go for another shape. Your total square footage is 120 x 40 = 4,800 square feet.
Avoiding wet feet
You have a fish pond on your property that’s a have a boat, a raft, or even waders. You’re a perfect circle with a diameter of 100 feet. In the clever sort, though, and you figure out how to exact center of your fish pond, there’s a flag – tie the rope to the flagpole without getting your pole. You want to tie a rope to that flagpole, but feet wet. How do you do it? your rope is only 101 feet long, and you don’t
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Twenty-five percent of 4,800 square feet is 1,200 square feet. To get the biggest pool possible, you want the area of the pool plus the deck to be 1,200 square feet (25 percent). Let the radius of your pool be represented by r. Add a 3-foot deck, and the total radius is R + 3. Now use the area formula to solve for R.
2
A = n (radius) 1,200 = N (R + 3)2
1, 200 2
—n – = (R + 3) /SP = R + 3
1,200
The radius of the pool should be about 16.54 feet. Add the 3-foot deck, and it makes the total 19.54 feet. The pool plus the deck just fit the width of the yard.
Creating a Poster
You see posters all over the place announcing events. Garage sale signs crop up every spring, and fundraising dinners are plastered all over the place before the big night. Creating an eye-catching yet economical poster is an art — which is why some people take courses to learn how to do it effectively. Color is important, but so is the use of white space. The problems in this section assume that you have a certain portion of the poster dedicated to the print and pictures and the rest of it is a plain, white border.
Starting with a certain amount of print
You’re in charge of printing up campaign posters for your favorite candidate. The person running for office has a lot to say and wants to devote 120 square inches of the poster to information about her position on the issues. You get to create the most economical size poster (use the least amount of print).
The Problem: What are the dimensions of the poster that uses the least amount of material if you have to include 120 square inches of print, 2-inch borders on the sides, and 3-inch borders on the top and bottom? (To make it more reasonable to solve, the measures all have to be whole numbers.)
Determine the different dimensions that rectangular areas of printed material can be in to get 120 square inches. Then determine the amount of border material that has to be added to each rectangle.
Figure 19-8 shows two possibilities for shapes of the resulting poster. Each poster has printed material in the middle taking up a total of 120 square inches. After you determine a height and width that gives you an area of 120 square inches, add 6 inches to the height (3 on the top and 3 on the bottom) and 4 inches to the width (2 and 2). Compute the new total area by multiplying the two new dimensions together. Look at Table 19-2 for the possibilities. (I haven’t included choices like 120 by 1 or 60 by 2 because they’re impractical.)
20
12
10
6
Figure 19-8:
Posters with a white border.
|
Table 19-2
|
Comparing Sizes of Posters
|
|
Printed Height
|
Printed Width
|
Height + 6
|
Width + 4
|
Total Area
|
|
40
|
3
|
46
|
7
|
322 Square inches
|
|
30
|
4
|
36
|
8
|
288 square inches
|
|
24
|
5
|
30
|
9
|
270 square inches
|
|
20
|
6
|
26
|
10
|
260 square inches
|
|

15
|
8
|
21
|
12
|
252 square inches
|
|
12
|
10
|
18
|
14
|
252 square inches
|
|
10
|
12
|
16
|
16
|
256 square inches
|
|
8
|
15
|
14
|
19
|
266 square inches
|
|
6
|
20
|

12
|
24
|
288 square inches
|
|
5
|
24
|
11
|
28
|
308 square inches
|
|
4
|
30
|
10
|
34
|
340 square inches
|
|

3
|
40
|
9
|
44
|
396 square inches
|
Two different sizes yield the least amount of material: either a poster that’s 21 by 12 or a poster that’s 18 by 14. Probably a rectangle that’s 18 by 14 is more aesthetically pleasing to the eye and would be the choice.
Climbing the ladder
Pete the painter is on a ladder leaning up against four rungs, down nine rungs, up three rungs and the wall that he’s working on at the time. He then up ten more rungs to reach the top bar of starts on the middle rung of the ladder, goes up the ladder. How many rungs are on this ladder?
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Working with a particular poster size
You’re to create a poster with a certain number of square inches. The white border around the printed material is to be 2 inches on each side and 3 inches across the top and bottom. Now you have to determine the amount of material left for pictures and words.
The Problem: You are to create a poster with a total area of 300 square inches. In the center of the poster you’ll put the printed material in a rectangular box with 2-inch borders on either side. You’ll also put 3-inch borders on the top and bottom. The material in the borders is included in the total area of 300 square inches. If the width of the poster is to be K the length, then what are the dimensions of the rectangular area containing the printed material? (Refer to Figure 19-9 to help you picture the situation.)
3
Figure 19-9:
Figuring out the dimensions.
2
2
3
J, VLA*
Let the dimensions of the print in the poster be L X W. Adding in the white borders, the dimensions of the overall poster are (l + 6) by (w + 4). The width of the poster is to be % the length, so replace the (w + 4) with % (l + 6). If the poster is to have an area of 300 square inches, then multiply the length times the width and set it equal to 300.
|
(L + 6)(
|
4(L + 6)) =
|
300
|
|
4 (l + 6)2 =
|
300
|
|
3 #
|
4 (l + 6)2 =
|
300 X
|
|
(l + 6)2 =
|
400
|
|
J(l + 6)2 =
|
7400
|
|
L + 6 =
|
20
|
L = 14
So the inside length of the poster (for the written part) is to be 14 inches. Add 6 to get the outside length of 20 inches. The width is to be % of the length, which makes it 15 inches. The inside width (for the printed material) is 15 inches less the border of 4 inches or 11 inches. To recap: The whole poster is to be 20 by 15 inches, giving you 300 square inches. The inside rectangle is 14 by 11 inches, giving you 154 square inches to put all the print and pictures.
Shedding the Light on a Norman Window
Houses have windows to let in the light and fresh air. The most common shape for a window is a rectangle, but you also see round windows and hexagonal windows and other creative shapes. A Norman window Is made up of two geometric shapes: a semicircle on top of a rectangle. One side of the rectangle is the same measure as the semicircle’s diameter. Figure 19-10 shows two Norman windows.
Figure 19-10:
The curve is on the top.
Maximizing the amount of light
A window lets in light and fresh air. Adding a semicircle to the top of a rectangular window not only lets in more light, but adds a decorative touch.
The Problem: You’re planning on putting in a Norman window in a room on the north side of your home. You want as much light as possible to be let in through the window. The rectangular part of the window is to be 6 feet by 4 feet, and the semicircle will sit on top. Which will let in more light: if the rectangular base is 4 feet and the sides 6 feet, or if the rectangular base is 6 feet with 4 foot sides? (Refer to Figure 19-10 for approximate figures.)
^VLA* Find the total area of each window. The rectangular part of each window is 24 square feet, so the main interest is in the respective areas of the semicircles on top. The window with a 4-foot base has a semicircle on top with a diameter of 4 feet — or a radius of 2 feet. The area of a circle is found with A = Nr2. The semicircle has half the area of the full circle, so the area of this
Semicircle is i n (2)2 = 2n. 6.28 square feet. The window with a 6-foot base 2
Has a semicircle on top whose radius is 3 feet. So the area of that semicircle is
I n (3)2 = 9 n. 14.13 square feet. Clearly, the wider window will let in more 22
Light.
The areas of the two Norman windows in the preceding problem are different by almost 8 square feet — a fairly large difference. You may be surprised to know that the perimeters of these two windows aren’t nearly so different.
The Problem: Which has the greater perimeter: a Norman window with a 4-by-6-foot base or a Norman window with a 6-by-4-foot base?
^VLA/y The perimeter of a Norman window consists of the three sides of the rectangle and the circumference of the semicircle. The circumference of a full circle is found by multiplying the diameter by n. Because you only want half that area, you multiply half the diameter (the radius) by n. The Norman window with the base of 4 feet and sides of 6 feet has a semicircle with radius 2 feet, so the total perimeter is 4 + 6 + 6 + 2n = 16 + 2n~ 22.28 feet. The Norman window with the base of 6 feet and sides of 4 feet has a semicircle with radius 3 feet, so the total perimeter is 6 + 4 + 4 + 3n = 14 + 3n~ 23.42 feet. The perimeters of these two windows differ by just a little over 1 foot.
Making the window proportional
What if you want the area of the rectangular part of a Norman window to be equal to the area of the semicircular part? This arrangement just may be more esthetically pleasing to you.
The Problem: How long should the base of a Norman window be if the two sides of the rectangle are 2 feet high and the area of the rectangle is to be equal to the area of the semicircle?
^VLAAf First, let the length of the base be represented by X. Because the area of a rectangle is length times width, the area of the rectangle is 2x. You want the area of the semicircle also to be equal to 2X. The area of the semicircle is found by taking half the area of a circle whose radius is half the base2 of the rectangle,
So the radius is X And the area of the semicircle is i n (X) . Set the area of 2 2 \ 2 /
The rectangle equal to the area of the semicircle and solve for X.
|
2x =
|
22
|
|
2X =
|
2 \ 4 /
|
|
2X =
|
Nx2 8
|
|
16x =
|
Nx2
|
|
0 =
|
Nx2- I6x
|
|
0 =
|
X (Nx – 16)
|
Two different values of X Satisfy the equation. Obviously, X = 0 Isn’t going to
Work — it’s extraneous. But setting the factor in the parentheses equal to 0, 1 F\
You get that X = ^j-, which is about 5.1 feet.
Fitting a Rectangular Peg into a Round Hole
An old adage says that you can’t fit a square peg into a round hole. (Or is it a round peg in a square hole?) Mathematicians must have found this statement to be a challenge. No, they still couldn’t fit the peg into the hole, but the statement opened up all sorts of questions and answers as to how large a round peg could fit into a square hole or how large a rectangular peg could fit into a semicircular hole. Oh, the possibilities!
Putting rectangles into circles

A rectangle has a set length and width and doesn’t fit very neatly into a circle. See Figure 19-11 for some examples of rectangles working their ways into circles.
Figure 19-11:
The circle has to be bigger than the
Rectangle.

Just how large a rectangle will fit into a circle, if you have some particular constraints?
The Problem: You want to fit a rectangle with length 8 inches into a circle with radius 5 inches. What width will the rectangle have if each of the vertices of the rectangle are to be on the circle?
^ylA/v Take advantage of the symmetry of a rectangle, the radius of the circle, and good old Pythagoras. First, a rectangle’s "center" is where the two diagonals (from opposite corners) intersect. The center of the rectangle is also where the center of the circle is. The radius of this circle is 5 inches, so that’s the distance from the center of the rectangle to any of its vertices, because the vertices are on the circle. The length of the rectangle is 8 inches, so a segment drawn from the center of the rectangle horizontally to one of the sides is half that, or 4 inches. Refer to the left circle in Figure 19-11 and you’ll see the radius and segment drawn in. The horizontal segment is perpendicular to the side of the rectangle — using the symmetry and angle measures of a rectangle — so the triangle formed is a right triangle. Solving for the length of the third side of the right triangle (I’ll call the length b) using the Pythagorean theorem, you get that 42 + B2 = 52 or B2 = 25 – 16 = 9. The length of the third side of the triangle is 3 inches, so the length of the entire side of the rectangle is 6 inches. You can fit an 8-by-6-inch rectangle into a circle whose radius is 5 inches.
A rectangle measuring 8 by 6 inches has an area of 48 square inches. A rectangle with these dimensions is not the largest rectangle that you can fit into a circle with a radius of 5 inches. In calculus, you prove that the largest rectangle that fits into a circle is really a square. So, if that’s the case, what size square fits into a particular circle?
The Problem: What is the length of any side of a square that fits into a circle whose radius is 5 inches?
V£?LA/v Use the diagonal of the square and the Pythagorean theorem to solve this
Problem. The diagonal of the square is the same as the diameter of the circle. A circle with a 5-inch radius has a 10-inch diameter. Let the lengths of the sides of the square be X Inches long. Because the angles of a square are all right angles, you have a right triangle whose hypotenuse is 10 inches long and whose sides are each X Units long. Filling in values in the Pythagorean theorem and solving for X,
X 2 _|_ X 2_
2X2= 100 X2= 50
X = /50 = 5 /2 . 7.07
So a square with sides measuring about 7.07 inches is the largest that will fit into the circle. The area of this square is about 49.98 square inches — larger than the rectangle measuring 8 by 6 inches.
Working with coordinate axes
Circles, rectangles, and squares are easily described using the coordinate axes and some points and equations. The distance formula for the coordinate plane allows you to solve for lengths of segments if you have the values of the coordinates at either end.
Using coordinate axes,
The equation of a circle with its center at the origin is X2 + Y2 = R2, Where R Is the radius of the circle.
The distance between the two points (x1,y1) and (x2,y2) is found with the formula D = J_X2 — X1) + _Y2 — Y 1) .
In Figure 19-12, you see a triangle drawn inside a semicircle. Even though a semicircle is only half a full circle, the equation for the circle is used when solving problems involving one of its semicircles.
Figure 19-12:
The base of the triangle rests on the axis.
The Problem: A rectangle is inscribed inside a semicircle (its vertices are all on some part of the semicircle). If the semicircle has its center at the origin and a radius of 5 units, then what are the coordinates of the vertices (x, y) In the first quadrant if the length of the rectangle is four times its height?
The Y Coordinate of the vertex you want also represents the height of the rectangle. The X Coordinate is actually half the length, because the X Is just the distance from the origin to the right. So the length of the rectangle is represented by 2x. Writing that The length is four times the height, The equation is 2x = 4y, which simplifies to X = 2y.
The equation of a circle with its center at the origin and with a radius of 5 is X + Y2 = 25. Replacing the X In this equation with 2y and solving for Y,
_2YI2_y2=25 4Y2_y2=25 5Y2= 25
Y2= 5
Y = ±y5
You use only the positive value for Y. Because X = 2y, then X Must be 2 and the coordinates of the vertex are
Any triangle that has two of its vertices at the endpoints of a diameter and the third vertex on the circle is a right triangle. You don’t prove this fact here, but the next problem allows you to see, from the coordinates of the points, that the rule holds.
The Problem: A triangle has one side along the flat part of a semicircle, with the endpoints of that side of the triangle at the endpoints of that diameter (refer to Figure 19-12). The radius of the semicircle is 10 units. What are the coordinates of the vertex of the triangle if it lies on the circle, and the distance from that vertex to the left endpoint is 2 /I0?
You see that the endpoints of the diameter lie along the X-axis. If the radius of the semicircle is 10 units, then the coordinates of the endpoints of the diameter are (-10,0) and (10,0). The left endpoint is at (-10,0). Use the distance formula, and set that distance equal to the value you get by substituting in the points (x, y) And (-10,0) into the formula.
2/T0 = \ (x — (—10))2 + (Y — 0 )2 = / (x + 10)2 + Y y
You see both XS and YS in the equation. Use the equation of the circle with its center at the origin and a radius of 10 by solving for Y2. Then you can substitute the equivalence for Y2 Into the equation and solve for X.
X2_y2=100 Y2=100 —x2
Substituting, squaring both sides, and solving for X,
|
2 /T0 =
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/(x + 10)2 + 100 — X2
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2
(2/10) =
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(7( X + 10)2 + 100 — X2
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4 X 10 =
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(x + 10)2 + 100 — X2
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|
40 =
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X2 + 20x + 100 + 100 —
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|
40 =
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20x + 200
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|
—160 =
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20X
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—8 =
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X
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The x-coordinate is -8. Substituting back into the equation for the circle, you get that Y2 = 100 – (-8)2 = 100 – 64 = 36. Taking the square root of each side, you get that Y Is either +6 or -6. The point you’re looking for is above the X-axis, so you want the +6. The coordinates of the vertex are (-8,6).
Back to the fact that you have a right triangle here, you can show that it’s true by using the Pythagorean theorem on the lengths of the sides of the triangle. The side along the X-axis is 20 units long. The side that goes from the point (-8,6) to (-10,0) is./40 units long. (You can check me by using the DistAnce formula.) And the side that goes from the point (-8,6) to (10,0) is /360 units long. Plugging these values into the Pythagorean theorem,
(740)2+(7360)2 = 202
40 + 360 = 400
Chapter 20