11. The perimeter of a square is 40 feet. What is its area? (Remember: P = 4s and A = s2.) 100 square feet.
Dividing each side of 40 = 4s by 4, you get 10 = s. Now find the area with A = s2. Yes, 100 square feet is correct.
12. You can find the area of a triangle with where the base (b) and the height (h) are perpendicular to one another. If a right triangle has legs of 10 inches and 24 inches and a hypotenuse of 26 inches, then what is its area? 120 square inches.
In a right triangle, the hypotenuse is always the longest side, and the two legs are perpendicular to one another. Use the measures of the two legs for your base and height, so .
13. You can find the volume of a box with . Find the height if the volume is 200 cubic feet and the square base is 5 feet on each side (length and width are each 5). 8 feet.
Replace the V with 200 and replace the l and w each with 5 to get . Dividing each side of by 25, you get . So the height is 8 feet.
14. The volume of a sphere (ball) is , where r is the radius of the sphere — the measure from the center to the outside. What is the volume of a sphere with a radius of 6 inches? 904.32 cubic inches.
Substituting in the 6 for r, you get .
15. You can find the volume of a right circular cylinder (soda pop can) with V = πr2h, where r is the radius and h is the height of the cylinder — the distance between the two circular bases (the top and bottom of the can). Which has the greater volume: a cylinder with a radius of 6 cm and a height of 9 cm or a cylinder with a radius of 9 cm and a height of 4 cm? Neither, the volumes are the same.
The first volume is .
The second volume is .
So they have the same volume, namely .
16. The volume of a cube is 216 cubic centimeters. What is the new volume if you double the length of each side? 1,728 cm3.
First, find the lengths of the edges of the original cube. If s = the length of the edge of the cube, then use the formula for the volume of a cube .
Doubling the length of the edge for the new cube results in sides of . So the volume of the new cube is .
17. If 60% of the class has the flu and that 60% is 21 people, then how many are in the class? 35 people.
Let x = the number of people in the class. Then 60% of x is 21, which is written . Divide each side by 0.60, and you get that x = 35 people.
18. How much simple interest will you earn on $4,000 invested at 3% for 10 years? What is the total amount of money the end of the 10 years? $1,200 and $5,200.
Use the formula I = Prt, where the principal (P) is 4,000, the rate (r) is 0.03, and the time (t) is 10: . Add this amount to the original $4,000 to get a total of $5,200.
19. How much money will be in an account that started with $4,000 and earned 3% compounded quarterly for 10 years? $5,393.39.
Use the formula for compound interest: . (Note: Compare this total amount with the total using simple interest, in problem 18. In that problem, the total is .) In the compound interest formula, .
So you do slightly better by letting the interest compound.
20. If you earned $500 in simple interest for an investment that was deposited at 2% interest for 5 years, how much had you invested? $5,000.
Use I = Prt where . Putting the numbers into the formula, . Divide each side of , and you get that , or the amount that was invested.
Chapter 18
Making Formulas Work in Basic Story Problems
In This Chapter
Working with the Pythagorean Theorem
Tackling geometric structures
Moving along with distance problems
Algebra students often groan and moan when they see story problems. You “feel their pain,” you say? It’s time to put the myth to bed that story problems are too challenging. You know you’re facing a story problem when you see a bunch of words followed by a question. And the trick to doing story problems is quite simple. Change the words into a solvable equation, solve that equation (now a familiar friend), and then answer the question based on the equation’s solution. Coming up with the equation is often the biggest challenge. However, some story problems have a formula built into them to help. Look for those first.
One technique for being successful with story problems is to start writing. Take your pencil and draw a picture of what’s going on. Next, read the last sentence. No, this isn’t a book where you don’t want to go to the end and spoil the rest of the story. The last sentence has the biggest clue; it tells you what you’re solving for. Take that pencil and start assigning letters (variables) to the different values in the problem. Just remember that a letter has to represent a number, not a person or thing.
Take a deep breath. Grab your pencil. Charge!
Applying the Pythagorean Theorem
Pythagoras, the Greek mathematician, is credited for discovering the wonderful relationship between the lengths of the sides of a right triangle (a triangle with one 90 degree angle). The Pythagorean theorem is a2 + b2 = c2. If you square the length of the two shorter sides of a right triangle and add them, then that sum equals the square of the longest side (called the hypotenuse), which is always across from the right angle. Right triangles show up frequently in story problems, so when you recognize them, you have an instant, built-in equation to work with on the problem.
Q. I have a helium-filled balloon attached to the end of a 500-foot string. My friend, Keith, is standing directly under the balloon, 300 feet away from me. (And yes, the ground is perfectly level, as it always is in these hypothetical situations — at least until you get into higher math.) These dimensions form a right triangle, with the string as the hypotenuse. Here’s the question: How high up is the balloon?
A. 400 feet up. Identify the parts of the right triangle in this situation and substitute the known values into the Pythagorean theorem. Refer to the picture.
The 500-foot string forms the hypotenuse, so the equation reads 3002 + x2 = 5002. Solving for x,
You only use the positive solution to this equation because a negative answer doesn’t make any sense.
1. Jack’s recliner is in the corner of a rectangular room that measures 10 feet by 24 feet. His television is in the opposite corner. How far is Jack (his head, when the recliner is wide open) from the television? (Hint: Assume that the distance across the room is the hypotenuse of a right triangle.)
Solve It
2. Sammy’s house is 1,300 meters from Tammy’s house — straight across the lake. The paths they need to take to get to one another’s houses (and not get wet) form a right angle. If the path from Sammy’s house to the corner is 1,200 meters, then how long is the path from the corner to Tammy’s house?
Solve It
3. A ladder from the ground to a window that’s 24 feet above the ground is placed 7 feet from the base of the building, forming a right triangle. How long is the ladder, if it just reaches the window?
Solve It
4. Calista flew 400 miles due north and then turned due east and flew another 90 miles. How far is she from where she started?
Solve It
Using Geometry to Solve Story Problems
Geometry is a subject that has something for everyone. It has pictures, formulas, proofs, and practical applications for the homeowner. The perimeter, area, and volume formulas are considered to be a part of geometry. But geometry also deals in angle measures, parallel lines, congruent triangles, polygons and similar figures, and so forth. The different properties that you find in the study of geometry are helpful in solving many story problems.
Story problems that use geometry are some of the more popular (if you can call any story problem popular) because they come with ready-made equations from the formulas. Also, you can draw a picture to illustrate the problem. I’m a very visual person, and I find pictures and labels on the pictures to be very helpful.
Q. The opposite angles in a parallelogram are equal, and the adjacent angles in the parallelogram are supplementary (add up to 180 degrees). If one of the angles of a parallel
ogram measures 20 degrees more than three times another angle, then how big is the larger of the two angles?
A. 140 degrees. The two angles have different measures, so they must be adjacent to one another. The sum of adjacent angles is 180. Let x represent the measure of the smaller angle. Then, because the larger angle is 20 degrees more than three times the smaller, you can write its measure as 20 + 3x. Represent this in an equation in which the two angle measures are added together and the sum is 180: x + 20 + 3x = 180. Simplifying and subtracting 20 from each side, you get 4x = 160. Divide by 4 to get x = 40. Putting 40 in for x in the measure 20 + 3x, you get 20 + 120 = 140.
Q. An isosceles triangle has two sides that have the same measure. In a particular isosceles triangle that has a perimeter of 27 inches, the base (the side with a different measure) is 1 foot more than twice the measure of either of the other two sides. How long is the base?
A. 14 inches. The perimeter of the isosceles triangle is x + x + 1 + 2x = 27. The x represents the lengths of the two congruent sides, and 1 + 2x is the measure of the base. Simplifying and subtracting 1 from each side of the equation, you get 4x = 26. Dividing by 4, you get inches. The base is equal to .
5. The sum of the measures of the angles of a triangle is 180 degrees. In a certain triangle, one angle is 10 degrees greater than the smallest angle, and the biggest angle is 15 times as large as the smallest. What is the measure of that biggest angle?
Solve It
6. A pentagon is a five-sided polygon. In a certain pentagon, one side is twice as long as the smallest side, another side is 6 inches longer than the smallest side, the fourth side is 2 inches longer than two times the smallest side, and the fifth side is half as long as the fourth side. If the perimeter of the pentagon is 65 inches, what are the lengths of the five sides?
Solve It
7. The sum of the measures of the angles in a quadrilateral (a polygon with 4 sides) adds up to 360 degrees. If one of the angles is twice as big as the smallest and the other two angles are both three times as big as the smallest, then what is the measure of that smallest angle?
Solve It
8. The sum of the measures of all the angles in any polygon can be found with the formula A = 180(n – 2), where n is the number of sides that the polygon has. How many sides are there on a polygon where the sum of the measures is 1,080 degrees?
Solve It
9. Two figures are similar when they’re exactly the same shape — their corresponding angles are exactly the same measure, but the corresponding sides don’t have to be the same length. When two figures are similar, their corresponding sides are proportional (all have the same ratio to one another). In the figure, triangle ABC is similar to triangle DEF, with AB corresponding to DE and BC corresponding to EF.
If side EF is 22 units smaller than side BC, then what is the measure of side BC?
Solve It
10. The exterior angle of a triangle lies along the same line as the interior angle it’s next to. See the following figure.
The measure of an exterior angle of a triangle is always equal to the sum of the other two interior angles. If one of the nonadjacent angles in a triangle measures 30 degrees, and if the exterior angle measures 70 degrees less than twice the measure of the other nonadjacent angle, then how big is that exterior angle?
Solve It
Putting Distance, Rate, and Time in a Formula
The distance-rate-time formula is probably the formula you’re most familiar with from daily life — even though you may not think of it as using a formula all the time. The distance formula is d = rt. The d is the distance traveled, the r is the speed at which you’re traveling, and the t is the amount of time spent traveling.
Examining the distance-rate-time formula
The only real challenge in using this formula is to be sure that the units in the different parts are the same. (For example, if the rate is in miles per hour, then you can’t use the time in minutes or seconds.) If the units are different, you first have to convert them to an equivalent value before you can use the formula.
Q. How long does it take for light from the sun to reach the earth? (Hint: The sun is 93 million miles from the earth, and light travels at 186,000 miles per second.)
A. About . Using d = rt and substituting the distance (93 million miles) for d and 186,000 for r, you get 93,000,000 = 186,000t. Dividing each side by 186,000, t comes out to be 500 seconds. Divide 500 seconds by 60 seconds per minute, and you get minutes.
Q. How fast do you go to travel 200 miles in 300 minutes?
A. Average 40 mph. Assume you’re driving over the river and through the woods, you need to get to grandmother’s house by the time the turkey is done, which is in 300 minutes. It’s 200 miles to grandmother’s house. Because your speedometer is in miles per hour, change the 300 minutes to hours by dividing by 60, which gives you 5 hours. Fill the values into the distance formula, 200 = 5r. Dividing by 5, it looks like you have to average 40 miles per hour.
11. How long will it take you to travel 600 miles if you’re averaging 50 mph?
Solve It
12. What was your average speed (just using the actual driving time) if you left home at noon, drove 200 miles, stopped for an hour to eat, drove another 130 miles, and arrived at your destination at 7 p.m.?
Solve It
Going the distance with story problems
Distance problems use, as you would expect, the distance formula, d = rt. There are two traditional types of distance problems. Recognizing them for what they are makes life so much easier.
One type of distance problem involves setting two distances equal to one another. The usual situation is that one person is traveling at one speed and another person is traveling at another speed, and they end up at the same place at the same time. The other traditional distance problem involves adding two distances together, giving you a total distance apart. That’s it. All you have to do is determine whether you’re equating or adding!
Q. Angelina left home traveling at an average of 40 mph. Brad left the same place an hour later, traveling at an average of 60 mph. How long did it take for Brad to catch up to Angelina?
A. 2 hours. This type of problem is where you set the distances equal to one another. You don’t know what the distance is, but you know that the rate times the time of each must equal the same thing. So let t represent the amount of time that Angelina traveled and set Angelina’s distance, 40t, equal to Brad’s distance, 60(t – 1). (Remember, he traveled one less hour than Angelina did.) Then solve the equation 40t = 60(t – 1). Distribute the 60 on the right, giving you 40t = 60t – 60. Subtract 60t from each side, resulting in –20t = –60. Divide by –20, and you get that t = 3. Angelina traveled for 3 hours at 40 mph, which is 120 miles. Brad traveled for 2 hours at 60 mph, which is also 120 miles.
Q. One train leaves Kansas City traveling due east at 45 mph. A second train leaves Kansas City three hours later, traveling due west at 60 mph. When are they 870 miles apart?
A. 10 hours after the first train left. This type of problem is where you add two distances together. Let t represent the time that the first train traveled and 45t represent the distance that the first train traveled. This is rate times time, which equals the distance. The second train didn’t travel as long; its time will be t – 3, so represent its distance as 60(t – 3). Set the equation up so that the two distances are added together and the sum is equal to 870: 45t + 60(t – 3) = 870. Distribute the 60 on the left and simplify the terms to get 105t – 180 = 870. Add 180 to each side to get 105t = 1050. Divide each side by 105, and you have t = 10 hours.
13. Kelly left school at 4 p.m. traveling at 25 mph. Ken left at 4:30 p.m., traveling at 30 mph, following the same route as Kelly. At what time did Ken catch up with Kelly?
Solve It
14. A Peoria Charter Coach bus left the bus terminal at 6 a.m. heading due north and traveling at an average of 45 mph. A second bus left the terminal at 7 a.m., heading due south and traveling at an average of 5
5 mph. When were the buses 645 miles apart?
Solve It
15. Geoffrey and Grace left home at the same time. Geoffrey walked east at an average rate of 2.5 mph. Grace rode her bicycle due south at 6 mph until they were 65 miles apart. How long did it take them to be 65 miles apart?
Algebra I Workbook For Dummies Page 17