Math A100 IUSB Sec. 2.4 detailed notes – Week 3 Perimeter and Area From time to time this semester, your HW will ask you to calculate the perimeter of a geometric figure, its area, or both. You will be expected to know the proper formulas for some basic geometric shapes (i.e. the formula will not be given you on quizzes or Unit Tests. I would strongly encourage you to reach out to your classmates for ideas of how they keep the different formulas straight. Perimeter – this is the distance around the outside of the geometric figure, i.e. add up the length of all the sides of the figure. One student told me he had been in the military and when on guard duty, he had to walk the perimeter of camp, i.e. around the outside of the camp. The units you measure perimeter in the single units, like inches, cm, feet, miles, etc. Area – this is the “space” inside the figure. This “space” is 2-dimensional, it has both length and width (or if you prefer, height and width). Since it has 2 dimensions, its measurement units also have 2 dimensions, i.e. ____2, such as inches2, cm2, ft2, miles2, etc. From formulas Basic geometric figure you will be expected to know: Rectangles: 2W Perimeter = 4 sides added together = Length + width + Length + width or or P = 2(L + W) Area = Length x Width or A = L*W P = 2L + or A = LW Squares: Perimeter = 4 sides added together. Since all sides are the same length, called “s”  P = s + s + s +s or P = 4s Area = Length x Width or A = s*s or A = s2 Circles: Perimeter  called the Circumference = pi x diameter or circumference = 2 x pi x radius C = 22πr or C= πd Area = pi x square of the radius or A = πr 2 NOTE: for any unusual geometric figures, you will be given the formulas for perimeter and area. You will only be expected to know the three above. Example 1: Find the perimeter and the area for a rectangle that is 5 inches long and 8 inches wide. Perimeter: P = 2L + 2W  P = 2(5 in.) + 2(8 in. )  P = 10 in. + 16 in. = 26 inches Perimeter = 26 inches Area = Length x Width = (5 in.)(8 in.) = 40 in2 Area = 40 in2 Example 2: Find the perimeter and the area for a square that is 7 feet on a side. Perimeter: P = 4s  P = 4(7 ft.)  P = 28 ft. Perimeter = 26 ft. Area = s2 = (7 ft.) 2 = 49 ft2 Area = 49 ft2 Exact Answers vs. Approximate answers MML ask you for the exact answer and the approximate answer for circumferences and areas of circles. As you see above, the formulas for both the Circumference an the Area include π, a.k.a. pi. Pi, π, is the relationship between the distance going around the outside of the circle and the circle’s diameter. Pi is an irrational number  it does not end or repeat: it is 3.1415…… We often use the value for pi of π = 3.14. That number is rounded off, so any calculations you make using 3.14 are only approximations, not exact answers. Similarly, the number using the π key on your calculator is also rounded off, so if you use the π key on you calculator, your answer is also an approximate answer. There is only one thing that represents the Exact value of pi, it is the symbol π. So, if you are asked for an exact answer, it must include the symbol π. Example 1: Find the exact and approximate circumference and the exact and approximate area for a circle that has a radius of 10 cm. Use 3.14 for π (i.e. π = 3.14) Circumference: C = 2πr Exact: C = 2(π)(10 cm)  C = 20π cm Approximate: C = 2(3.14)(10 cm)  C = 62.8 cm Area = πr 2 A = (π)(10 cm.) 2  A = 100π cm2 Exact: Approximate: A = (3.14)(10 cm.) 2 = (3.14)(100 cm2)  A = 314 cm2 Solving for an Indicated Variable  Re-writing formulas These often gives students trouble simply because you are given formulas with very few, if any, actual numbers in them. Normally, you are given an equation and asked to find the value of the variable that makes the equation work (i.e. to solve for the variable  get the variable by itself. You will be doing the same thing here, only the answers will not be a number; they will be the same letters and numbers but mixed around into a different order to get the chosen variable by itself. Believe it or not, you already know how to do these ---- Undo what they gave you!!!!!!! Example 1: a+b=c solve for “a” (i.e. you want to get “a” by itself) They gave you addition, so use subtraction to undo it. a + b -b = c – b subtract “b” from both sides, giving you: a=c–b and yes, this is the answer! Example 2: d = rt solve for “r” (i.e. you want to get “r” by itself) They gave you multiplication, so use division to undo it. d/t = rt/t divide both sides by “t”, giving you: d/t = r and yes, this is the answer! Example 3: P = 2L + 2W solve for “W” (i.e. you want to get “W” by itself) First, you have addition, so use subtraction to undo it. P- 2L = 2L + 2W - 2L subtract “2L” from both sides, giving you: P – 2L = 2W (P – 2L)/2 = 2W/2 you have multiplication, so divide both sides by “2”, giving you: (P – 2L)/2 = W this is the answer Example 4: I = Prt solve for “t” (i.e. you want to get “t” by itself) They gave you multiplication (both P and r are being multiplied by “t”), so use division to undo it. I = Prt I/Pr = Prt/PR I/Pr = t divide both sides by both “P” & “r” at the same time, giving you: this is the answer NOTE: you must use exactly the same letters in your answer that they gave you in the question. So, if they used a capital letter, then you must use the same capital letter. If you answered i/pr = t  it would be incorrect, as each letter means something specific. For example, P stands for Perimeter, but p stands for pressure. Example 5: I = Prt solve for “t” (i.e. you want to get “t” by itself) They gave you multiplication (both P and r are being multiplied by “t”), so use division to undo it. I = Prt I/Pr = Prt/PR I/Pr = t Example 6: divide both sides by both “P” & “r” at the same time, giving you: this is the answer A = P(1 + r) solve for “r” (i.e. you want to get “t” by itself) They gave you multiplication (P is being multiplied by the ( ), so use division to undo it. A = P(1 + r) A/P = P(1 + r)/P divide both sides by “P”, giving you: A/P = 1 + r now subtract 1 from both side, giving you: A/P – 1 = r this is the answer Example 7: V = 1/3b2h solve for “h” (i.e. you want to get “h” by itself) They gave you multiplication (both 1/3 and b2 are being multiplied by “h”), so use division to undo it. V = 1/3b2h V/(1/3) = 1/3b2h/(1/3) divide both sides by 1/3, giving you: 3V = b2h 3V/b2 = 1/3b2h/b2 now divide both sides by b2, giving you: 3V/b2 = h this is the answer NOTE: you can make a “guide” for yourself if you are not sure what to do next by replacing the all the letters except the one you are solving for with numbers. This may give you a hint on what the next step should be. HOWEVER, don’t actually do any math with the replacement number – just use them to see what needs to come next. For example: solve for “r” A = P(1 + r) Guide Actual Questions Replace the A which 20 Replace the P with 4 20 = 4(1 + r) divide both sides by 4  A = P(1 + r) 20/4 = 1 + r so divide both sides by P A/P = 1 + r 20/4 - 1 = 1 + r -1 subtract 1 from both sides  A/P -1 = 1 + r -1 so subtract 1 from both sides 20/4 – 1 = r r is by itself, so you are done  A/P – 1 = r r is by itself, so you are done AND NOW the examples you will be using a LOT the rest of the semester: Example 8: solve for “y” (i.e. you want to get “y” by itself) 3x + 2y = 8 You have both addition and multiplication here 3x + 2y = 8 3x + 2y – 3x = 8 – 3x subtract 3x from both sides, giving you: 2y = 8 – 3x 2y/2 = (8 – 3x)/2 now divide both sides by 2, giving you: y = (8 – 3x)/2 does this look familiar? Maybe try dividing 8 and -3x individually by 2 y = 8/2 – 3x/2 y = 4 – 3/2x does this look familiar? Try rearranging the 4 and -3/2x y = -3/2x + 4  final answer does this look familiar? IT BETTER, because it is y = mx + b Thanks right, this leads up to changing an equation into y = mx + b form!! (a.k.a. the slope-intercept form) Example 9: 3x - 5y = 10 4x - 5y = 10 put in slope-intercept form (a.k.a. y = mx + b form) subtract 4x from both sides, giving you: -5y = -4x + 10 divide by sides by -5, giving you: y = -4x/-5 + 10/-5 which simplifies to: y = 4/5x – 2  final answer