Mth 7, Unit 9: Mesurement: Two-Dimensionl Figures Notes Precision nd Accurcy Syllus Ojective: (6.) The student will red the pproprite mesurement tool to the required degree of ccurcy. We often use numers tht re not exct. Mesurements re pproximte there is no such thing s perfect mesurement. The precision of numer refers to its exctness to the level of detil to which the tool cn mesure. Mesurements cnnot e more precise thn the mesuring tool. This is very importnt in science! Exmple: To wht degree of precision cn you mesure length using this ruler? To the nerest 1 of n inch 8 To the nerest mm The smller the unit of mesurement, the more precise the mesure. Consider some mesures of time, such s 15 seconds nd 15 hours. A mesure of 15 seconds implies it is precise to the nerest second, or time intervl etween 14.5 nd 15.5 seconds. The time of 15 hours is fr less precise: it suggests time etween 14.5 nd 15.5 hours. The potentil error in the first intervl is 0.5 seconds; the potentil error in the 15 hour scenrio is 0.5 hour or 1800 seconds. Becuse the potentil for error is greter, the 15-hour-mesure is less precise. Exmple: Choose the more precise mesurement in the given pir. () 3 m, 35 km 3 m is more precise (meters re smller thn km) () 1 inches, 1 foot 1 inches (inches re smller thn foot) (c) 1 pound, 1 ounce 1 ounce (n ounce is smller thn pound) The numer of deciml plces in mesurement cn lso ffect precision. Using time gin, mesure of 5.1 seconds is more precise thn 5 seconds. The 5.1 mesurement implies mesure precise to the nerest tenth of second. The potentil error in 5.1 seconds is 0.05 seconds, compred to the potentil error of 0.5 seconds with the mesure of 5 seconds. Exmple: Choose the more precise mesurement in the given pir. () 5.4 m, 5.67 m 5.67 is more precise (hundredths of meter is smller thn tenths) Mth 7 Notes Unit 9: Mesurement: Two-Dimensionl Figures Pge 1 of 17
() 3 yrds, 3.6 yrds 3.6 is more precise (tenths of yrd is more precise thn yrds) Syllus Ojective: (1.9) The student will round to the pproprite significnt digit. This leds us to discussion out significnt digits. All the digits tht re known with certinty re clled significnt digits. Below re the rules for determining significnt digits. The only tricky digits re the zeros: All non-zero digits re significnt digits. o 3 hs one significnt digit o.5 hs two significnt digits o 356.491 hs six significnt digits Zeros tht occur etween significnt digits re significnt digits. o 07 hs 3 significnt digits o 6.005 hs 4 significnt digits o 0.006 hs 5 significnt digits Zeros to the right of the deciml point AND to the right of non-zero digit re significnt digits. o 0.10 hs significnt digits (the 0 efore the deciml is not significnt while the 0 to the right of the deciml point nd the digit 1 re significnt) o 0.0040 hs significnt digits (just the lst two) o 4.60 hs 3 significnt digits o 460 hs significnt digits (zero is to the left of the deciml point) o 46.00 hs 4 significnt digits o 460.00 hs 5 significnt digits (the two zeros to the right of the deciml point re significnt this mkes the zero to the left of the deciml point significnt ecuse it lies etween significnt digits) Exmple: Determine the numer of significnt digits in ech mesurement. () 3.75 () 43.03 (c) 0.040 (d) 0.007 4 significnt digits (ll nonzero digits) 5 significnt digits (zero is etween significnt digits) 3 significnt digits (zero fter the lst nonzero digit nd to the right of the deciml point is significnt) 1 significnt digit Differing levels of precision cn cuse us prolem when deling with rithmetic opertions. Suppose I wish to dd 11.1 seconds to 13.47 seconds. The nswer, 4.57 seconds, is misleding. Tht is: 11.1 seconds implies the time is etween 11.05 nd 11.15 seconds 13.47 seconds implies the time is etween 13.465 nd 13.475 seconds The sum should imply the time is etween 4.515 nd 4.65 seconds But the sum 4.57 seconds im plies the time is etween 4.565 nd 4.575, which is more precise thn the ctul res ult. So it is generlly ccepted tht when you dd or sutrct, you report your nswer to the sme precision s the lest precise mesure. In other words, the nswer should hve the sme numer of digits to the right of the deciml point s the mesurement with the lest numer of digits to Mth 7 Notes Unit 9: Mesurement: Two-Dimensionl Figures Pge of 17
the right of the deciml point. In our exmple, we would report 4.57 s 4.6 seconds. Exmple: Clculte. Use the correct numer of significnt digits. () () 4.5 +.17 6.67, 6.7 15 5.6 9.4 9 Answer rounded to 1 digit to the right of the deciml (tenths) Answer rounded to no digits to the right of the deciml (units) Multiplyin g or dividing mesures cretes different type of prolem. For instnce, I wnt to find the re of rectngle tht mesures.7 cm y 4.6 cm. When I multiply, I otin the nswer 1.4 cm. However,.7 implies.65 cm to.75 cm 4.6 implies 4.55 cm to 4.65 cm The product should imply 1.0575 cm to 1.7875 cm But t he product 1.4 cm implies 1.415 cm to 1.45 cm, which is more precise thn the ctul result. The ccepted prctice when multiplying or dividing is to report the result using the fewest numer of significnt digits in the originl mesures given. Or, when you multiply or divide mesurements, the nswer should hve the sme numer of significnt digits s the mesurement with the lest numer of significnt digits. In our exmple, there re two significnt digits in.7 cm nd 4.6 cm, so the result is rounded to two significnt digits, 1cm. Exmple: Clculte. Use the correct numer of significnt digits. () Find the re of prllelogrm with height of 1.1 inches nd se of 6 inches. Are (1.1)(6), which is 7.6. However, the lowest numer of significnt digits is one, so we would round the nswer to 70 squre inches. () 14. 0.05 14. 0.05 84. However, the lowest numer of significnt digits is one, so we would round the nswer to 300. Another concept tht hs to do with mesurement is ccurcy. Mny people think precision nd ccurcy re the sme thing. They re not! The ccurcy of mesurement refers to how close the mesured vlue is to the true or ccepted vlue. For exmple, if you re in l nd you otin weight mesurement of 4.7 kg for n oject, ut the ctul or known weight is 10 kg, then your mesurement is not ccurte (your mesurement is not close to the ccepted vlue). However, if you weigh the oject five times, nd get 4.7 kg ech time, then your mesurement is precise ech mesurement ws the sme s the previous. Precision is independent of ccurcy. In this cse, you were very precise, ut inccurte. Another wy to explin it: Imgine sketll plyer shooting skets. If the plyer shoots with precision, his im will lwys tke the ll to the sme loction, which my or my not e close to the sket. If the plyer shoots with ccurcy, his im will lwys tke the ll close to or into the sket. A Mth 7 Notes Unit 9: Mesurement: Two-Dimensionl Figures Pge 3 of 17
good plyer will e oth precise nd ccurte: shoot the ll the sme wy ech time nd mke the sket. If you re soccer plyer, nd you lwys hit the left gol post (insted of scoring), wht cn you conclude? You re precise, ut not ccurte! A drtord nlogy is often used to help us understnd the difference etween ccurcy nd precision. Imgine person throwing drts, trying to hit the ull s eye. There re 4 scenrios: Not Precise/Not Accurte It is rndom pttern: drts re not clustered nd re not ner ull s eye. Precise/Not Accurte Drts re clustered together ut did not hit the ull s eye. Not Precise/Accurte Drts re not clustered together, ut their verge hit the ull s eye. Precise/Accurte Drts re clustered together nd their verge position hit the ull s eye. It is esy to confuse precision nd ccurcy. The tool tht you use ffects oth the precision nd ccurcy of your mesurement. Mesuring with millimeter tpe llows greter precision thn mesuring with n inch tpe. Becuse the error using the millimeter tpe should e less thn the inch tpe, ccurcy lso improves. Suppose tht tpe mesure is used to mesure the dimeter of two circles. Let s suppose you mesure the first circle to e 15 cm, nd second circle to e 01 cm. The two mesures re eqully precise (oth mesured to the nerest cm). However, their ccurcy my e quite Mth 7 Notes Unit 9: Mesurement: Two-Dimensionl Figures Pge 4 of 17
different. Let s further suppose tht the ccepted vlues for the mesurements re 16 cm nd 0 cm. The errors for these mesurements re 1 0.065 16 or 6.5% nd 1 0.0049504 or out 0 0.5%. The second mesurement is more ccurte ecuse the error is smller. You cn view short video on this topic: http://videos.howstuffworks.com/hsw/13176-discovering-mth-precision-nd-significnt-digits- video.htm Here is powerpoint regrding this mteril: http://www.cced.net/octcomhs/mth/cc_prec/ccurcy_printfile.pdf One more wy to think of this: ccurcy implies tht mesurement is siclly right, given mrgin of error. Precision is the level of detil; or typiclly the numer of digits fter deciml point. For instnce, I sk how fr it is to the store. You could tell me out 3 miles, while GPS device might tell me.85 miles. Aout 3 miles is pretty ccurte, ut.85 is oth ccurte nd precise. Now, you could hve told me 15.345 miles, which would mke you very precise, ut not ccurte. Most of the time in our everydy life, we wnt ccurcy; precision is not s useful. But in science nd engineering, oth precision nd ccurcy re importnt. Syllus Ojective: (3.3) The student will evlute formuls nd lgeric expressions for given vlues of vrile. Perimeter The perimeter of polygon is the sum of the lengths of the segments tht mke up the sides of the polygon. m Exmple: Find the perimeter of the regulr pentgon. Since the pentgon is regulr, we know ll five sides hve mesurement of meters. So we simply multiply 5 for n nswer of 10 meters for the perimeter. Exmple: Find the perimeter for rectngle with length 7 feet nd width feet. Since opposite sides of rectngle re equl in length, ( ) ( P 7+ )or 18 feet. Are of Prllelogrms, Tringles nd Trpezoids One wy to descrie t he size of room is y nming its dimensions. A room tht mesures 1 ft. y 10 ft. would e descried y sying it s 1 y 10 foot room. Tht s esy enough. Mth 7 Notes Unit 9: Mesurement: Two-Dimensionl Figures Pge 5 of 17
There is nothing wrong with tht description. In geometry, rther thn tlking out room, we might tlk out the size of rectngulr region. For instnce, let s sy I hve closet with dimensions feet y 6 feet. Tht s the size of the closet. ft. 6 ft. Someone else might choose to descrie the closet y determining how mny one foot y one foot tiles it would tke to cover the floor. To demonstrte, let me divide tht closet into one foot squres. ft. 6 ft. By simply counting the numer of squres tht fit inside tht region, we find there re 1 squres. If I continue mking rectngles of different dimensions, I would e le to descrie their size y those dimensions, or I could mrk off units nd determine how mny eqully sized squres cn e mde. Rther thn descriing the rectngle y its dimensions or counting the numer of squres to determine its size, we could multiply its dimensions together. Putting this into perspective, we see the numer of squres tht fits inside rectngulr region is referred to s the re. A shortcut to determine tht numer of squres is to multiply the se y the height. The re of rectngle is equl to the product of the length of the se nd the length of height to tht se. Tht is A h. Most ooks refer to the longer side of rectngle s the length (l), the shorter side s the width (w). Tht results in the formul A lw. The nswer in n re prolem is lwys given in squre units ecuse we re determining how mny squres fit inside the region. Exmple: Find the re of rectngle with the dimensions 3 m y m. A lw A 3 A 6 The re of the rectngle is 6 m. Mth 7 Notes Unit 9: Mesurement: Two-Dimensionl Figures Pge 6 of 17
Exmple: Find the re of the rectngle. 9 ft. yd. Be creful! Are of rectngle is esy to find, nd students my quickly multiply to get n nswer of 18. This is wrong ecuse the mesurements re in different units. We must first convert feet into yrds, or yrds into feet. yrds 1 x feet 3 9 We now hve rectngle with dimensions 3 yd. y yd. The re of our rectngle is 6 squre yrds. A lw A A 6 ( 3)( ) 9 3x 3 x If I were to cut one corner of rectngle nd plce it on the other side, I would hve the following: height height se se W e no w hve prllelogr m. Notic e, to form prllelogr m, w e cut piece of rectngle fro m one side nd plced it on the other side. Do you thin k w e chnge d th e re? The nswer is no. All we did ws rerrnge it; the re of the new figure, the prllelogrm, is the sme s the originl rectngle. So we hve the re of prllelogrm h. h Exmple: The height of prllelogrm is twice the se. If the se of the prllelogrm is 3 meters, wht is its re? Mth 7 Notes Unit 9: Mesurement: Two-Dimensionl Figures Pge 7 of 17
First, find the height. Since the se is 3 meters, the height would e twice tht or (3) or 6 m. To find the re, A h A 36 A 18 The re of the prllelogrm is 18 m. We hve estlished tht the re of prllelogrm is A h. Let s see how tht helps us to understnd the re formul for tringle nd trpezoid. h se h se For this prllelogrm, its se is 4 units nd its height is 3 units. Therefore, the re is 4 3 1 units. If we dr w digonl, it cuts the prllelogrm into tringles. Tht mens one tringle would hve one-hlf of the re or 6 units. Note the se nd height sty the sme. So for tringle, 1 1 ( )( ) A h, or 4 3 6 units h se h 1 1 se For this prllelogrm, its se is 8 units nd its height is units. Therefore, the re is 8 16 units. If we drw line strtegiclly, we cn cut the prllelogrm into congruent trpezoids. One trpezoid would hve n re of one-hlf of the prllelogrm s re (8 units ). Height remins the sme. The se would e written s the sum of nd. For trpezoid: 1 1 1 A 1+ h + nits ( ), or ( 6) 8 u Mth 7 Notes Unit 9: Mesurement: Two-Dimensionl Figures Pge 8 of 17
Circles: Circumference nd Are A circle is defined s ll points in plne tht re equl distnce (clled the rdius) from fixed point (clled the center of the circle). The distnce cross the circle, through the center, is clled the dimeter. Therefore, dimeter is twice the length of the rdius, or d r. We clled the distnce round polygon the perimeter. The distnce round circle is clled the circumference. There is specil reltionship etween the circumference nd the dimeter of circle. Let s get visul to pproximte tht reltionship. Tke cn with 3 tennis lls in it. Wrp string round the cn to pproximte the circumference of tennis ll. Then compre tht mesurement with the height of the cn (which represents three dimeters). You will discover tht the circumference of the cn is greter thn the three dimeters (height of the cn). You cn mke n exercise for students to discover n pproximtion for this circumference/dimeter reltionship which we cll π. Hve students tke severl circulr ojects, mesure the circumference (C) nd the dimeter (d). Hve students determine C d for ech oject; hve groups verge their results. Agin, they should rrive t nswers little igger thn 3. This should help convince students tht this rtio will e the sme for every circle. C We cn then introduce tht π d or C πd. Since d r, we cn lso write C πr. Plese note tht π is n irrtionl numer (never ends or repets). Mthemticins use to represent the exct vlue of the circumference/dimeter rtio. Exmple: If circle hs dimeter of 4 m, wht is the circumference? Use 3.14 to pproximte π. Stte your nswer to the nerest 0.1 meter. Using the formul: C πd C (3.14)(4) C 1.56 The circumference is out 1.6 meters. Mny stndrdized tests (including the CRT nd the district common exms) sk students to leve their nswers in terms of π. Be sure to prctice this! Exmple: If circle hs rdius of 5 feet, find its circumference. Do not use n pproximtion for π. C πr Using the formul: C π 5 C 10π The circumference is out 10π feet. Exmple: If circle hs circumference of 1π inches, wht is the rdius? Mth 7 Notes Unit 9: Mesurement: Two-Dimensionl Figures Pge 9 of 17
Using the formul: C πr 1π πr 1π πr π π 6 r The rdius is 6 inches. Exmple: A circle hs circumference of 4 m. Using π 3.14, find the dimeter. Round your nswer to the nerest whole numer. Using the formul: C πd 4 (3.14) d 4 (3.14) d 3.14 3.14 7.6 d The dimeter is out 8 meters. You cn demonstrte the formul for finding the re of circle. First, drw circle; cut it out. Fold it in hlf; fold in hlf gin. Fold in hlf two more times, creting 16 wedges when you unfold the circle. Cut long these folds. Rerrnge the wedges, lternting the pieces tip up nd down (s shown), to look like prllelogrm. rdius (r) This is ½ of the distnce round the circle or ½ of C. We know tht C π r, so 1 1 C π r 1 C πr Mth 7 Notes Unit 9: Mesurement: Two-Dimensionl Figures Pge 10 of 17
The more wedges we cut, the closer it would pproch the shpe of prllelogrm. No re hs een lost (or gined). Our prllelogrm hs se of πr nd height of r. We know from our previous discussion tht the re of prllelogrm is h. So we now hve the re of circle: A h A (π r)( r) A πr πr rdius (r) Exmple: Find the re of the circle to the nerest squre meter if the rdius of the circle is 1 m. Use π 3.14. Using the formul: A πr A (3.14)(1) A 45.16 The re of the circle is out 45 squre meters. Exmple: Find the re of the circle if the dimeter is 10 inches. Leve your nswer in terms of π. Using the formul: A πr A π ( 10) A 100π The re of the circle is 100π squre inches. Exmple: If the re of circle is 70 squre meters, find the rdius to the nerest meter. Using the formul: A πr ( ) ( 3.14) 70 3.14 r 70 r 3.14 3.14.3 r.3 r 4.7 r The rdius of the circle is out 5 meters. Mth 7 Notes Unit 9: Mesurement: Two-Dimensionl Figures Pge 11 of 17
Students should lso prctice finding the re of irregulr figure y reking it up into fmilir figures. Exmple: The dimensions of church window re shown elow. Find the re of the window to the nerest squre foot. 8 feet First, find the re of the rectngle. A h 11 feet A 11 8 A 88 Next, we hve hlf of circle. We re given the dimeter, so the rdius would e hlf of the 11 feet or 5.5 feet. To find the re of hlf of circle with rdius 5.5, 1 π A r 1 A 3.14 5.5 A 47.495 ( )( ) To find the totl re we dd the two res we found: 88 + 47.495 135.495 The re of the church window is out 136 squre feet. Squres nd Squre Roots The squre root of numer n is numer m such tht m n. The rdicl sign,, represents the nonnegtive squre root. The symol ±, red plus or minus, refers to oth the positive nd negtive squre root. Therefore, the squre roots of 36 re 6 nd 6, ecuse 6 36 nd ( 6) 36. Also, 36 6, 36 6, nd ± 36 ± 6. We understnd tht will e the positive vlue. We refer to this s the principl squre root. Students should memorize the vlues of the squres for 1 through 15. These perfect squres (squres of integers) re listed elow: Perfect Squres: 1 1 4 3 9 4 16 5 5 6 36 7 49 8 64 9 81 10 100 11 11 1 144 13 169 14 196 15 5 Mth 7 Notes Unit 9: Mesurement: Two-Dimensionl Figures Pge 1 of 17
Simplifying expressions such s the 5 nd 64 re pretty stright forwrd. 5 5 nd 64 8 But wht if the expression is lrge, like 576? Consider the perfect squres of few multiples of 10: 10 100 0 400 30 900 40 1600 nd so on... Now to find 576 : 400 < 576 < 900 Identify perfect squres closest to 576. 400 < 576 < 900 Tke positive squre root of ech numer. 0 < 576 < 30 Evlute squre root of ech perfec t squre tht I know. Vlues we now need to consider re 1,, 3, 4, 5, 6, 7, 8, nd 9. But wit! Since we know n even even even, our nswer must e even. So now our nswer choices re reduced to, 4, 6, nd 8. We now need to sk ourselves which of those nswer choices will give us 6 in the one s plce ( 576 ). There re only two: 4 (since 4 16 ) nd 6 (since 6 36 ). Since 576 is closer to 400 thn it is to 900, we quickly know tht the nswer is 4. A quick check will identify 4 s the nswer: 576 4. Let s look t nother exmple, finding 15. 900 < 15 < 1600 Identify perfect squres closest to 15. 900 < 15 < 1600 Tke positive squre root of ech numer. 30 < 15 < 40 Evlute squre root of e ch perfect squre tht I know. The vlues we need to consider re 31, 3, 33, 34, 35, 36, 37, 38, nd 39. Since our rdicnd is odd, we condense those considertions to 31, 33, 35, 37, nd 39. However, if we look t the rdicnd 15, we cn quickly recognize tht the only numer whose squre will give us 5 is 35. We hve our nswer very quickly! 15 35. The question tht we need to consider next is wht hppens if we wnt to tke the squre root of numer tht is not perfect squre. Mth 7 Notes Unit 9: Mesurement: Two-Dimensionl Figures Pge 13 of 17
Approximting Squre Roots You cn use perfect squres to pproximte the squre root of numer. Exmple: Approximte 53 to the nerest integer. First, find the perfect squre tht is closest ut less thn 53. Tht would e 49. The perfect squre closest to 53 ut greter thn 53 is 64. So, 53 is etween 49 nd 64. 49 < 53 < 64 Identify perfect squres closest to 53. 49 < 53 < 64 Tke positive squre root of ech numer. 7 < 53 < 8 Evlute squre root of ech perfect squre. Becuse 53 is closer to 49, 53 is closer to 7. Therefore, 53 7. One could lso use clcultor or tle to pproximte squre root. Another wy to mke the estimte is shown elow: Exmple: Approximte 53. Strt s we did with the previous exmple. 49 < 53 < 64 49 53 4 15 64 Determine the difference etween 49 (smller perfect squre) nd 53 (the rdicnd). The difference is 4. Then find the difference etween 49 (smller perfect squre) nd 64 (the lrger perfect squre). The difference is 15. We know tht 53 flls etween 7 nd 8. To pproximte the deciml, tke 4 15 0.6. We would estimte 53 7.7. Using clcultor, we find 53 7.8, which is very close to our estimte! Mth 7 Notes Unit 9: Mesurement: Two-Dimensionl Figures Pge 14 of 17
The Pythgoren Theorem Syllus Ojective: (6.4 )The student will model the Pythgoren Theorem using vriety of methods. Syllus Ojective: (6.5)The student will solve for the hypotenuse using the Pythgoren Theorem. In right tringle, the side opposite the right ngle is clled the hypotenuse. The legs re the sides tht form the right ngle. We typiclly lel the legs nd, while the hypotenuse is c. c hypotenuse legs If we look t enough right tringles nd experiment little, we will notice reltionship developing. Drw right tringle on piece of grph pper, with leg mesurements of 3 units nd 4 units. Drw squre on ech side of the tringle s shown. Cut out squres A nd B nd plce them side y side. Squre A Squre B Tpe the repositioned tringles. Compre your new squre with squre C. Squre A Squre C Drw two tringles s shown; cut long the hypotenuses nd slide s shown. Squre A Squre B Squre B Squre A Squre B It ppers tht the sum of the res of the squres formed y the legs is equl to squre formed y the hypotenuse. the re of the Here is nother wy to look t this reltionship: Begin with right tringle, legs nd with hypotenuse c. c Mth 7 Notes Unit 9: Mesurement: Two-Dimensionl Figures Pge 15 of 17
Crete squre tht hs mesure + on side, using 4 congruent tringles s shown. c The re of this new squre cn e expressed two wys: 1) A ( ) + (length of side is +, so multiply sides to get re) 1 ) A 4 + c (4 tringles, ech with re of 1 ; dd to the re of the squre with side c) c Setting these two vlues equl to ech other, we get ( ) + + c c 1 4 + + + + c This is n importnt reltionship in mthemtics. Since it is importnt, we will give this reltionship nme, the Pythgoren Theorem. Pythgoren Theorem: If tringle is right tringle, then + c. Exmple: Find the unknown length c in simplest form. + c 5 1 c Mth 7 Notes Unit 9: Mesurement: Two-Dimensionl Figures Pge 16 of 17
5 + 1 c 5 + 144 c 169 c 169 c 13 c Plese note tht this 5-1-13 is one of our Pythgoren triples. A Pythgoren triple is set of three positive integers,,, nd c such tht + c. Another triple worth noting is 3-4-5. You cn crete other Pythgoren triples y multiplying the originl triple y fctor. For instnce, 3-4-5 ecomes 6-8-10 y multiplying ll the sides y. One might crete nother, 10-4-6, y multiplying the 5-1-13 y. This skill is very vlule when working with prolems on the Nevd CRT nd NV High School Proficiency Exm. It cn sve the student time nd frustrtion! Exmple: Find the unknown length. 15 5 If we recognize the common fctor of 5, we cn think: 5 3 15 5 5 5 This is leg nd the hypotenuse of the 3-4-5. So the missing side would e 5 4, which is 0. We hd 3-4-5 tht ws multiplied y fctor of 5 to produce 15-0-5. So 0. Exmple: A 1-foot ldder is plced ginst uilding. The foot of the ldder is 5 feet from the se of the uilding. How high up the side of the uilding does the ldder rech? Give your nswer to the nerest foot. + c 100 < 119 < 11 5 + 1 100 < 119 < 11 1 5 + 144 10 < 119 < 11 119 Since 119 is closer to 11, 119 is closer to 11. 119 Therefore, the ldder reches pproximtely 5 ft. 11 feet up the side of the uilding. Mth 7 Notes Unit 9: Mesurement: Two-Dimensionl Figures Pge 17 of 17