M A T H 3 1 0 U Se lf-T est M EASU RE U

1. Find th e exact perimeters of, and areas inside of, f igures a and b.

a. b. c. d. 8 m ) | 8mm | . . . . . . . . . . . . . . . 6 m

»Assume . . . . . . . . . . . . . . .

8mm arcs are . . . . . . . . . . . . . . . 7m

semi-c ircular . . . . . . . . . . . . . . . ) . . . . . . . . . . . . . . .

| 6mm |

1c. Estimat e th e area inside t he cu rve sho w n abov e right (at 1 c), u sing t he “ w hole & half squares” proc edure.

1d. Refer to box. I f each dimension is doubled, are the volume & surface area doubled? Explain.

I f each d imens ion is increased by 1m, is the volume increased by 1m3 ? Explain.

2 . Demo nst rate w ith an appro priat e sket ch t hat th e area of a parallelogram is th e same as t hat of a rect angle

of equivalent height and width.

3 . W ith out havin g t o look up any fo rmu las, (a-c) f ind t he area enc losed by a:

a. trapezoid of height h w ith bases a and b. b. c ircle of radius r c. x o sect or of a circle o f rad ius r

& fin d: d . circ um feren ce of a circle o f rad ius r e. vo lum e and f . surf ace area of a sphere of radius r

4 . a. Sket ch t he poin ts (!4, 5) an d (4 ,1 ) in th e plane. Find th e dist ance b etw een t he poin ts.

b. Do th e point s (!1,4), (2,0 ), and (!2 ,!3) l ie at the vert ices of a r ight tr iangle? How do you know ?

5 . Find t he area inside a sq uare of side 1 2 c m. but out side of th e inscrib ed circ le.

6 . Find the volume of a prism 20 m eters high, given

th e oct agonal b ase has area 4 00 square m eters.

7 . Find t he vo lum e of a pyram id w ith height 20 cm . & base:

a. a 3 0 c m. by 3 0 c m. square.

b. an 80 cm 2 pentagon.

8 . Find the volume and lateral surface area of

a. a cone of height 25 cm . w i th base radius 5 cm.

b. a c ircular cylinder of height 25 cm. w i th base radius 5 cm.

9 . Find the surface area and volume of a w edge of brie cut from

a wheel 3 cm. high and 30 cm in diameter, given

a. t he w edge is on e-sixt h of th e original w heel.

b. t he cen tral an gle of th e w edge is 3 0 o.

1 0 . Find the total surface area and volume of an ice cream cone, f i lled and topped with a hemisphere of ice

cream, given these facts: the diameter of the top of the cone is 10 cm. and the height of the cone is 12 cm.

1 1 . Using t he f act th at w ater f reezes at 32oF & 0oC, & boils at 212oF & 100oC, det ermin e a fo rmu la th at relat es

temperatures in degrees Fahrenheit to degrees Celsius. Sketch a conversion graph, wit h oF on one scale, andoC on the other. Use your formula to convert each temperature to i ts counterpart . a. 72oF b. 30oC c. 98.6oF

1 2 . From t he list , select th e mo st ap prop riate met ric un its o f m easurem ent fo r each it em.

kg m mm cm km cc ml kl L t g (gm)

a. w idt h of a fin ger b. leng th of a fin ger c. height of a bridge d. d istan ce f rom N.Y .C. t o M iami

e. mass of a book f. mass of a r ing g. m ass of a fly h. m ass of a car

i. dro pperf ul of medic ine j. gaso line f or a car k. drink of w ater l . wat er in a swimming pool

1 3 . Conv ert 1 kilolit er of w ater t o lit ers; (t hen) t o m illiliters; to cc' s; t o gram s* ; t o kilo gram s; t hen t o m etric to ns.

1 4 . a. Carpet cost s $3 2. 00 a square y ard; w ood flo oring $4 per squar e fo ot . W hich cost s less ?

b. How many square inches comprise a square foot? How many square feet comprise a square yard?

c. How ma ny cu bic inc hes are in a cu bic fo ot ? .. .c ub ic f eet in a c ub ic y ard ? .. .c ub ic i nc hes in a c ub ic y ard ?

d. How many cc' s (cm3) in a cu bic me ter? Ho w ma ny mi llili t ers in a cu bic me ter?

e. How many gram s of w ater in a cub ic m eter? (at 4 oC) f . 1300 mL = daL

g . 30 0 dam = cm h . 20 0 cm2 = m

2i . 328 dL = ___ cm

3j . 1 2 . 5 m

3 = dm

3

1 5 . John must w rap a new baton, 26" long, for h is orchest ra leader . Wi ll the baton f i t in a 24" by 7" by 7" box?

1 6 . A photo is enlarged to 2.5 t imes or ig inal height & w idth. The area was 100 cm 2. What is the area now?

M A T H 3 1 0 U MEASURE Self-Test ANSW ERS U

1 . a. (½)CB(4u)2 + [(8u)2 ! B(2u)2] = (4B+ 6 4 ) u2 (see diagram `) 1 a .

Perimet er = B4 u + 2 C8 u + 2 C ½ C2B(2u) = (8 B+ 1 6 ) u ²(½)B(2u)2

b . 7 .5 u2 [ 4 C6 – 4 C3 C½ – 1 C3 C½ – 6 C3 C½ = 7 . 5 ] (work f rom outside in) ²(½)B(2u)2

Perimeter = ( 5 + r &1&&0& + r &4&&5& ) units (Use the Pythagorean theorem.) ü (½)CB(4u)2 (8u)2 — Thatü

c. w hole + ½ partial = 10 u2 + (½ )(10u 2) = 1 5 u2 . . . . . . . .

d. 2lC2wC2h = 8lwh = 8 C(original v olum e) . . . . . . . .

1b. . . . . . . . .

3 . a. h @(a + b) /2 b. Br2 c. (x o/ 3 6 0o)CBr2 . . . . . . . .

d. C = 2Br e . V = 4Br3 f . SA = 4Br2 . . . . . . . .

3 ))))))) )

4 . a. r 4 2 + 8 2 u = 4 r 5 uni ts. 2 .

b. D[(!1,4),(2,0)] = 5; D[(2,0),(!2 ,!3)] = 5; D[(!1,4),(!2 ,!3 ) ] = 5 0 1/2 .

Yes: d12 + d2

2 = d32.

5 . Area in square – Area in c irc le = (12 cm)2 ! B(6 cm)2 = ( 1 4 4 ! 3 6B)cm2

6 . V o f p rism o r cy linder = (Area o f b ase)C(Height) = 8 0 0 0 m3

7 . a. (a)C(30 cm)220 cm = 6000 cm 3 Volume of pyramid or cone = (1/3) (area of base)(height)

b. (a)C( 8 0 cm 2)C20cm = 1600 /3 cc. ( “cc” is an al ternate abbreviat ion for cm3 – cub ic cen tim eters)

8 . a . V = (a)CB(5 cm)2C2 5 c m = 6 2 5 B/3 cm 3 lateral SA = ( 1/ 2 )CB(10 cm )(5 r &2&&6& c m ) = 2 5 r &2&&6& Bcm 2

b . V = B(5 cm)2C2 5 c m = 6 2 5B cc º BTW: Tot al SA = ( 25B + 25r &2&&6& B ) cm2

lateral SA = B(10cm)C2 5 c m = 2 5 0Bcm 2 BTW: Tot al SA = ( 2C25B + 250B ) cm2 ³ radial faceþ þpart of Circumference

9a. SA = top+ bottom + radial faces + str ip of C V = one-six th of th e V o f t he origin al w heel

SA = 2 C( 1 5 cm C3cm) + 2 C( 1 /6 )CB(15cm)2 + 3 C3 0Bcm 2/6 V = ( 1 /6 )CB(15cm)2C3 c m = 2 2 5B/2cm 3

9 b . S A = 2 C( 1 5 cm C3cm) + 2 C( 3 0 /3 6 0 )CB(15cm)2 + (1/ 12 )3 C3 0Bcm 2

V = ( 3 0 /3 6 0 )CB(15cm)2C3 c m = 2 2 5B/4 cm 3 30 ° is 1/1 2 of the c ircle, so V is half V of part a

1 0 . SA = (1/2)C4B(5cm)2 + ( 1 /2 )CB(10cm)C1 3 c m = 1 1 5Bcm 2 ³ V = V hemisphere + V cone = ( 1/ 2 )C( 4 /3 )B(5cm)3 + ( 1 /3 )CB(5cm)2

C( 1 2 cm ) = ( 55 0B/3)cm3 ³

1 1 . The relationship between oC and oF is a l ine going from (0,32) to (100,212 );

that makes a r ise of 100 oC equal to a r ise of 180 oF: i .e. 100 oC = 1 8 0 oF, so ( 1 00 oC /1 8 0 oF ) = 1! .

1 0 0 oC /1 8 0 oF reduc es to 5 oC/9oF. An d, y es, 5 oC/9oF = 1 also. So oC = (5 oC/9oF)C(oF ! 3 2 oF).

(Not ice oF must be adjust ed dow n t o 0 bef ore m ult iplyin g, so th at 3 2 oF wil l end up being 0 oC.).

C = (5/9)(F ! 3 2 ) a . 7 2 oF = 22 oC (18 °Reaumur) c. 98.6 oF = 37 oC

F = (9/5)C + 32 b . 30 oC = 86 oF

1 2 . a. cm b. cm c. m d. km e. kg f . g g. g (act ually! mg , bu t t hat ' s not in t he list )

h. t i . mL or cc j . L k. mL or cc l . kL

1 3 . 1 k L = 1 0 0 0 L = 1 0 0 0 C1 0 0 0 m L = 1 0 6m L = 1 0 6cc = j 1 0 6g = 100 0 kg = 1 metric ton. OR:

1 kL = 1 kL C 1 0 0 0 L C 1000 mL C 1 cc C 1 g* C 1 kg C 1 metric ton = 1 metric ton

1 kL 1 L 1mL cc 1000 g 1000 kgj val id ON LY f or w at er at 4°C

1 4 . a. Carpet cost s $32/yd 2; w o o d co st s $ 4 / f t2 = $ 4 / f t2C3f t /yd C3f t /yd = $36/yd 2; carpet is ch eaper.

b . 1 f t 2 = (12 in)2 = 1 4 4 in2; 1 yd2 = ( 3 f t )2 = 9 f t 2 (Use the method shown in #13.)

c . (1 f t )3 = (12 in)3 = 1 7 2 8 in3; (1 yd)3 = ( 3 f t )3 = 2 7 f t 3; 1 yd3 = 2 7 f t 3 = 27 C1 7 2 8 i n3 = 466 56 in 3

d . 1 m 3 = ( 1 0 0 c m )3 = 1000 000 cm 3; (see # 13 )

e. (see #13; t he conversion at * is val id for water only!)

f . 3 0 0 dam g . 20 0 cm 2 = 200 cm 2 C 1m C 1 m. = . 0 2m 2 h. 3 28 dl = 32 8d l C100mlC1cc = 3 0 0 0 m 100cm.C100 cm d l m l = 3 0 0 0 0 cm = 3 2 8 0 0 cc

1 5 . Just barely n ot . Th e longest dimen sion in a 2 4" x7 " x7 " box is th e ext reme d iagonal, (67 4) ½ “ Ñ 2 5. 96 "

1 6 . New area = 100cm 2 C 2 . 5 C 2.5 = 625 cm 2

M AT H 3 10 Self -Test M EASURE ext ended solut ions w ith com ments

1a. We assume arcs are semi-circular.

| 8mm | ) We v iew this as a half-circular region[I] (wit h diameter 8mm)

8mm joined t o a rectangular region [II] (8mm by 8mm) from w hich tw o small semicircular regions [III] have been removed )

| 6mm | radius of th e small c ircles m ust be t he dif feren ce 8 mm – 6m m= 2m m .(And must also be ½ of 8mm/2 .)

Area I: radius is ½ of 8mm, or 4mm. Area of half -circular disc is (½ )B(4mm)2 Area II: area of 8m by 8mm square is 64mm 2

Area III: area of t w o half-circular “cutout s” [diameter 8mm'2 = 4mm]. .. is 2C½ CB(2mm)2 Thus Tot al Area of f igure is (½ )B(4mm)2 + 64mm 2 — 2C½ CB(2mm)2 = (4B + 64)mm2

1b. . . . . . . . . We start w it h the smallest vert ical-horizont al bounded rectangle that . . . . . . . . contains the polygon given (w hich happens to be a t riangle, but that ’s . . . . . . . . not important, this w orks f or any polygon w ith vert ices on lat t ice pts). . . . . . . . . Area within polygon = area of rect angle – areas of “ take-away” parts: . . . . . . . . (4u)(6u) – ½ (4u)(3u) – ½ (3u)(1u) – ½ (6u)(3 ) = 24 – 33/2 = 7.5 u2

1c. . . . . . . . . We cover the region w ith squares. . . . . . . . . We count the squares w hich lie entirely w ithin t he region outlined. . . . . . . . . We count the squares w hich lie partially w ithin t he region outlined. . . . . . . . . We then make the ESTIMATE:

. . . . . . . . Area w ith in out line . 1 0 u2 + (½ )(10)u 2 = 1 5 u2 (approx imat e area)

(This is based on the assumption t hat the squares which are partiallyenclosed average half inside, half out side)

1d. 8m 6m The volume of t he box is (area of base)(height) = lengthCw idthCheight

7m = (6m C 8m)C7m = 336 m3

Doubling al l t he dimensions, w e can compute: NEW V= (12m C16m)C14m = 2688 m3 ... that is, 8C336 m3

In f act , w e could have ant icipat ed this, since V = 2 l C2 w C2 h = 8 l w h = 8C(original volume)

If each dimension is increased by 1m, is the volume increased by 1m3 ? Explain. 8m + 1 m

NO! The increase in volume is much, much great er than t hat ! 7 m Just consider, first the eff ect of increasing just ONE dimension– !!

e.g. increase the length from 8m by 1m. 6 m As il lustrated at lef t , t his w ould add a 1m by 6m by 7m “ slab” to one end

of t he box. That alone, then, adds 42 m3 of volume to the box.

Here w e illust rate the addit ion of a 1m extension to the depth of the box.Just f or the old box, this extension results in an additional 1m C8mC7m or56m3, and that does not even take into account the extension of the box From 8m length to 9m! To account for that, w e need another 1m by 1m by 7m piece in the corner!

Extending t he height an addit ional 1m, from 7m t o 8m results in a newslab added to the top of t he box, 1m by 6m by 8m t o cover the original box,but 1m x 7m x 9m t o cover the box w ith it s new ext ended length and depth.(³Picture it yourself!)

Here’ s the ent ire dif ference, seen algebraically: Original volume l w h, NEW V = (l+a)( w+a)(h+a) = l w h + l w a + l a h + a w h + l a a + a w a + h a a+ a a aDif ference: l w a + l a h + a w h + l a a + a w a + h a a+ a a a (Replace a by 1m )

4. a. Sketch the point s (!4,5) and (4 ,1 ) in the plane. Find the distance betw een the points. ( -4,5)

. . . . . . . . . . The change in x = 5 ! 1 = 4

. . . . . . . . . . The change in y = 4 !(!4) = 8

. . . . . . . . . .

. . . . . . . . . . 42 + 82 = c2

. . . . . . . . . . ( 4 ,1 ) c2 = 80

. . . . . . . . . . c = sqrt (80 ) = sqrt (16 C5) = 4sqrt (5)

. . . . . . . . . .

b. Do t he point s (!1,4), (2 ,0), and (!2,!3) lie at the vertices of a right triangle? How doyou know ?

The solut ion is based on the Pythagorean Theorem: the sum of the squares of t he tw o shortlegs is equal t o the square of the long leg i f , and ONLY if , t he t riangle is a right triangle. Sochecking the distances bet w een the three pairs: 5, 5 , and sqrt (50), w e see the py thagoreanrelat ionship holds true, so t he t riangle must be a right triangle.

16 . The scale fact or applies to bot h dimensions that cont ribute to area: height and w idth. Thescale factor is the ratio of new lengths to old, so the ratio of (New height)/(Old height) = 2.5 andthe ratio of (New w idth)/ (Old width) is also 2.5 . New Area = (Old Area) (Ratio of New height to Old height ) (Ratio of New w idth to Old w idth) = (Old area) (scale factor) (scale factor)

If you t hink in terms of the simplest area (think rectangle), t his is obv ious.

x by y 2.5x by 2.5y It becomes even more obvious w hen scaled up by a factor w hich is a whole number, say 3:

x by y 3x by 3y: Area is 9 t imes as great.