Property Transactions: Capital Gains and Losses

() 1. Convert each of the follow ing units, show ing your w ork. a. 0.052 km = .052 km b. 1000 m 1000 mm @ ))))))) @ ))))))) 1 km 1m .052 @1000 @1000 mm = = 52000 mm kg = 250 g 1kg @ )))))) 1000 g 0.53 m3 = 0.53 m 3 = d. .052km 250 g = 250 g c. = mm = 250 1000 ))))) kg = .250 kg cm3 0.53 m 3 100 cm 100cm 100cm = .53 @100@100@100 cm 3 @ )))))) @ )))))) @ )))))) 1m 1m 1m 2500 mL w ater (at 4EC) = 2500 mL = 2500 mL = 530,000 cm3 kg. 1 g* 1 kg @ )))))) @ )))))) 1 mL 1000 g = 2500 1000 ))))) kg = 2.5 kg * ( this equivalence good only for substances w hich have the same density as w ater at 4EC ) () 2. Place the follow ing measures in increasing order: 0.5 in .5 in < 2dm < 50 cm < 2 ft < 1m < Sm allest () 3a. 1.5 yd Largest Draw a sketch w hich illustrates the relationship betw een square inches and square feet. 1 ft = 12 in ² 1 in2 3b. 1m 2 ft 50 cm 2 dm 1.5 yd Use t he f act t hat 1 y d is close t o 1 m t o “ ballpark” t hese lengt hs! So, f or inst ance, 1 m < 1 .5 y d and 5 0 cm = ½ m < 2 f t (since 2 f t = by d . bm ) Which gold leaf is less expensive? X charges $20/sq ft; Y charges $1/sq in Which should Jean buy? Show w hy. To buy 1 square foot at $20 per ft 2 , cost is $20 To buy one sq ft at $1 per in2 , cost is: 1 ft = 12 in 1 ft 2 = 1 ft 2 @ $1 ))2 @ 12in )) @12in )) = $144 in ft ft Notice how dimensional analysis handles ALL units above. X, at $20 per square foot, is far less expensive! ONE SQUARE FOOT = 144 square inches ! Another w ay to figure this– takes some realization of the relative sizes of in2 and ft 2 : At $1/in2 , $20 buys 20 in2 , a lot less gold leaf than one ft 2 . People often do not picture a square w hen they talk about square feet (or square yards, or square meters, etc.) Unscrupulous retailers can take advantage of that. Here’ s another example: Look at the classroom floor tiles! Each tile covers 1 ft 2 – area of a square 1 ft on each side. As w e previously discussed, a square yard is the area of a square 1 yard on each side. It takes NINE of those 1-sq-ft tiles to cover a square yard! (Not THREE !) () 4. A building has the form of a 10 m cube, topped by a 12 m high square-based pyramid ( show n at right ). Find the V OLUM E of this building. Show your w ork in a clear, orderly manner. VOLUME of building = VOLUME OF CUBE + VOLUME OF PYRAMID 12m = ( length of side)3 = (10 m )3 + (a) (10 m)2 12m = 1000 m3 + (a) 100m2 12m = 1000 m 3 + 400 m 3 + (a) (area of base) (height) 5m 10m 1400 m3 = () 5. Find the SURFA CE A REA of this right triangular prism. 10m 6m The SURFA CE A REA of this prism consists of the areas of the triangular bases, and its three rectangular faces. 18m 8m Note the dimensions (6-8-10) show base is a right triangle. Areas of the tw o congruent bases add up to 10m 2 @ (½) base @ height 2 @ (½) @ 8m @ 6m 6m 48 m 2 8m Areas of the rectangular faces add up to 8m @ 18m + 6m @ 18m + 10m @ 18m 144 m2 + 108 m2 18m + 180 m2 10m 432 m 2 or (8m + 6m + 10m) @ 18m = 24 m = 432 m2 6m (One long “ strip” ) 18m @ 18 m The tot al surface area of this prism is 8m 48 m2 + 432 m2 = 480 m2 () 6. Find the PERIM ETER and A REA of the demilune* w indow show n at right. * ( rectangular w ith a semicircular w indow on top ) Perimeter = rectangular parts length + length of semicircular arc on top = 6 ft + 4 ft + 6 ft + (½) ð(diameter) = + (½) ð (4 ft) = Area = = 7. Find the 6 ft ( 16 + 2 ð ) ft = area of rectangular part + area of semicircular top = () 16 ft A REA ( 6 ft ) ( 4 ft ) 24 ft + 2 + 4ft (½) ð (2 ft )2 2 ð ft 2 2 ð) ft 2 ( 24 + of the shaded sector of the circle given. Area of entire circle w ould be ð (radius )2 ... and radius is 30 cm 30 cm 40E ð ( 30 cm ) 2 Area of this sector = (appropriate fraction) of area of entire circle 3 6 0 E– 4 0 E = 3 2 0 E Or... 4 0 /3 6 0 = 1 /9 , so t he rest is 8 /9 of t he circ le () 8. 320E ð ( 30 cm )2 360E = ))) = ))) = 8 9 ð @ 900 cm2 800 ð cm2 The oblique cylinder show n at right has a base w ith area 77 m 2 . Find its volume. Volume of any cylinder = (area of base) (height) 10 m 12m ... and height is alw ays perpendicular to the base ! Volume of this cylinder = ( 77 m2 ) ( 10 m ) = 770 m3 () 9. Write a formula w hich gives the volume of a right circular cylinder w ith base radius r and height h. Volume = (area of base) (height) = ð @ r2 @ h What happens to the volume of the cylinder if the base radius is doubled and the height is tripled? New V= ð @(2r)2 @ 3h = ð @ 4r2 @ 3h F 12 times as great. = 12 ð @ r2 @ h = 12 (original volume) () 10. Write a formula for the surface area of a sphere in terms of its radius. 4 ð r2 It’ s round! It’ s area! Write a f ormula f or t he volume of a sphere in terms of the radius of the sphere. 4 3 ) 3 ð r It’ s round! It’ s volume! () 11. Find the distance betw een the points (– 5,– 1) and (7,4). . . . . . . . . . Distance is length of line draw n; call it “ c” . c 2 = 12 2 + 5 2 c 2 = 169 c = 13 () . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . c. . . . . . . . . . . . . . 5. . . . . . . . . . . . . . . . . 12. . . . . . . 12. Find the area of an equilateral triangle w hose sides are 2 cm long. Area of triangle = (½) w idth @ height Height of triangle is h w here h2 + (1 cm)2 = (2 cm)2 h2 = 4 cm 2 – 1 cm 2 = (½) 2 cm @ %& 3 cm 2cm h = %& 3 cm2 h2 = 3 cm 2 1 cm h () 1 cm %& 3 cm = 13. Use the sketch at right to prove the pythagorean theorem. Given– the entire figure is a square, a+ b on each side.... (Hint: area of the entire figure; areas of parts; how do you know the c by c figure in the center is a square?) a b 1 3 b 2 a c 1 c 2 c a b 2 2 b c a 2 Area of the entire (large) square is ( a + b) = a + 2ab + b Areas of the parts = Areas of the 4 Îs + area of inner square* = 4 (½) a b + c 2 = 2ab + c 2 The area must be the same either w ay it is calculated, (The w hole is the sum of its parts) so: a2 + 2ab + b2 = 2ab a2 = c2 + b2 + c2 ... subtracting 2ab from each side yields QED * As for the question as to w hether the c by c by c by c rhombus in the center is really a square: The four a-b-c triangles are congruent right triangles. Because they are right triangles, their angles marked 1 and 2 are complementary, thus total 90E. Therefore the angle marked 3 must be a right angle (because, as show n at the top of the diagram, angles marked 1 and 2 and 3 together form a line, a 180E angle).