Basic Concepts of Geometry. Test Measure Answers

(6)1a. Draw a sket ch w hich illust rat es t he relat ionship bet w een square yards and square f eet. 1 yard (or 3 feet) 1b. Show the dim ensional analysis f or t he conversion of 60 square yards t o square f eet. Complete the conversion. In t he sket ch w e can see t hat 1 square y ard is 9 square f eet. 1 y ard (or 3 feet) Dimensional analysis says t his: (12) 2. a. 1 yd2 5.2 km = 60 yd2 = ))))) x 60 yd2 = 520000 cm ))))) x ))))) 3 ft yd 3 ft yd = 9 sq. ft 3 ft yd 3 ft yd = 540 ft 2 ))))) x ))))) cm 5.2 km 1000m 100cm ))))) x )))))) x ))))) = km m = 5.3 m3 = 5.3 m3 c. ))))) x Convert each of the following, showing your work. 5.2 km b. 1 yd2 = 520000 cm 5300 000 cm 3 cm3 5.3 m3 ))))) x 100cm 100cm 10 0cm )))))) x ))))) x ))))) = m m m 25 50 mL w ater (at 4°C) = _ 5300 000 cm 3 _____ kg. 1 kg 2550 mL = 2550 cc = 2550 g = 2550 g x ))))) = 2.550 kg 1000 g (5) 3. How many liters of w at er ar e needed to f ill a t ank 1 met er w ide, 1 met er high and 1 met er deep? Show t he dimensional analysis that leads to your answer. 1 m3 = 1 m3 ))))) x 10dm m )))))) 10dm 10dm x ))))) x ))))) = m m 1000 dm3 = 1000 L OR 10 00 dm3 = 10 00 dm3 x 10 cm ‰ 1 dm  )))))) 3 = 1000 000cm 3 = 1000000 g = 100 000 0 mL = 1000 L If w e know : 1 cc = 1 g = 1 mL (o f W A T ER a t 4°C ) If w e know 1 dm 3 = 1L (10) 4. Find t he volume of the py ramid show n at right , gi ven that the base of the py ramid has area 84 0 m 2 and the height (h ) is 60m. Show your work. Volume of Cone or Pyramid is 1/3 t he Volume of t he corresponding cylinder. Thus Volume = (1/3 ) (Area ofbase) (height) = (1/3) ( 840 m2 ) ( 60 m ) 3 = 16,800 m (10) 5. The Flatiron Building (oldest remaining sky scraper in NYC) is shaped like a triangle, as seen from above. (Well, OK, a slightly rounded triangle.) The dimensions are as shown. The height of the fairly flat -topped building is 285 ft (from ground level). Using t his inf ormation, f ind the volume (as near as possible) of t he Flatiron Building (th at is abov e ground lev el). 173ft The Flatiron Building, then, is a triangular-based prism. Thus it s volum e is: 190 ft Volume = (Area of base) (height) = (87 ft x 173 ft /2) (285 ft) 87 ft = (½) (87)(173) (285) ft 3 = 2144767.5 ft 3 1903 picture of Flatiron Building from... w w w .hellonewy ork.com (See a good modern pi ct ure of the Flat iron Building at : htt p:// w w w .budow le.pl/budow le/grafika/ameryka_polnocna/flatiron_building.jpg ) 6. (5) Find the ent ire surf ace area of the solid object illustrat ed below. Assum e all angles betw een connected segments are right angles. 20m 20m 22m The “roof ” , as seen from straight above, is a 50 m x 5 0m square Surface A rea = SA(roof + base) + SA(Big Laterals) + SA (small lat erals) = 2 C (50m)2 + 4 C50mC60m + 4 C20mC22m 50 m 60 m = 2 C 250 0m = 187 60 m 50 m The “ roof ” as seen f rom abov e: 2 2 + 4 C300 0m 2 + 4 C440 m 2 7. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . In all the above, the area shown here is one square unit. (5) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9. . . . . . . . . . . . . . . . . . . . . . . . . (-4, -3) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ( 4 ,3 ) . . . 6 . . .(4, – 3) 8 7. Estim ate the area of t he figure in #7. # of w hole square units w ithin region + (½) #of partial units w ithin region + (½) (18 unit 2 ) 14 unit 2 23 unit 2 (5) 8. Find th e area enclosed by th e figure in #8. (Curve t urns at point s: (1 ,1 ) & (6 ,7 ) & (1 ,7 ) & (2 ,5 ) ) 5 C 6 u2 30 u2 — (½) 6 C1 u2 —(½) 5 C6 u2 — 3 u2 — 15 u2 12 u2 (8) 9. Find the PERIMETER of t he triangle in figure #9 above Perimeter = Dist ance from (– 4, – 3) t o (4, -3) + Distance to (4,3) + Length of Hypot enuse = 4– (– 4) = = + 3 – (– 3) 8 + + 6 + square root (8 2 + 6 2 ) 10 24 (unit s) 82 + 64 + (8) 10 . 6 2 = c2 36 = c 2 100 = c 2 Find the AREA of the figure at right, given all arcs are semicircular. 7m m 1 0 m m 7 m m First w e look at the region, and det ermine it is just a semic irc ular regio n w it h a semicircular “ bit e” taken out ! Thus t he area cont ained is: Area of t he larger semicircular region —area of t he smaller semicircular region .. .w here R is t he radius of the = ½ of B R 2 — ½ of B r 2 outer semicircle, and r is the radius of the inner semi cir cle. = (½ ) B (12mm) 2 — (½ ) B (5mm) 2 = = (½) 144 B mm 2 (½) 119 B mm 2 — (½ ) 25 B mm 2 (5) 11. Find the area of a 60° sector of a circle w ith radius 3 m. The area of a 60 ° sect or of a circle is just 60 /3 60 , or 1/ 6 of the area of the ent ire circ le. (1/6 ) of B R2 = (1/6 ) of = (1/6 ) of B (3m) 2 B (3m) 2 /6 = (4) B R2 = B (3/ 2) m 2 = 3 B/2 m2 OR 1.5 B m2 12. Write an expression for the Volume of a sphere of radius r: (4/3) Write an expression for the Surface Area of a sphere of radius r: B r3 4 B r2 Volume MUST be 3-dimensional. Area MUST be 2-dimensional. (6) 13. Find the area of th e shaded region (w it hin t he square, out side t he circle). AREA of SQUARE — AREA of CIRCLE Side2 — B R2 (2m)2 — B(1m) 2 4 m2 — B m2 (4 – B) 2m 2m m2 Ñ .858 m 2 Addit ional M easure Questions (1 6 pts) Name: 1. A circular cylinder was designed to hold 500mL. If t he diameter of a cylinder is doubled, and the height is tripled, w hat is the new capacity ? V = 500 mL B r2 V= h If t he diameter is doubled, then t he radius is doubled; so t he new radius = 2r If t he height is t ripled, then t he new height is 3 h. NV = B (2r) 2 NV = B 4r 2 (3h) 3h = 12 B r2 h = 12 C(original volume) So the volume is multiplied by 1 2. (Increased tw elve-fold) 2. A very t all (Oops, make that v ery SHORT) right circular cone has dimensions measured as show n: Using this inform ation, det ermine the volume of t he cone, showing any intermediate steps and computat ions needed. 3. Find t he complete surf ace area of the cone. 25cm 2. Circumf erence of base = 49 B cm tells us the radius of the base is 24.5 (ugh!) (Sinc e 2 B r = 49 B , r = 49 B /(2 B ) ) Then the height of the cone is the vertical arm of the right t riangle (...t he SHORT right tr iangle, c ont rary t o t he illust rati on!): 25 cm H 24.5 cm H2 H2 + 24.5 2 = 25 2 + 600.25 = 625 C = 49 B cm H2 = 24.75 H = square root of 24.75 Ñ 4.974937 So now w e know th e height of th e cone is (square root of 24 .7 5) and the radius of the cone is 24.5 cm and the slant height of the cone is 25cm. Volume cm Ñ 4.97 4937 cm = (1/ 3) of Ar ea of Base CHeight of c one = (1/ 3) B r 2 H Ñ (1/ 3) B (24.5cm) 2 (4.974 937 cm) Ñ (1/ 3) B (24.5cm) 2 (4.974 937 cm) Ñ 3127.15 cm 3 3. Surf ace A rea of cone = Area of Base + area of the lateral surface + (½) Perimeter of base CSlant height = B r2 2 = B(24.5 cm) + (½) 49 B cm C25 cm = 600.25 B cm 2 + 612.5 B cm 2 = 1212.75 B cm2 Ñ 3809 .97 cm 2