Geometry / Mr. Hansen 5/5/2005

Name: ________________________

Solutions to Selected Chapter 12 Review Problems

	Chapter 12 Review (pp. 594-597)
4.	Imagine folding the big left flap rightward to make a box. If you can’t visualize this, try tracing the shape and cutting it out of scratch paper. Since the dimensions of the box are 5 by 9 by 2, V = lwh = (5)(9)(2) = 90 cu. units.
8.(a) (b) (c)	LSA = 4Atriangle = 4(½bh) = 4(½)(16)(10) = 320 sq. units TSA = LSA + Abase = 320 + s^2 = 320 + 16^2 = 576 sq. units Marked rt. D is 6-8-10 with h = 6; V = ßh/3 = 16^2(6)/3 = 512 cu. units
10.(a)	LSA = 3Atriangle = 3(½bh) = 3(½)(10)(12) = 180 sq. units TSA = LSA + Abase = 180 + s^2(Ö3)/4 = 180 + 10^2(Ö3)/4 = (180 + 25Ö3) sq. units
(b)	LSA = prl = p(6)(10) = 60p sq. units TSA = LSA + pr^2 = 60p + p(6^2) = 96p sq. units
(c)	LSA = 2prh = 2p(6.5)(15) = 195p sq. units TSA = LSA + 2pr^2 = 195p + 2p(6.5^2) = 279.5p sq. units
11.(a)	TSA = Ahemisphere + Acircle = ½(4pr^2) + pr^2 = 3pr^2 = 3p(5^2) = 75p sq. units
(b)	TSA = A“hemicone” + Asemicircle + Atriangle = ½(prl) + ½pr^2 + ½bh = ½p(5)(13) + ½p(5^2) + ½(10)(12) = 32.5p + 12.5p + 60 = (45p + 60) sq. units
15.	Since each base is an isosceles D formed from a pair of 7-24-25 rt. Ds joined together, ß = ½bhtriangle = ½(14)(24) = 168 sq. units. Therefore, V = ßhprism = (168)(30) = 5040 cu. units.
17.	Vdesired = Vblock – Vcylinder           = lwhblock – ßhcylinder           = (8)(5)(6) – p(1^2)(8)           = 240 – 8p » 215 cu. in.
18.	By Hero’s Formula on Smokey, ß = 10Ö2 sq. units. Therefore, V = ßhprism = (10Ö2)(7) = 70Ö2 cu. units.
20.	Sketch circle O with sector AOB: Do you see that circular segment AB (i.e., the sliver that equals sector AOB minus DAOB) is the cross section of the log? Since central ÐO = 60° and ÐA @ ÐB (base Ðs of isosc. D), DAOB is equilateral with area s^2(Ö3)/4 = 10^2(Ö3)/4 = 25Ö3. Asegment AB = Asector AOB – ADAOB              = (60/360)pr^2 – 25Ö3              = (1/6)p10^2 – 25Ö3              = 100p/6 – 25Ö3 Note here that we used the fact that r = 10, which follows from the cross-section diagram above. Because the log is an extruded shape (i.e., a generalized cylinder), Vlog = ßh       = (Asegment AB) h       = (100p/6 – 25Ö3) (30)       = 3000p/6 – 750Ö3       = 500p – 750Ö3 cu. units TSAlog = Afront cap + Aback cap + Aflat top + Acurved bottom           = Asegment AB + Asegment AB + (10 · 30) + (60/360)LSAcylinder           = 2(100p/6 – 25Ö3) + 300 + (1/6) Ch           = 200p/6 – 50Ö3 + 300 + (1/6)(20p)(30)           = 200p/6 – 50Ö3 + 300 + 600p/6           = 800p/6 – 50Ö3 + 300           = (400p/3 – 50Ö3 + 300) sq. units Again, note that the circumference C uses the fact that the log has diameter 20, which is clear from the cross-section diagram above.
21.	The trick is to extend the cone upward to apex A: By similar Ds, ratio GE : FD is the same as ratio AG : AF (be careful not to say AG : GF). But that is also the same as AE : AD (be careful not to say AE : ED). Let AE = x. Then x/(x + 6) = GE/FD = 9/12. Solving gives 12x = 9(x + 6) 12x = 9x + 54 3x = 54 x = 18 If AD = 24 and FD = 12, then by inspection, rt. Ds AGE and AFD are not only similar but also 30°-60°-90°. (You wouldn’t necessarily have this, but it’s helpful to notice that it happens in this particular problem.) Thus AG = 9Ö3, AF = 12Ö3. Vfrustum = Vcone ACD – Vcone ABE           = (1/3)ßbighbig – (1/3)ßsmallhsmall           = (1/3)(p12^2)(12Ö3) – (1/3)(p9^2)(9Ö3)           = (1728p/3)(Ö3) – (729p/3)(Ö3)           = (576p)(Ö3) – (243p)(Ö3)           = 333pÖ3 cu. units
22.(a)	The given triangle is 30°-60°-90° with hypotenuse 8. Thus the short leg is 4, and when spun, the figure makes a cone with slant height 8, radius 4, and height 4Ö3. Vcone = (1/3)ßh        = (1/3)(p4^2)(4Ö3)        = (64/3)(pÖ3) cu. units
(b)	When spun, the figure forms a tube with outer radius 4 and inner radius 3 (i.e., a short pipe with walls of thickness 1, outer diameter 8, and inner diameter 6). Another way to think about this is as a large cylinder with a cylindrical hole removed from its center. Vdesired = Vouter cyl. – Vinner cyl.           = ßouterh – ßinnerh           = (p4^2)(5) – (p3^2)(5)           = 80p – 45p           = 35p cu. units