Answers:
21. The other leg is 19.8 inches
22. The triangles are not similar
23. The three sides of triangle XYZ are XY = 21; YZ = 42; XZ = 49
24. The fourth angle is 195 degrees
25. The area is 91 square feet
26. False; Change "any vertex" to "midpoint of any side"
27. The segment QS is 10.4 units long
28. Arc RQU is 155 degrees
29. The area of the shaded sector is about 32.7 square cm
30. The surface area is exactly 152 square meters
31. The volume of the solid is roughly 1389 cubic inches
32. The volume of the cone is about 3.5 cubic feet
Work Shown:
Problem 21)
LL = longer leg, SL = shorter leg
(SL of triangle1)/(SL of triangle2)=(LL of triangle2)/(LL of triangle2)
(9)/(SL of triangle2)=(10.5)/(23.1)
(9)/(SL of triangle2)=0.45454545454546
9=0.45454545454546*(SL of triangle2)
0.45454545454546*(SL of triangle2)=9
SL of triangle2=9/0.45454545454546
SL of triangle2=19.7999999999998
SL of triangle2=19.8
Problem 22)
12/5 = 15/4
2.4 = 3.75
Equation is false; triangles are not similar
You have the correct answer
Problem 23)
Perimeter of ABC = AB+BC+AC
Perimeter of ABC = 9+18+21
Perimeter of ABC = 48
scale factor = (perimeter of XYZ)/(perimeter of ABC)
scale factor = 112/48
scale factor = 7/3
So that means
XY = (7/3)*(AB) = (7/3)*9 = 21
YZ = (7/3)*(BC) = (7/3)*18 = 42
XZ = (7/3)*(AC) = (7/3)*21 = 49
In summary:
XY = 21
YZ = 42
XZ = 49
Problem 24)
(angle1+angle2+angle3) = 280
angle4 = 3*x
angle5 = x
S = 180*(n-2)
S = 180*(5-2)
S = 180*(3)
S = 540
All five angles of any pentagon add to 540
angle1+angle2+angle3+angle4+angle5 = 540
(angle1+angle2+angle3)+angle4+angle5 = 540
280+angle4+angle5 = 540
280+3x+x = 540
280+4x = 540
280+4x-280 = 540-280
4x = 260
4x/4 = 260/4
x = 65
If x = 65, then
angle4 = 3*x
angle4 = 3*65
angle4 = 195
Problem 25)
A = 0.5*h*(b1+b2)
A = 0.5*7*(11+15)
A = 0.5*7*26
A = 3.5*26
A = 91
Problem 26)
False. An apothem goes from the center to the midpoint of any side. Changing "any vertex" to "midpoint of any side" will correct the statement to make it true.
Problem 27)
All radii of the circle are congruent, so PS = PR.
QS is a diameter so it's twice that of the radius
QS = 2*PS
QS = 2*PR
QS = 2*5.2
QS = 10.4
Problem 28)
Angle TPU = 25
Angle SPT = 80
Angle RPQ = angle SPT
Angle RPQ = 80
Angle SPU = Angle SPT + angle TPU
Angle SPU = 80+25
Angle SPU = 105
Angle QPU+angle SPU = 180
Angle QPU+105 = 180
Angle QPU+105-105 = 180-105
Angle QPU = 75
Angle RQU = Angle RPQ + angle QPU
Angle RQU = 80 + 75
Angle RQU = 155
Arc RQU = angle RQU = 155
Problem 29)
A = Area of full circle
A = pi*r^2
A = pi*5^2
A = 25pi
S = area of shaded sector
S = (150/360)*A
S = (150/360)*25pi
S = 32.7249234748937
S = 32.7
Problem 30)
L = 8
W = 6
H = 2
SA = 2*(L*W+W*H+L*H)
SA = 2*(8*6+6*2+8*2)
SA = 152
Problem 31)
VB = volume of cube block
VB = L*W*H
VB = 12*12*12
VB = 1728
VC = volume of cylinder
VC = pi*r^2*h
VC = pi*3^2*12
VC = 108*pi
VC = 339.292006587698
VF = volume of finished product (after cylinder is removed)
VF = VB - VC
VF = 1728-339.292006587698
VF = 1388.7079934123
VF = 1389 (rounded to the nearest whole number)
Problem 32)
d = diameter = 3
r = radius = d/2 = 3/2 = 1.5
h = height = 1.5
V = volume of cone
V = (1/3)*pi*r^2*h
V = (1/3)*pi*1.5^2*1.5
V = 1.125pi
V = 3.53429173528851
V = 3.5 (rounded to the nearest tenth)
