The document is a review game for Chapter 11 geometry concepts that includes: 1) Formulas for calculating the area of basic shapes like rectangles, polygons, circles, etc. 2) Practice problems calculating missing measures and areas using the formulas. 3) Finding surface areas and geometric probabilities of various shapes.

1. Ch 11 Review Game March 12, 2012
Study for Test 1

2. Ch 11 Review Game March 12, 2012
State the Formula for Each Figure:
1) Rectangle: O: (1 pt)
D: Steal
A = bh
2) Regular Polygon:
O: (1 pt)
A = 1/2asn D: Blocked Shot
3) Kite/Rhombus:
O: (1 pt)
1
A = /2d1d2 D: Steal w/layup
4) Parallelogram: O: (2 pt)
A = bh D: Steal
5) Trapezoid:
O: (2 pt)
A = 1/2h(b1 + b2) D: Steal w/layup
6) Circle:
O: (2 pt)
A = πr2 D: Fast break, missed the layup
O: (2 pt)
D: Fast Break w/3 pt
Study for Test 2

3. Ch 11 Review Game March 12, 2012
Find the missing measure
a) b = 8.5 cm b) A = 30 m2
6.5m
6.2 cm
12m x
b
4m
A = ________
x = ________
O: (2 pt) O: (2 pt)
D: Blocked Shot D: Fast break for field goal
A= 1/2bh
A = A b =8.5 h = 6.2 A= 1/2h(b1 + b2)
A = 1/2(8.5)(6.2) A = 30 h =4 b1 = 12 b2 = x
A = 26.35 cm2 30 = 1/2(4)(12 +x)
30 = 24 + 2x
6 = 2x
x = 3 cm
Study for Test 3

4. Ch 11 Review Game March 12, 2012
30 in
c) d)
20 in
l
9 in
l
22 in
18.5 in
A = ________
A = ________
C = ________
O: (2 pt) O: (3 pt)
D: Steal w/ 3 pts D: Steal w/layup
A= 1/2h(b1 + b2) A= πr2
A = A h =11 b1 = 30 b2 = 22 A= A r = 9.25
A = 1/2(11)(30 + 22) A = 9.252π
A = 286 A = 85.5625π in2
A= bh
A = A b= 11 h = 9 C= 2πr
A = 11(9) C= C r = 9.25
A = 99 C = (2)9.25π
C = 18.5π in
TSA = 286 + 99
TSA = 385 in2
Study for Test 4

5. Ch 11 Review Game March 12, 2012
Find the area of the shaded region
a) b)
24 in
9 in
A = ________ Area of the Square is 121m2
O: (2 pt) O: (2 pt)
D: Blocked Shot D: Fast Break w/3 pt
A = πr2 A = πr2
Small Circle Large Circle A = A r = 5.5
A = A r = 5.5 A = A r = 12 A = 5.52π
A = 92π A = 122π A = 30.25π
A = 81π A = 144π
Shaded Area: 121 ­ 30.25π m2
Shaded Area: 63π
Study for Test 5

6. Ch 11 Review Game March 12, 2012
Find the area of the regular polygon.
a = 21 in, r = 23.2 in, A = _____
The area of a 15­gon is 4095 cm2.
The sides are 22.2 cm each.
The radius is 27 cm.
a = ____________
O: (3 pt) O: (2 pt)
D: Steal w/layup D: Steal w/ 3 pts
4095 = 1/2 (a)(15)(22.2)
s 4095/(1/2*15*22.2) = a
A = 1/2 ans /2 = 23.22 ­ 212 = 9.82
A = 1/2 (21)(5) s s = (9.8 + 9.8) = 19.6 24.59 = a
A = 1/2 (21)(5)(19.6) or
A = 1029 in2 a = 272 ­ 11.12 = 24.62
4095 = 1/2 (24.6)(15)(22.2)
Study for Test 6

7. Ch 11 Review Game March 12, 2012
Find the geometric probability.
Find the probability that a point chosen Find the probability that a point chosen
randomly will be in the shaded region. randomly on AE will be on CD.
A B C D E
12 m
0 6 15 26 28
164
O: (2 pt) O: (1 pt)
D: Blocked Shot D: Steal w/layup
2
Asector = 164/360* π12 CD = 26­15 = 11
2
total Area = π12 AE = 28 ­ 0 = 28
Asector = 206.09 Probability = 11/28 = 0.393, 39.3%
total area = 452.39
Probability = 0.455, 45.5%
Study for Test 7

8. Ch 11 Review Game March 12, 2012
Find the surface area of each figure
in
in
15 in
3
25
in
22
5 in
20 in
O: (3 pt)
O: (2 pt)
D: Fast break for field goal
D: Fast break, missed layup
SA = πr2 + πrL SA = PH + 2(1/2asn)
SA = SA r = 10 L = 25 SA = SA P=30 H = 22
SA = π102 + π(10)(25) a=3 s=5 n=6
SA = 100π + 250π SA = 30(22) + 2(1/2)(3)(5)(6)
SA = 350π SA = 660 + 90
SA = 750 in2
Study for Test 8