Geo_Book_Answers
Geo_Book_Answers
Geo_Book_Answers
You also want an ePaper? Increase the reach of your titles
YUMPU automatically turns print PDFs into web optimized ePapers that Google loves.
<strong>Answers</strong> to Exercises<br />
LESSON 11.5<br />
1. 18 cm2 2. 18 3. 5; 10<br />
4. 27 cm2 7. <br />
1<br />
5. 49<br />
6. 1:3<br />
m<br />
n2 2<br />
or m 2<br />
n<br />
8. Possible answer: Assuming the ad is sold by<br />
area, Annie should charge $6000.<br />
9. 5000 tiles<br />
11.<br />
9<br />
10. 1<br />
12a. a(x) 2x<br />
Area<br />
x in cm 2<br />
1 2<br />
2 4<br />
3 6<br />
4 8<br />
5 10<br />
6 12<br />
12b. A(x) 2x 2<br />
Area<br />
x in cm 2<br />
1 2<br />
2 8<br />
3 18<br />
4 32<br />
5 50<br />
6 72<br />
12c. The equation for a(x) is linear, so the graph is<br />
a line. The equation for A(x) is quadratic, so the<br />
graph is a parabola.<br />
13. Possible proof: The area of the first rectangle<br />
is bh. The area of the dilated rectangle is rh rb,or<br />
r2bh. The ratio of the area of the dilated rectangle<br />
to the area of the original rectangle is r2bh<br />
, bh<br />
or r2 .<br />
14. 16<br />
, <br />
3<br />
20<br />
<br />
3<br />
15. 8<br />
16a. (i) 40°; 50°; 40° (ii) 60°; 30°; 60°<br />
(iii) 22°; 68°; 22°<br />
Conjecture: similar; right triangle<br />
s h<br />
16b. (i) h<br />
(ii) <br />
x (iii) h b<br />
<br />
m<br />
or a <br />
p h<br />
Conjecture: <br />
h q<br />
<br />
16c. h pq <br />
Area in cm 2<br />
Area in cm 2<br />
10<br />
5<br />
70<br />
60<br />
50<br />
40<br />
30<br />
20<br />
10<br />
130 ANSWERS TO EXERCISES<br />
y<br />
y<br />
5<br />
5<br />
x<br />
x<br />
9<br />
16d. x 9, y 16, 12<br />
12<br />
,<br />
16<br />
h (9)(16 ) <br />
144 12<br />
17. 7.1 m 14.2 m 14.2 m 6.5 m<br />
18.<br />
42 m<br />
L<br />
10 m<br />
E<br />
O V M<br />
A<br />
Consider similar triangles LVE and MTH with<br />
corresponding altitudes EO and HA . Show<br />
EO EL<br />
that LOE MAH, then HA<br />
. HM<br />
You know<br />
that L M. Because EO and HA are altitudes,<br />
LOE and MAH are both right angles, and<br />
LOE MAH.<br />
EO EL<br />
So, LOE MAH by AA. Thus HA<br />
, HM<br />
which shows that the corresponding altitudes<br />
are proportional to corresponding sides.<br />
19. x 92°; Explanations will vary but should<br />
reference properties of linear pairs, isosceles<br />
triangles, and alternate interior angles formed<br />
by parallel lines.<br />
20. 105.5 cm2 21. 60 ft2 22. true<br />
23. Two pairs of angles are congruent, so the<br />
triangles are similar by the AA Similarity<br />
Conjecture. However, the two sets of corresponding<br />
sides are not proportional 6<br />
0<br />
<br />
80<br />
105<br />
,so 135<br />
the<br />
triangles are not similar.<br />
24.<br />
Top<br />
Front<br />
42 m<br />
Right side<br />
Top: 6 square units; front: 4 square units; right side:<br />
4 square units. The sum of the areas is half the<br />
surface area. The volume of the original solid is<br />
8 cubic units. The volume of the enlarged solid is<br />
512 cubic units. The ratio of volumes is 64<br />
.<br />
H<br />
T<br />
1
Change language