Core Connections Integrated II, 2015
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Core Connections Integrated II, 2015 View details
2. Section 3.2
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Exercise 74 Page 179

Practice makes perfect
a The two triangles are similar, which means that the ratio between corresponding sides is the same. Note that AC is the greater of the known side lengths in △ ABC. Additionally, △ OPQ is larger than △ ABC, meaning QP has to correspond to BC and not AC. If AC and PQ were corresponding side lengths, the two triangles would instead be congruent.
Let's solve for f in the equation.
f/6=6/4
Solve for f
f=6/4* 6
f=36/4
f=9
b Comparing the two figures, we see that EF and ST are corresponding sides, as are HG and VU. Since the ratio between corresponding sides in similar figures is the same, we can write an equation using these sides, which have been labeled.
HG/VU=EF/ST ⇒ g/15=24/20 Let's solve for g in the equation.
g/15=24/20
Solve for f
g/15=6/5
g=6/5* 15
g=90/5
g=18
c Comparing the two figures, we see that ZW and MJ are corresponding sides, as are ZY and ML. Since the ratio between corresponding sides in similar figures is the same, we can write an equation using these sides.
ZW/MJ=ZY/ML ⇒ h/7=10/3 Let's solve for h.
h/7=10/3
Solve for f
h=10/3* 7
h=70/3
h=23.33333...
h≈ 23.3
d Examining the diagram, we see that the triangles have different positions, rotations, and sizes. Therefore, we would need to perform a translation, a rotation, and a dilation to make △ ABC map onto △ OPQ. Let's first perform a translation to make two corresponding vertices map onto each other.

Next, we will perform a rotation to make two of the corresponding sides map onto each other.

Finally, we will dilate △ A''B''C' to make all corresponding sides map onto each other.