Core Connections Integrated II, 2015
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Core Connections Integrated II, 2015 View details
2. Section 5.2
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Exercise 114 Page 302

Practice makes perfect
a

In similar figures the ratio of two corresponding sides is always equal. By identifying corresponding sides, we can set up a proportion.

Let's substitute the side lengths in our proportion and solve for x.

JK/AB=JM/AD
x/12=9/6
â–¼
Solve for x
x/12=3/2
x=3/2* 12
x=36/2
x=18

b

Like in Part A, we will identify corresponding sides of the triangles and set up a proportion. Notice that the way the similarity statement is written tells us which vertices are corresponding.

â–³ NOP ~ â–³ XYZThis must mean that OP~ YZ and NP~ XZ. With this information, we can set up our proportion.

Let's substitute the side lengths in our proportion and solve for x.

YZ/OP=XZ/NP
w/5=12/3
w/5=4
w=20

c

Like in Part B, we will identify corresponding sides of the triangles and set up a proportiona. Again, the way the similarity statement is written tells us which vertices are corresponding.

â–³ GHI ~ â–³ PQRThis must mean that GI~ PR and GH~ PQ. With this information, we can set up our proportion.

Let's substitute the side lengths in our proportion and solve for x.

PR/GI=PQ/GH
n/3=16/7
â–¼
Solve for n
n=16/7* 3
n=48/7
n=6.85714...
n≈6.86

d

Like in Part B, we will identify corresponding sides of the triangles and set up a proportion. Again, the way the similarity statement is written tells us which vertices are corresponding.

â–³ ABC ~ â–³ XYZThis must mean that AB~ XY and AC~ XZ. With this information, we can set up our proportion.

Let's substitute the side lengths in our proportion and solve for x.

XY/AB=XZ/AC
m/7=11/10
â–¼
Solve for m
m=11/10* 7
m=77/10
m=7.7