Core Connections Integrated II, 2015
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Core Connections Integrated II, 2015 View details
1. Section 7.1
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Exercise 51 Page 391

Practice makes perfect
a We know a pair of congruent sides and a pair of congruent angles. If we can find a second pair of congruent angles, we have enough information to claim similarity. Examining the diagram, we notice that both triangles share ∠ S as an angle which means we have one more pair of congruent angles.
Since △ SRH and △ SKA have two pairs of congruent angles, we know that they are similar by the AA Similarity condition.
b From Part A we know that △ RHS and △ KAS are similar triangles. We also know that SR=KS which means KS is twice as long as SR.
In similar triangles, the ratio between corresponding sides is always the same. With the information from the diagram, we can determine this ratio.

SR/SK=x/2x=1/2 This ratio is the same between our other corresponding sides as well, HR and AK and between SH and SA. HR/AK=1/2 and SH/SA=1/2 Since SH is half that of SA, then SH and HA must have equal lengths.

c From Part B, we know that sides in △ SKA are twice the length of their corresponding sides in △ SRH. Therefore, if RH is 8 units, its corresponding side KA is 16 units.
Now we have enough information to calculate SA by using the Pythagorean Theorem.
a^2+b^2=c^2
16^2+b^2= 20^2
Solve for b
256+b^2=400
b^2=144
b=± 12

b > 0

b=12
If SA is 12 units then HA must be half of this which is 6 units.