Core Connections Geometry, 2013
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Core Connections Geometry, 2013 View details
1. Section 6.1
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Exercise 26 Page 355

Practice makes perfect
a The following is true about similar and congruent shapes.

Similar shapes:& Same shape Congruent shapes:& Same shape and size We have been given two angles in both triangles. If the two triangles have at least two pairs of congruent angles, we know they have the same shape and are similar. Let's find the last angle in both triangles by using the Triangle Angle Sum Theorem. m∠ D+70^(∘)+60^(∘)&=180^(∘) ⇔ m∠ D=50^(∘) m∠ K+70^(∘)+50^(∘)&=180^(∘) ⇔ m∠ K=60^(∘) The triangles have the same angles which means they have the same shape and are similar. To determine if the triangles are congruent, we have to identify corresponding sides.

From the diagram, we see that DE and JL are corresponding sides because they are included between congruent angles. Since the sides have the same length, the common ratio of these sides must equal 1. Therefore, the triangles are congruent. △ DEF ≅ △ LJK When we wrote our congruence statement, it is important to write the statement in the correct order. On the left-hand side, D comes first which means its corresponding vertex in the second triangle, L, must come first on the right-hand side of the statement, and so on.

b The following series of transformations is just one example. Let's first translate â–³ DEF so that two corresponding vertices map onto each other.
Next, we have to rotate â–³ DEF so that the corresponding sides, EF and JK, line up.

Finally, we will reflect â–³ DEF in JK to get all three corresponding vertices to line up.

A series of transformations that maps the two triangles onto each other is a translation, a rotation and a reflection, not necessarily in that order.

c The two triangles are congruent. Therefore, if we find the length of DF, we also find the length of the corresponding side, KL. We can use the Law of Sines to determine DF, which we will label x. The Law of Sines equates the quotients of the sine of an angle and the length of the opposite side.

Let's solve this equation for x.

sin 60^(∘)/4=sin 70^(∘)/x
â–¼
Solve for x
sin 60^(∘)/4 * x=sin 70^(∘)
x sin 60^(∘)=4sin 70^(∘)
x=4sin 70^(∘)/sin 60^(∘)
x=4.34025...
x≈ 4.34

Therefore, KL is about 4.34 units long.