Core Connections Geometry, 2013
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Core Connections Geometry, 2013 View details
2. Section 7.2
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Exercise 54 Page 421

Practice makes perfect
a We are given that BD divides ∠ ABC equally into two smaller angles, ∠ ABD and ∠ DBC.
This allows us to relate the measure of ∠ ABC to the measures of the two smaller angles. m ∠ ABD + m ∠ DBC = m ∠ ABC Since BD bisects ∠ ABC, the measures of ∠ ABD and ∠ DBC are the same. m∠ ABD= m∠ DBC Let's use this to simplify our equation for the measure of ∠ ABC.
m ∠ ABD + m ∠ DBC = m ∠ ABC
m ∠ ABD + m ∠ ABD = m ∠ ABC
2 m ∠ ABD = m ∠ ABC
Finally, now that we have an equation relating the measures of angles ABD and ABC, we can substitute the given values to write an equation that we can solve for x. Let's do it!
2 m ∠ ABD = m ∠ ABC
2( 5x - 10 ^(∘)) = 65 ^(∘)
10x - 20 ^(∘) = 65 ^(∘)
10x = 85 ^(∘)
x = 8.5 ^(∘)
b Let's add the given lengths to the diagram.
We are told that point M is the midpoint of EF. This means that it divides EF into two smaller but equal lengths. EM &= MF &⇓ 4x-2 &= 3x+9 Let's solve the resulting equation for x.
4x-2 = 3x+9
x-2=9
x=11
c Let's add the given measures to the diagram of the parallelogram.
Notice that the angles W and Z are consecutive angles. There is a theorem that says that if a quadrilateral is a parallelogram, then its consecutive angles are supplementary. From this, we know that the measures of our consecutive angles add up to 180^(∘). m ∠ W + m ∠ Z = 180 ^(∘) ⇓ 9x-3^(∘) + 3x+15^(∘) = 180 ^(∘) Let's solve for x!
9x-3 ^(∘) + 3x +15 ^(∘) = 180 ^(∘)
12x + 12^(∘) = 180 ^(∘)
12x = 168 ^(∘)
x = 14^(∘)