Exercise 33 Page 383

Practice makes perfect

a We are given that the ball hits the first wall at an angle of 63^(∘), and we want to evaluate m∠ AFE. Let's take a look at the given diagram.

Since ∠ AEF≅∠ BEH, the measure of ∠ AEF is also 63^(∘). Now let's notice that △ EAF is a right triangle with a right angle ∠ FAE. Using this information and the Triangle Angle Sum Theorem, we can write an equation. m∠ AEF+m∠ AFE+m∠ FAE=180^(∘) Let's substitute appropriate angle measures to find m∠ AFE.

m∠ AEF+m∠ AFE+m∠ FAE=180^(∘)

SubstituteII

m∠ AEF= 63^(∘), m∠ FAE= 90^(∘)

63^(∘)+m∠ AFE+ 90^(∘)=180^(∘)

AddTerms

Add terms

153^(∘)+m∠ AFE=180^(∘)

SubEqn

LHS-153^(∘)=RHS-153^(∘)

m∠ AFE=27^(∘)

The measure of ∠ AFE is 27^(∘).

b In this part we want to explain why m∠ FGD. To do this we will use the fact that m∠ AFE= 27^(∘). This indicates that m∠ DFG is also 27^(∘).

Looking at the diagram, we can see that △ FDG is a right triangle with a right angle ∠ FDG. Therefore, by using the Triangle Angle Sum Theorem we can write and solve an equation to find m∠ FGD.

m∠ DFG+m∠ FDG+m∠ FGD=180^(∘)

SubstituteII

m∠ DFG= 27^(∘), m∠ FDG= 90^(∘)

27^(∘)+ 90^(∘)+m∠ FGD=180^(∘)

AddTerms

Add terms

117^(∘)+m∠ FGD=180^(∘)

SubEqn

LHS-117^(∘)=RHS-117^(∘)

m∠ FGD=63^(∘)

We showed that the measure of ∠ FGD is 63^(∘). Notice that this also means that m∠ CGH=63^(∘).

c Looking at the diagram we can see that △ AFE and △ CHG are similar, as they both are right angles and m∠ AEF=m∠ GHC, as we showed in the previous part.

Keeping in mind that similar triangle's corresponding angles are congruent, we can say that ∠ AFE and ∠ CHG. This means that m∠ GHC is equal to 27^(∘). This also indicates that m∠ EHB=27^(∘), as the ball bounces off at the same angle at which it hits the wall.

d Finally let's notice that since ∠ EFG and ∠ EHG are both adjacent to two congruent angles that together add up to 180^(∘), they will have the same measure.

By using the same logic we can deduct that ∠ HEF and ∠ HGF are also congruent.

Since we found that quadrilateral EFGH has pairs of opposite angles that are congruent, this figure is a parallelogram by the Parallel Opposite Angles Converse Theorem.

Subchapter links

Communicate Your Answer p.375

Monitoring Progress p.377-380

Exercises p.381-384

Exercise 33 Page 383

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