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McGraw Hill Integrated II, 2012

MH

McGraw Hill Integrated II, 2012 View details

2. Congruent Triangles

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Exercise 20 Page 349

Begin by using the Triangle Angle-Sum Theorem. Then, consider the Third Angles Theorem, to write a system of two equations.

x=4
y=1

Practice makes perfect

Let's add some labels to the given diagram.

Notice that ∠ B and ∠ E are marked as right angles, so they measure 90^(∘). We can use the Triangle Angle-Sum Theorem to find the measure of ∠ F.

m∠ D+m∠ E+m∠ F=180

m∠ D= 52, m ∠ E= 90

52+ 90+m∠ F=180

Add terms

142+ m∠ F=180

LHS-142=RHS-142

m∠ F=38

With this measure and the given expression for ∠ F, we can write an equation. 6x+14y=38 Next, notice that ∠ C and ∠ D are marked as congruent angles, so their measures are the same. Using this relationship, we can write another equation. m∠ C = m∠ D ⇔ 15x-8y=52 From here, we can create a system of two equations. 15x-8y=52 & (I) 6x+14y=38 & (II) To solve this system, we will use the Elimination Method.

15x-8y=52 6x+14y=38

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(II): Solve for y

(II): LHS * (- 5)=RHS* (- 5)

15x-8y=52 - 30x-70y=- 190

(I): LHS * 2=RHS* 2

30x-16y=104 - 30x-70y=- 190

(II): Add (I)

30x-16y=104 - 30x-70y+( 30x-16y)=- 190+ 104

(II): Add and subtract terms

30x-16y=104 - 86y=- 86

(II): .LHS /(- 86).=.RHS /(- 86).

30x-16y=104 y=1

Now that we found the value of y, we can substitute it into the first equation to find the value of x.

30x-16y=104 y=1

(I): y= 1

30x-16( 1)=104 y=1

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(I):Solve for x

(I): Multiply

30x-16=104 y=1

(I): LHS+16=RHS+16

30x=120 y=1

(I): .LHS /30.=.RHS /30.

x=4 y=1

Subchapter links

Exercises p.347-352