Exercise 37 Page 579

Let's analyze the given figure. Since we are given a triangle with two lines that are parallel to one of its sides, we can use the Triangle Proportionality Theorem.

At first we will apply it to △ QSW and RX. The lengths of the segments intercepted by the parallel line are proportional. Let's write a proportion using the expressions for the lengths of the segments. QR/RS=QX/XW ⇕ 2/RS=QX/12Since we are given the lengths of XW and QW, we can find the length of QX.

QX+XW=QW

SubstituteII

XW= 12, QW= 15

QX+ 12= 15

SubEqn

LHS-12=RHS-12

QX=3

Now we can substitute the missing length into our proportion. 2/RS=QX/12 ⇕ 2/RS=3/12 We will solve it for RS.

2/RS=3/12

▼

Solve for RS

MultEqn

LHS * RS=RHS* RS

2=3/12RS

ReduceFrac

a/b=.a /3./.b /3.

2=1/4RS

MultEqn

LHS * 4=RHS* 4

8=RS

RearrangeEqn

Rearrange equation

RS=8

Similarly, we can apply the Triangle Proportionality Theorem to △ QTV and SW. Let's write a proportion using the expressions for the lengths of the segments. QS/ST=QW/WV ⇕ QS/5=15/WV Since we know the lengths of QR and RS, we can find the length of QS.

QR+RS=QS

SubstituteII

QR= 2, RS= 8

2+ 8=QS

▼

Solve for QS

AddTerms

Add terms

10=QS

RearrangeEqn

Rearrange equation

QS=10

Now we can substitute the missing length into our proportion. QS/5=15/WV ⇕ 10/5=15/WV Let's solve it for WV.

10/5=15/WV

▼

Solve for WV

CalcQuot

Calculate quotient

2=15/WV

MultEqn

LHS * WV=RHS* WV

2WV=15

DivEqn

.LHS /2.=.RHS /2.

WV=15/2

CalcQuot

Calculate quotient

WV=7.5

Finally, we found that RS=8 and WV=7.5.