Exercise 55 Page 426

We are given that PQRS is a square and PLMS is similar to LMRQ. We want to find the exact value of x.

Notice that since PQRS is a square, it has four congruent sides.

Moreover, since LMRQ is a rectangle, LM equals to 1 as well.

From here we will use the similarity statement PLMS ~ LMRQ to find the exact value of x. Let's find the value of x.

Finding the Value of x

Remember that when two polygons are similar their corresponding side lengths are proportional. PLMS ~ LMRQ → PL/LM=LM/MR To apply this proportion we need to find MR. To do so we will use Segment Addition Postulate.

SM=SR+MR

SubstituteII

SM= x, SR= 1

x=1+MR

SubEqn

LHS-1=RHS-1

x-1=MR

RearrangeEqn

Rearrange equation

MR=x-1

Now, we can use the proportion to find the value of x.

PL/LM=LM/MR

SubstituteValues

Substitute values

x/1=1/x-1

▼

Simplify

MultEqn

LHS * (x-1)=RHS* (x-1)

x(x-1)/1=1

DivByOne

a/1=a

x(x-1)=1

Distr

Distribute x

x^2-x=1

SubEqn

LHS-1=RHS-1

x^2-x-1=0

Next, we will use the Quadratic Formula.

x=-b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

x=-(-1)±sqrt((- 1)^2-4(1)(- 1))/2(1)

▼

Solve for x

NegNeg

- (- a)=a

x=1±sqrt((- 1)^2-4(1)(- 1))/2(1)

NegBaseToPosBase

(- a)^2=a^2

x=1±sqrt(1^2-4(1)(- 1))/2(1)

BaseOne

1^a=1

x=1±sqrt(1-4(1)(- 1))/2(1)

Multiply

x=1±sqrt(1+4)/2

AddTerms

Add terms

x=1±sqrt(5)/2

StateSol

State solutions

lcx= 1+sqrt(5)2 & (I) x= 1-sqrt(5)2 & (II)

Since x is a measure of a length, it cannot be negative. Therefore, we will only consider x= 1+sqrt(5)2, and we will check whether it satisfies the equation.

x/1=1/x-1

Substitute

x= 1+sqrt(5)/2

\dfrac{{\color{#0000FF}{\frac{1+\sqrt{5}}{2}}}}{1}\stackrel{?}=\dfrac{1}{{\color{#0000FF}{\frac{1+\sqrt{5}}{2}}}-1}

▼

Simplify left-hand side and right-hand side

DivByOne

a/1=a

\dfrac{1+\sqrt{5}}{2}\stackrel{?}=\dfrac{1}{\frac{1+\sqrt{5}}{2}-1}

WriteAsFrac

Write as a fraction

\dfrac{1+\sqrt{5}}{2}\stackrel{?}=\dfrac{1}{\frac{1+\sqrt{5}}{2}-\frac{1}{1}}

ExpandFrac

a/b=a * 2/b * 2

\dfrac{1+\sqrt{5}}{2}\stackrel{?}=\dfrac{1}{\frac{1+\sqrt{5}}{2}-\frac{2}{2}}

SubFrac

Subtract fractions

\dfrac{1+\sqrt{5}}{2}\stackrel{?}=\dfrac{1}{\frac{\sqrt{5}-1}{2}}

DivOneByFracSL

.1 /a/b.=b/a

\dfrac{1+\sqrt{5}}{2}\stackrel{?}=\dfrac{2}{\sqrt{5}-1}

Because we do not want a root in the denominator, we will expand the fraction by the conjugate of the denominator.

\dfrac{1+\sqrt{5}}{2}\stackrel{?}=\dfrac{2}{\sqrt{5}-1}

ExpandFrac

a/b=a * (sqrt(5)+1)/b * (sqrt(5)+1)

\dfrac{1+\sqrt{5}}{2}\stackrel{?}=\dfrac{2(\sqrt{5}+1)}{(\sqrt{5}-1)(\sqrt{5}+1)}

▼

Simplify right-hand side

(a-b)(a+b)=a^2-b^2

\dfrac{1+\sqrt{5}}{2}\stackrel{?}=\dfrac{2(\sqrt{5}+1)}{(\sqrt{5})^2-1^2}

BaseOne

1^a=1

\dfrac{1+\sqrt{5}}{2}\stackrel{?}=\dfrac{2(\sqrt{5}+1)}{(\sqrt{5})^2-1}

CalcPow

Calculate power

\dfrac{1+\sqrt{5}}{2}\stackrel{?}=\dfrac{2(\sqrt{5}+1)}{5-1}

SubTerm

Subtract term

\dfrac{1+\sqrt{5}}{2}\stackrel{?}=\dfrac{2(\sqrt{5}+1)}{4}

ReduceFrac

a/b=.a /2./.b /2.

\dfrac{1+\sqrt{5}}{2}\stackrel{?}=\dfrac{\sqrt{5}+1}{2}

CommutativePropAdd

Commutative Property of Addition

1+sqrt(5)/2=1+sqrt(5)/2 ✓

Thus, x= 1+sqrt(5)2 satisfies the equation. By using a calculator we can rewrite the value of x as 1.618, which is the golden ratio. From here, we will show that the similar rectangles in the diagram are golden rectangles.

Showing the Golden Rectangles

Note that in golden rectangles the ratio of the length to the width is approximately 1.618. Thus, for rectangle PLMS we will find the ratio of PL=x to LM=1. In the same way, for rectangle LMRQ we will find the ratio of LM=1 to MR=x-1.

Ratio	x≈ 1.618	Result (approximately)
PL/LM=x/1	x/1≈1.618/1	1.618
LM/MR=1/x-1	1/x-1≈1/1.618-1	1.618

The ratio of the length to the width is equal to the golden ration for both rectangle PLMS and rectangle LMRQ. With this information, we conclude that these similar rectangles are golden rectangles.

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Finding the Value of x

Showing the Golden Rectangles