Question

dklm and ptrs are shown, where dklm~ptrs with md = 20, dk = 6z + 6, kl = 8w + 1, lm = 40, pt = 12, tr = 4w - 2, and rs = 16.

1. determine the value of each variable.

w =
z =

1. determine the value of each side.

dk =
sp =
tr =

1. what is the perimeter of dklm?

consider additional measurements of angles for dklm and ptrs where m∠dkl = 150°, m∠mdk = (15x - 25)°, m∠spt = (12x + 5)°, m∠dml = (5y - 5)°, m∠psr = (4y + 4)°.

1. determine the value of each variable.

x =
y =

1. determine the value of each angle.

m∠mdk =
m∠ptr =
m∠psr =

Explanation:

Step1: Use the property of similar polygons (ratio of corresponding sides) for \(w\)

Since \(DKLM\sim PTRS\), \(\frac{KL}{TR}=\frac{LM}{RS}\). Substitute \(KL = 8w + 1\), \(TR=4w - 2\), \(LM = 40\), \(RS = 16\)

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Step2: Use the property of similar polygons (ratio of corresponding sides) for \(z\)

Also, \(\frac{MD}{SP}=\frac{DK}{PT}\). First, find \(SP\). Since \(\frac{LM}{RS}=\frac{MD}{SP}\), \(\frac{40}{16}=\frac{20}{SP}\), \(40SP=16\times20\), \(SP = 8\). Then \(\frac{DK}{PT}=\frac{MD}{SP}\), substitute \(DK = 6z+6\), \(PT = 12\), \(MD = 20\), \(SP = 8\)

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Step3: Find the lengths of sides

For \(DK\): Substitute \(z = 4\) into \(DK=6z + 6\), \(DK=6\times4+6=30\)
For \(TR\): Substitute \(w = 3\) into \(TR = 4w-2\), \(TR=4\times3-2 = 10\)

Step4: Find \(x\) using the property of similar polygons (corresponding angles)

Since \(DKLM\sim PTRS\), \(\angle MDK=\angle SPT\). So \(15x-25=12x + 5\)

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Step5: Find \(y\) using the property of similar polygons (corresponding angles)

Since \(DKLM\sim PTRS\), \(\angle DML=\angle PSR\). So \(5y-5=4y + 4\)

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Step6: Find the measures of angles

For \(m\angle MDK\): Substitute \(x = 10\) into \(m\angle MDK=(15x - 25)^{\circ}\), \(m\angle MDK=(15\times10-25)^{\circ}=125^{\circ}\)
For \(m\angle PTR\): In polygon \(DKLM\), the sum of interior angles of a quadrilateral is \((4 - 2)\times180^{\circ}=360^{\circ}\). \(\angle DKL = 150^{\circ}\), \(\angle MDK=125^{\circ}\), \(\angle DML=(5\times9-5)^{\circ}=40^{\circ}\), so \(\angle KLM=360-(150 + 125+40)=45^{\circ}\). Since \(DKLM\sim PTRS\), \(m\angle PTR=\angle KLM = 45^{\circ}\)
For \(m\angle PSR\): Substitute \(y = 9\) into \(m\angle PSR=(4y + 4)^{\circ}\), \(m\angle PSR=(4\times9+4)^{\circ}=40^{\circ}\)

Step7: Find the perimeter of \(DKLM\)

\(Perimeter=MD+DK+KL+LM\). \(MD = 20\), \(DK = 30\), \(KL=8\times3+1=25\), \(LM = 40\). \(Perimeter=20 + 30+25+40=115\)

Answer:

1. \(w = 3\), \(z = 4\)
2. \(DK = 30\), \(SP = 8\), \(TR = 10\)
3. \(115\)
4. \(x = 10\), \(y = 9\)
5. \(m\angle MDK=125^{\circ}\), \(m\angle PTR = 45^{\circ}\), \(m\angle PSR=40^{\circ}\)