Calculating measurements of quadrilaterals.

Problem 1

Given

• ΔMAN ~ ΔWOM
• AM = 8
• AN = 7
• MN = 10
• OM = 14

Finding the measurements of OW and MW.

Since ΔMAN ~ ΔWOM, then AM ~ OW and AN ~ OM.

AN / OM = AM / OW

7 / 14 = 8 / OW

112 = 7OW

112 / 7 = 7OW / 7

OW = 16

Since ΔMAN ~ ΔWOM, then NM ~ MW and AN ~ OM.

AN / OM = NM / MW

7 / 14 = 10 / MW

140 = 7MW

140 / 7 = 7MW / 7

MW = 20

Problem 2

Calculating the lengths of the parallelogram.

Given

• ▱LEAD ~ ▱SORP
• m∠L = 120°
• EL = 9
• EA = 6
• SO = 24

Since ▱LEAD ~ ▱SORP, then LE ~ SO, LE ~ PR, LD ~ SP, LD ~ OR, DA ~ SO, DA ~ PR.

Both are parallelograms, so LE ≅ DA, LD ≅ EA, SO ≅ PR and SP ≅ OR. LE ≅ DA, if LE = 9 then DA = 9. LD ≅ EA, if EA = 6, then LD = 6. SO ≅ PR, if SO = 24 the PR = 24.

LE / SO = EA / OR

9 / 24 = 6 / OR

144 = 9OR

144 / 9 = 9OR / 9

OR = 16

OR ≅ SP, if OR = 16 then SP = 16

Finding the angles of the parallelogram.

Opposite angles are ≅, m∠L ≅ m∠A, if m∠L = 120° then m∠A = 120°.

Since ▱LEAD ~ ▱SORP, then m∠L ≅ m∠S ≅ m∠R, if m∠L = 120° then m∠S = 120° and m∠R = 120°

Consecutive angles are supplementary, so m∠L + m∠E = 180° and m∠D + m∠A = 180°

180 = m∠L + m∠E

180 = 120 + m∠E

180 - 120 = m∠E

m∠E = 60°

Opposite angles are ≅, m∠E ≅ m∠D, if m∠E = 60° then m∠E = 60°.

Since ▱LEAD ~ ▱SORP, then m∠E ≅ m∠P ≅ m∠O, if m∠E = 60° then m∠P = 60° and m∠O = 60°.

Problem 3

Given

• Trapezoid CEBU and OTAP are isoceles.
• CE = 6
• CU = 15
• OT = 9
• m∠A = 70°
• CEBU ~ OTAP

Calculating the lengths of the trapezoid.

Since Trapezoid CEBU and OTAP are isoceles, then CU ≅ EB, if CU = 15 then EB = 15.

Since CEBU ~ OTAP, then CE ~ OT and CU ~ OP.

CE / OT = CU / OP

6 / 9 = 15 / OP

135 = 6OP

135 / 6 = 6OP / 6

OP = 22.5

Since Trapezoid CEBU and OTAP are isoceles, then OP ≅ TA, if OP = 22.5 then TA = 22.5.

Finding the angles of the trapezoid.

Consecutive angles are ≅ so m∠A ≅ m∠P, if m∠A = 70°, then m∠P = 70°

Since CEBU ~ OTAP, then m∠A ≅ m∠B ≅ m∠U, if m∠A = 70°, then m∠B = 70° and m∠U = 70°

Opposite Angles are supplementary, so m∠A + m∠O = 180° and m∠B + m∠C = 180°

180 = m∠A + m∠O

180 = 70 + m∠O

180 - 70 = m∠O

m∠O = 110°

Consecutive angles are ≅ so m∠O ≅ m∠T, if m∠O = 110°, then m∠T = 110°

Since CEBU ~ OTAP, then m∠O ≅ m∠C ≅ m∠E, if m∠O = 110°, then m∠C = 110° and m∠E = 110°

Finding the length of UB and PA.

Since there is no stated height or altitude, the only way to find the longer base is by using trigonometry.

Before you can even solve, follow these steps.

• Construct two vertical lines between the trapezoid.
• You just have to solve one triangle, since the trapezoid is isoceles, the other side is congruent.
• Set the midpoints as X and Y.

Proof that ΔCXU ≅ ΔEYB

Let X and Y be points of U and B. Since CEBU and OTAP are isoceles, we can say that UX ≅ YB, CX ≅ EB. Because UX ≅ YB, CX ≅ EY therefore CU ≅ EB. Since UX ≅ YB, CX ≅ EY and CU ≅ EB, therefore ΔCXU ≅ ΔEYB by SSS Postulate.

Solving UX and YB.

Given

• Hypothenuse (CU) = 15
• Angle of the Leg = 70°
• UX = ?

We will use cosine as the function because we are finding the adjacent of theta and the hypothenuse is given.

cos 70 = UX / CU

cos 70 = UX / 15

15cos 70 = UX

UX = 5.13

Since CEBU and OTAP are isoceles, we can say that UX ≅ YB, if UX = 5.13 then YB = 5.13

Solving UB and PA.

Since CEBU and OTAP are isoceles, we can say that CE ≅ XY, if CE = 6 then XY = 6

UB = UX + XY + YB

UB = 6 + 5.13 + 5.13

UB = 16.26

Since CEBU ~ OTAP, then CE ~ UB and OT ~ PA

CE / UB = OT / PA

6 / 16.26 = 9 / PA

146.34 = 6PA

146.34 / 6PA = 6PA / 6

PA = 24.39

I hope you appreciate my answer. This is the longest answer I made on Brainly so far. It took me more than 2 hours. Stay Safe and God Bless.

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