**#4.) in rectangle quad, ud = 8x - 2 cm and ub = 2x + 3 cm. determine the length of $\overline{qa}$.

**#5.) abcd is a parallelogram
a) determine the perimeter of abcd.
b) determine the m$\angle b$.

#6.) in parallelogram abcd, dp = 2y, db = 3y + 6, ap = 3x + 3 and cp = 4x - 2.
a) **determine the length of $\overline{ac}$.
b) determine the length of $\overline{pb}$.

Question

**#4.) in rectangle quad, ud = 8x - 2 cm and ub = 2x + 3 cm. determine the length of $\overline{qa}$.

**#5.) abcd is a parallelogram
  a) determine the perimeter of abcd.
  b) determine the m$\angle b$.

#6.) in parallelogram abcd, dp = 2y, db = 3y + 6, ap = 3x + 3 and cp = 4x - 2.
  a) **determine the length of $\overline{ac}$.
  b) determine the length of $\overline{pb}$.

Accepted Answer

### Problem #4
# Explanation:
## Step1: Recall rectangle diagonals property
In a rectangle, diagonals are equal and bisect each other. So $ UD = 2 \times UB $ (since $ B $ is the midpoint of $ UD $ and also of $ QA $, and diagonals $ UD $ and $ QA $ are equal).
Set up the equation: $ 8x - 2 = 2(2x + 3) $

## Step2: Solve for $ x $
Expand the right side: $ 8x - 2 = 4x + 6 $
Subtract $ 4x $ from both sides: $ 4x - 2 = 6 $
Add 2 to both sides: $ 4x = 8 $
Divide by 4: $ x = 2 $

## Step3: Find $ UB $ length
Substitute $ x = 2 $ into $ UB = 2x + 3 $: $ UB = 2(2) + 3 = 7 $ cm

## Step4: Find $ QA $ length
Since diagonals in a rectangle are equal and $ UD = QA $, and $ UD = 2 \times UB $, so $ QA = 2 \times 7 = 14 $ cm

# Answer:
The length of $ \overline{QA} $ is $ \boldsymbol{14} $ cm

### Problem #5 (a)
# Explanation:
## Step1: Recall parallelogram side properties
In a parallelogram, opposite sides are equal. So $ AB = CD $ and $ AD = BC $.
From $ AB = CD $: $ y + 8 = 5y $
From $ AD = BC $: $ 4x^2 = 3y $ (wait, no, angle and side? Wait, $ AD $ and $ BC $ are sides, but also consecutive angles in a parallelogram are supplementary, and opposite angles are equal. Wait, first solve for $ y $ from $ AB = CD $:

## Step2: Solve for $ y $
$ y + 8 = 5y $
Subtract $ y $: $ 8 = 4y $
Divide by 4: $ y = 2 $

## Step3: Find side lengths
$ AB = y + 8 = 2 + 8 = 10 $, $ CD = 5y = 10 $
$ BC = 3y = 6 $, $ AD $: Wait, also, consecutive angles: $ \angle D + \angle A = 180^\circ $, but $ \angle A = 4x^2 $ and $ \angle D = (2x - 6)^\circ $. Wait, no, in a parallelogram, opposite angles are equal, so $ \angle A = \angle C $ and $ \angle B = \angle D $, and consecutive angles are supplementary. Wait, maybe I misread: $ AD $ has length $ 4x^2 $? No, wait the diagram: $ AD $ is labeled with angle $ (2x - 6)^\circ $ and side? Wait, no, the side $ AD $ and $ BC $: Wait, the sides: $ AB = y + 8 $, $ BC = 3y $, $ CD = 5y $, $ DA $: maybe the angle is a typo? Wait, no, in a parallelogram, opposite sides are equal, so $ AB = CD $ (so $ y + 8 = 5y $) and $ AD = BC $. Wait, the $ 4x^2 $ and $ (2x - 6)^\circ $: maybe $ AD $ has length related, but also angle $ \angle D = (2x - 6)^\circ $ and $ \angle A = 4x^2 $ (angle). In a parallelogram, consecutive angles are supplementary: $ \angle A + \angle D = 180^\circ $, and opposite angles are equal. Wait, first solve $ y $:

From $ AB = CD $: $ y + 8 = 5y \implies y = 2 $ (as before). Then $ AB = 10 $, $ CD = 10 $, $ BC = 3y = 6 $, so $ AD = BC = 6 $ (since opposite sides equal). Wait, but also check angles: $ \angle A = 4x^2 $, $ \angle D = (2x - 6)^\circ $, and $ \angle A + \angle D = 180^\circ $. But maybe $ 4x^2 $ is a side? Wait, no, the diagram: $ A $ to $ D $ is labeled $ 4x^2 $ (side length) and angle $ (2x - 6)^\circ $. Wait, maybe it's a typo, and $ AD = 3y $? No, let's proceed with $ y = 2 $, so $ BC = 3(2) = 6 $, $ AB = 10 $.

## Step4: Calculate perimeter
Perimeter of parallelogram is $ 2(AB + BC) = 2(10 + 6) = 32 $

# Answer:
The perimeter of $ ABCD $ is $ \boldsymbol{32} $ units (assuming length units)

### Problem #5 (b)
# Explanation:
## Step1: Recall parallelogram angle properties
In a parallelogram, consecutive angles are supplementary, and opposite angles are equal. We found $ y = 2 $, now solve for $ x $. From angle $ \angle A = 4x^2 $ and $ \angle D = (2x - 6)^\circ $, and $ \angle A + \angle D = 180^\circ $ (consecutive angles). Wait, but $ \angle B = \angle D $ (opposite angles). Wait, first, we know $ y = 2 $, so let's find $ x $. Wait, maybe $ 4x^2 $ is a side, but no, angle is in degrees. Wait, maybe the angle $ \angle D = (2x - 6)^\circ $ and $ \angle A = 4x^2 $ (angle), but that can't be. Wait, maybe it's $ \angle A = (2x - 6)^\circ $ and $ \angle D = 4x^2 $? No, the diagram: $ A $ has $ 4x^2 $, $ D $ has $ (2x - 6)^\circ $. Wait, maybe it's a mistake, and $ AD $ length is $ 4x $ (not squared). Let's assume $ 4x $ instead of $ 4x^2 $ (typo). Then $ AD = BC = 3y = 6 $, so $ 4x = 6 \implies x = 1.5 $, but angle: $ \angle D = (2x - 6)^\circ = (3 - 6) = -3^\circ $, impossible. So maybe the angle is $ \angle A = (2x - 6)^\circ $ and $ \angle D = 4x^\circ $. Then $ \angle A + \angle D = 180^\circ $: $ 2x - 6 + 4x = 180 \implies 6x = 186 \implies x = 31 $. But this is confusing. Wait, original problem: $ ABCD $ is a parallelogram, with $ AB = y + 8 $, $ BC = 3y $, $ CD = 5y $, $ DA $: angle $ (2x - 6)^\circ $, and $ \angle A = 4x^\circ $ (maybe). Wait, no, the user's diagram: $ A $ has $ 4x^2 $ (maybe area? No, side). Wait, perhaps the angle is $ \angle D = (2x - 6)^\circ $ and $ \angle A = 4x^\circ $, and since $ \angle A = \angle C $, $ \angle B = \angle D $, and $ \angle A + \angle B = 180^\circ $. But we already found $ y = 2 $, so $ BC = 6 $, $ AB = 10 $. Now, for angles: in a parallelogram, opposite angles are equal, so $ \angle A = \angle C $, $ \angle B = \angle D $. Also, consecutive angles are supplementary. So $ \angle A + \angle D = 180^\circ $. If $ \angle A = 4x $ and $ \angle D = (2x - 6) $, then $ 4x + 2x - 6 = 180 \implies 6x = 186 \implies x = 31 $. Then $ \angle B = \angle D = 2(31) - 6 = 56^\circ $. Wait, but maybe the $ 4x^2 $ is a typo. Alternatively, maybe the side $ AD = 4x $ and $ BC = 3y $, and angle $ \angle D = (2x - 6)^\circ $. Since $ AD = BC = 6 $ (from $ y = 2 $, $ 3y = 6 $), so $ 4x = 6 \implies x = 1.5 $, then $ \angle D = 2(1.5) - 6 = -3^\circ $, invalid. So perhaps the angle is $ \angle D = (2x + 6)^\circ $, then $ 2(1.5) + 6 = 9^\circ $, still small. This is confusing. Wait, maybe the original problem has $ AD = 4x $ (not squared) and angle $ \angle D = (2x - 6)^\circ $, and since $ AD = BC = 3y = 6 $, then $ 4x = 6 \implies x = 1.5 $, and $ \angle D = 2(1.5) - 6 = -3 $, which is impossible. So maybe the angle is $ \angle A = (2x - 6)^\circ $ and $ \angle D = 4x^\circ $, and $ \angle A = \angle C $, $ \angle B = \angle D $, $ \angle A + \angle B = 180 $. Then $ 2x - 6 + 4x = 180 \implies 6x = 186 \implies x = 31 $, so $ \angle B = \angle D = 4(31) = 124^\circ $? Wait, no, $ \angle B = \angle D $, and $ \angle A + \angle D = 180 $. If $ \angle A = 2x - 6 $, $ \angle D = 4x $, then $ 2x - 6 + 4x = 180 \implies 6x = 186 \implies x = 31 $, so $ \angle D = 124^\circ $, so $ \angle B = 124^\circ $? No, consecutive angles: $ \angle A $ and $ \angle B $ are consecutive, so $ \angle A + \angle B = 180 $. Wait, I'm overcomplicating. Wait, the problem says "Determine the $ m\angle B $". In a parallelogram, $ \angle B = \angle D $ (opposite angles), and $ \angle A + \angle D = 180^\circ $. Also, from sides: we found $ y = 2 $, so $ AB = 10 $, $ CD = 10 $, $ BC = 6 $, $ AD = 6 $ (since opposite sides equal). Now, the angle at $ D $ is $ (2x - 6)^\circ $, and angle at $ A $ is $ 4x^\circ $ (assuming $ 4x $ instead of $ 4x^2 $). Then $ \angle A + \angle D = 180 \implies 4x + 2x - 6 = 180 \implies 6x = 186 \implies x = 31 $. Then $ \angle D = 2(31) - 6 = 56^\circ $, so $ \angle B = \angle D = 56^\circ $? Wait, no, $ \angle B $ and $ \angle D $ are opposite? No, $ \angle B $ is at vertex $ B $, $ \angle D $ at $ D $, so they are opposite? Wait, in parallelogram $ ABCD $, vertices are $ A, B, C, D $ in order, so $ \angle A $ and $ \angle C $ are opposite, $ \angle B $ and $ \angle D $ are opposite. So yes, $ \angle B = \angle D $. So if $ \angle D = 56^\circ $, then $ \angle B = 56^\circ $. But let's check with $ x = 31 $, $ \angle A = 4(31) = 124^\circ $, and $ 124 + 56 = 180 $, which works (consecutive angles supplementary). So:

## Step1: Solve for $ x $
From $ \angle A + \angle D = 180^\circ $: $ 4x + (2x - 6) = 180 $
$ 6x - 6 = 180 $
$ 6x = 186 $
$ x = 31 $

## Step2: Find $ m\angle B $
$ \angle B = \angle D = 2x - 6 = 2(31) - 6 = 56^\circ $

# Answer:
The measure of $ \angle B $ is $ \boldsymbol{56^\circ} $

### Problem #6 (a)
# Explanation:
## Step1: Recall parallelogram diagonal properties
In a parallelogram, diagonals bisect each other. So $ DP = PB $ and $ AP = CP $.
From $ AP = CP $: $ 3x + 3 = 4x - 2 $

## Step2: Solve for $ x $
Subtract $ 3x $: $ 3 = x - 2 $
Add 2: $ x = 5 $

## Step3: Find $ AP $ length
Substitute $ x = 5 $ into $ AP = 3x + 3 $: $ AP = 3(5) + 3 = 18 $

## Step4: Find $ AC $ length
Since diagonals bisect each other, $ AC = 2 \times AP $ (or $ 2 \times CP $). So $ AC = 2 \times 18 = 36 $

# Answer:
The length of $ \overline{AC} $ is $ \boldsymbol{36} $

### Problem #6 (b)
# Explanation:
## Step1: Recall parallelogram diagonal bisect property
In a parallelogram, diagonals bisect each other, so $ DP = PB $ and $ DB = 2 \times DP $. Also, $ DP = 2y $, $ DB = 3y + 6 $. So $ DB = 2 \times DP \implies 3y + 6 = 2(2y) $

## Step2: Solve for $ y $
Expand: $ 3y + 6 = 4y $
Subtract $ 3y $: $ 6 = y $

## Step3: Find $ DP $ length
Substitute $ y = 6 $ into $ DP = 2y $: $ DP = 12 $

## Step4: Find $ PB $ length
Since $ DP = PB $ (diagonals bisect each other), $ PB = DP = 12 $

# Answer:
The length of $ \overline{PB} $ is $ \boldsymbol{12} $

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QUESTION IMAGE

Question

Explanation:

Problem #4

Step1: Recall rectangle diagonals property

Step2: Solve for \( x \)

Step3: Find \( UB \) length

Step4: Find \( QA \) length

Step1: Recall parallelogram side properties

Step2: Solve for \( y \)

Step3: Find side lengths

Step4: Calculate perimeter

Step1: Recall parallelogram angle properties

Answer:

Problem #5 (a)