Question

1. perimeter = 34 in

y
x+2
2x-3
x=

y=
2.
16x+6
5y+2
18x-8
x=

y=
3.
9x-2
65°
y
7x+14
x=

y=
4.
ap = 6x-2
cp = 3x+4
bd = 14
bp = 3y-2
a
d
b
p
c
x=

y=

Explanation:

Step1: Solve for \( x \) in Problem 1 (Parallelogram Perimeter)

In a parallelogram, opposite sides are equal. So \( x + 2 = 2x - 3 \). Solve:
\( x + 2 = 2x - 3 \)
Subtract \( x \): \( 2 = x - 3 \)
Add 3: \( x = 5 \).

Step2: Find \( y \) using Perimeter

Perimeter \( = 2[(x + 2) + y] = 34 \). Substitute \( x = 5 \):
\( 2[(5 + 2) + y] = 34 \)
\( 2(7 + y) = 34 \)
Divide by 2: \( 7 + y = 17 \)
Subtract 7: \( y = 10 \).

Step3: Solve for \( x \) in Problem 2 (Parallelogram Opposite Sides)

Opposite sides: \( 16x + 6 = 18x - 8 \). Solve:
\( 16x + 6 = 18x - 8 \)
Subtract \( 16x \): \( 6 = 2x - 8 \)
Add 8: \( 14 = 2x \)
Divide by 2: \( x = 7 \).

Step4: Find \( y \) (Assume it's a Parallelogram, so adjacent sides? Wait, no—wait, maybe it's a rhombus? Wait, no, opposite sides equal. Wait, maybe it's a parallelogram, so \( 5y + 2 \) is equal to... Wait, no, maybe it's a rhombus? Wait, no, the first side is \( 16x + 6 \), second is \( 5y + 2 \). Wait, maybe it's a rhombus? Wait, no, in a parallelogram, opposite sides are equal. Wait, maybe the figure is a rhombus (all sides equal). So \( 16x + 6 = 5y + 2 \). Substitute \( x = 7 \):

\( 16(7) + 6 = 5y + 2 \)
\( 112 + 6 = 5y + 2 \)
\( 118 = 5y + 2 \)
Subtract 2: \( 116 = 5y \)
Wait, that can't be. Wait, maybe it's a parallelogram, so \( 16x + 6 \) and \( 18x - 8 \) are opposite, which we did. Then \( 5y + 2 \) is equal to the other pair? Wait, maybe the figure is a parallelogram, so two pairs: \( 16x + 6 \) & \( 18x - 8 \), and \( 5y + 2 \) & the other side. Wait, maybe I misread. Let's recheck. Problem 2: the figure has sides \( 16x + 6 \), \( 5y + 2 \), and \( 18x - 8 \). So opposite sides: \( 16x + 6 = 18x - 8 \) (solved \( x = 7 \)), then the other pair: \( 5y + 2 \) and the remaining side (which is equal to \( 16x + 6 \) or \( 18x - 8 \))? Wait, maybe it's a parallelogram, so perimeter \( = 2[(16x + 6) + (5y + 2)] \). But we don't have perimeter. Wait, maybe it's a rhombus (all sides equal). So \( 16x + 6 = 5y + 2 = 18x - 8 \). We know \( 16x + 6 = 18x - 8 \) gives \( x = 7 \), so \( 16(7) + 6 = 118 \), so \( 5y + 2 = 118 \) → \( 5y = 116 \) → \( y = 23.2 \). But that seems odd. Maybe the figure is a rectangle? No, the angles aren't right. Wait, maybe the problem is a parallelogram, so \( 5y + 2 \) is equal to \( 16x + 6 \) (if it's a rhombus). But maybe I made a mistake. Let's move to Problem 3.

Step5: Problem 3 (Kite? Or Rhombus? The figure has a diagonal, angles \( 65^\circ \), sides \( 9x - 2 \) and \( 7x + 14 \). In a kite, two pairs of adjacent sides equal. Wait, or a rhombus (all sides equal). So \( 9x - 2 = 7x + 14 \). Solve:

\( 9x - 2 = 7x + 14 \)
Subtract \( 7x \): \( 2x - 2 = 14 \)
Add 2: \( 2x = 16 \)
Divide by 2: \( x = 8 \).

Step6: Find \( y \) (Angle in Kite/Rhombus)

In a rhombus, adjacent angles are supplementary? Wait, no, the diagonal bisects the angle. Wait, the angle is \( 65^\circ \), and \( y \) is equal? Wait, maybe it's a rhombus, so the diagonal bisects the angle, so \( y = 65^\circ \)? Or maybe it's a kite with two equal angles. Wait, the figure has a diagonal, so maybe it's a rhombus, so \( y = 65^\circ \)? Not sure. Let's do Problem 4.

Step7: Problem 4 (Parallelogram Diagonals)

In a parallelogram, diagonals bisect each other. So \( AP = CP \) and \( BP = PD \).

Substep7a: Solve \( AP = CP \)

\( 6x - 2 = 3x + 4 \)
Subtract \( 3x \): \( 3x - 2 = 4 \)
Add 2: \( 3x = 6 \)
Divide by 3: \( x = 2 \).

Substep7b: Solve \( BP = PD \)

Diagonal \( BD = 14 \), so \( BP + PD = 14 \). Since \( BP = PD \) (bisected), \( 2BP = 14 \) → \( BP = 7 \). So \( 3y - 2 = 7 \)
Add 2: \( 3y = 9 \)
Divide by 3: \( y = 3 \).

Answer (for Problem 4):

\( x = 2 \), \( y = 3 \)

(Note: For Problem 1: \( x = 5 \), \( y = 10 \); Problem 2: \( x = 7 \), \( y = 23.2 \) (or recheck); Problem 3: \( x = 8 \), \( y = 65^\circ \) (assumed); Problem 4: \( x = 2 \), \( y = 3 \))

Answer:

(for Problem 4):
\( x = 2 \), \( y = 3 \)

(Note: For Problem 1: \( x = 5 \), \( y = 10 \); Problem 2: \( x = 7 \), \( y = 23.2 \) (or recheck); Problem 3: \( x = 8 \), \( y = 65^\circ \) (assumed); Problem 4: \( x = 2 \), \( y = 3 \))