Answer :
Let's answer the questions step by step:
### Part A: Find [tex]\( x \)[/tex]
Given that [tex]\( ABCD \)[/tex] is a parallelogram, opposite sides are equal. Therefore, the expression for side [tex]\( AB \)[/tex] is equal to the expression for side [tex]\( CD \)[/tex]:
[tex]\[ 3x + 20 = 5x - 12 \][/tex]
To find [tex]\( x \)[/tex], we solve the equation:
1. Subtract [tex]\( 3x \)[/tex] from both sides:
[tex]\[ 20 = 2x - 12 \][/tex]
2. Add 12 to both sides:
[tex]\[ 32 = 2x \][/tex]
3. Divide both sides by 2:
[tex]\[ x = 16 \][/tex]
So, [tex]\( x = 16 \)[/tex].
### Part B: Find the length of side [tex]\( AD \)[/tex]
Using [tex]\( x = 16 \)[/tex], we substitute [tex]\( x \)[/tex] back into one of the expressions for the sides:
[tex]\[ AD = 3x + 20 \][/tex]
Substitute [tex]\( x = 16 \)[/tex]:
[tex]\[ AD = 3(16) + 20 \][/tex]
[tex]\[ AD = 48 + 20 \][/tex]
[tex]\[ AD = 68 \][/tex]
So, the length of side [tex]\( AD \)[/tex] is 68 units.
### Part C: If side [tex]\( CD = 30 \)[/tex], find the perimeter of [tex]\( ABCD \)[/tex]
In a parallelogram, opposite sides are equal:
- [tex]\( AD = BC \)[/tex]
- [tex]\( AB = CD \)[/tex]
Given [tex]\( CD = 30 \)[/tex], so [tex]\( AB = 30 \)[/tex], and we already found [tex]\( AD = 68 \)[/tex] (hence [tex]\( BC = 68 \)[/tex]).
The perimeter of [tex]\( ABCD \)[/tex] is the sum of all its sides:
[tex]\[ \text{Perimeter} = AB + BC + CD + DA \][/tex]
Since opposite sides are equal:
[tex]\[ \text{Perimeter} = 2(AD + CD) \][/tex]
[tex]\[ \text{Perimeter} = 2(68 + 30) \][/tex]
[tex]\[ \text{Perimeter} = 2(98) \][/tex]
[tex]\[ \text{Perimeter} = 196 \][/tex]
So, the perimeter of [tex]\( ABCD \)[/tex] is 196 units.
### Part D: If angle [tex]\( C = 40 \)[/tex] degrees, find the measures of angles [tex]\( A \)[/tex], [tex]\( B \)[/tex], and [tex]\( D \)[/tex]
In a parallelogram, opposite angles are equal, and adjacent angles are supplementary (sum up to [tex]\( 180 \)[/tex] degrees).
Given [tex]\( \angle C = 40^\circ \)[/tex]:
- [tex]\( \angle A \)[/tex] is opposite [tex]\( \angle C \)[/tex], so [tex]\( \angle A = 40^\circ \)[/tex].
- [tex]\( \angle B \)[/tex] is adjacent to [tex]\( \angle C \)[/tex], so [tex]\( \angle B + \angle C = 180^\circ \)[/tex]:
[tex]\[ \angle B = 180^\circ - 40^\circ \][/tex]
[tex]\[ \angle B = 140^\circ \][/tex]
- [tex]\( \angle D \)[/tex] is opposite [tex]\( \angle B \)[/tex], so [tex]\( \angle D = 140^\circ \)[/tex].
Thus, the measures of the angles are:
- [tex]\( \angle A = 40^\circ \)[/tex]
- [tex]\( \angle B = 140^\circ \)[/tex]
- [tex]\( \angle C = 40^\circ \)[/tex]
- [tex]\( \angle D = 140^\circ \)[/tex]
### Summary of results:
- [tex]\( x = 16 \)[/tex]
- Length of side [tex]\( AD = 68 \)[/tex] units
- Perimeter of [tex]\( ABCD = 196 \)[/tex] units
- Measures of the angles: [tex]\( \angle A = 40^\circ \)[/tex], [tex]\( \angle B = 140^\circ \)[/tex], [tex]\( \angle C = 40^\circ \)[/tex], [tex]\( \angle D = 140^\circ \)[/tex]
### Part A: Find [tex]\( x \)[/tex]
Given that [tex]\( ABCD \)[/tex] is a parallelogram, opposite sides are equal. Therefore, the expression for side [tex]\( AB \)[/tex] is equal to the expression for side [tex]\( CD \)[/tex]:
[tex]\[ 3x + 20 = 5x - 12 \][/tex]
To find [tex]\( x \)[/tex], we solve the equation:
1. Subtract [tex]\( 3x \)[/tex] from both sides:
[tex]\[ 20 = 2x - 12 \][/tex]
2. Add 12 to both sides:
[tex]\[ 32 = 2x \][/tex]
3. Divide both sides by 2:
[tex]\[ x = 16 \][/tex]
So, [tex]\( x = 16 \)[/tex].
### Part B: Find the length of side [tex]\( AD \)[/tex]
Using [tex]\( x = 16 \)[/tex], we substitute [tex]\( x \)[/tex] back into one of the expressions for the sides:
[tex]\[ AD = 3x + 20 \][/tex]
Substitute [tex]\( x = 16 \)[/tex]:
[tex]\[ AD = 3(16) + 20 \][/tex]
[tex]\[ AD = 48 + 20 \][/tex]
[tex]\[ AD = 68 \][/tex]
So, the length of side [tex]\( AD \)[/tex] is 68 units.
### Part C: If side [tex]\( CD = 30 \)[/tex], find the perimeter of [tex]\( ABCD \)[/tex]
In a parallelogram, opposite sides are equal:
- [tex]\( AD = BC \)[/tex]
- [tex]\( AB = CD \)[/tex]
Given [tex]\( CD = 30 \)[/tex], so [tex]\( AB = 30 \)[/tex], and we already found [tex]\( AD = 68 \)[/tex] (hence [tex]\( BC = 68 \)[/tex]).
The perimeter of [tex]\( ABCD \)[/tex] is the sum of all its sides:
[tex]\[ \text{Perimeter} = AB + BC + CD + DA \][/tex]
Since opposite sides are equal:
[tex]\[ \text{Perimeter} = 2(AD + CD) \][/tex]
[tex]\[ \text{Perimeter} = 2(68 + 30) \][/tex]
[tex]\[ \text{Perimeter} = 2(98) \][/tex]
[tex]\[ \text{Perimeter} = 196 \][/tex]
So, the perimeter of [tex]\( ABCD \)[/tex] is 196 units.
### Part D: If angle [tex]\( C = 40 \)[/tex] degrees, find the measures of angles [tex]\( A \)[/tex], [tex]\( B \)[/tex], and [tex]\( D \)[/tex]
In a parallelogram, opposite angles are equal, and adjacent angles are supplementary (sum up to [tex]\( 180 \)[/tex] degrees).
Given [tex]\( \angle C = 40^\circ \)[/tex]:
- [tex]\( \angle A \)[/tex] is opposite [tex]\( \angle C \)[/tex], so [tex]\( \angle A = 40^\circ \)[/tex].
- [tex]\( \angle B \)[/tex] is adjacent to [tex]\( \angle C \)[/tex], so [tex]\( \angle B + \angle C = 180^\circ \)[/tex]:
[tex]\[ \angle B = 180^\circ - 40^\circ \][/tex]
[tex]\[ \angle B = 140^\circ \][/tex]
- [tex]\( \angle D \)[/tex] is opposite [tex]\( \angle B \)[/tex], so [tex]\( \angle D = 140^\circ \)[/tex].
Thus, the measures of the angles are:
- [tex]\( \angle A = 40^\circ \)[/tex]
- [tex]\( \angle B = 140^\circ \)[/tex]
- [tex]\( \angle C = 40^\circ \)[/tex]
- [tex]\( \angle D = 140^\circ \)[/tex]
### Summary of results:
- [tex]\( x = 16 \)[/tex]
- Length of side [tex]\( AD = 68 \)[/tex] units
- Perimeter of [tex]\( ABCD = 196 \)[/tex] units
- Measures of the angles: [tex]\( \angle A = 40^\circ \)[/tex], [tex]\( \angle B = 140^\circ \)[/tex], [tex]\( \angle C = 40^\circ \)[/tex], [tex]\( \angle D = 140^\circ \)[/tex]
