6. if pqrs is a parallelogram, find the values of x and y.

7. if abcd is a parallelogram, find m∠c.

8. which quadrilaterals always have diagonals that bisect opposite angles? select all that that apply.
- parallelograms
- rectangles
- rhombi
- squares

9. if wxyz is a square with wz = 27, find each measure.
a) zy =
b) wy =
c) rx =
d) m∠wrz =
e) m∠xyz =
f) m∠zwy =

Question

6. if pqrs is a parallelogram, find the values of x and y.

7. if abcd is a parallelogram, find m∠c.

8. which quadrilaterals always have diagonals that bisect opposite angles? select all that that apply.
    - parallelograms
    - rectangles
    - rhombi
    - squares

9. if wxyz is a square with wz = 27, find each measure.
    a) zy =
    b) wy =
    c) rx =
    d) m∠wrz =
    e) m∠xyz =
    f) m∠zwy =

Accepted Answer

### Problem 6
# Explanation:
## Step1: Use parallelogram side property
In parallelogram $PQRS$, opposite sides are equal. So $PS = QR$ and $PQ = SR$. For $PS$ and $QR$: $13x + 15 = 19x - 9$.
## Step2: Solve for $x$
$13x + 15 = 19x - 9$
$15 + 9 = 19x - 13x$
$24 = 6x$
$x = 4$.
## Step3: Use parallelogram angle property
Adjacent angles in a parallelogram are supplementary? No, wait, opposite angles are equal, and consecutive angles are supplementary. Wait, the angles given: $\angle Q=(4y + 7)^\circ$ and $\angle R=(10y - 37)^\circ$. In parallelogram, consecutive angles are supplementary? Wait, no, $PQ \parallel SR$, so $\angle Q$ and $\angle R$ are consecutive? Wait, $PQRS$: $P - Q - R - S - P$. So $PQ \parallel SR$, so $\angle Q$ and $\angle R$ are same - side interior angles, so they are supplementary? Wait, no, in parallelogram, opposite angles are equal, and consecutive angles are supplementary. Wait, but also, $\angle Q$ and $\angle S$ are opposite, $\angle P$ and $\angle R$ are opposite. Wait, maybe the angles given are adjacent? Wait, the diagram: $P$ to $Q$ to $R$ to $S$ to $P$. So $\angle Q$ is at $Q$, between $P$ and $R$; $\angle R$ is at $R$, between $Q$ and $S$. So $PQ \parallel SR$, so $\angle Q + \angle R = 180^\circ$? Wait, no, in parallelogram, consecutive angles are supplementary. Wait, but also, if we consider that $\angle Q$ and $\angle R$ are consecutive, then $ (4y + 7)+(10y - 37)=180$? Wait, no, maybe they are equal? Wait, no, maybe I made a mistake. Wait, in a parallelogram, opposite angles are equal, and consecutive angles are supplementary. Wait, but the angles given: $\angle Q=(4y + 7)^\circ$ and $\angle R=(10y - 37)^\circ$. Wait, maybe they are opposite? No, $Q$ and $S$ are opposite, $R$ and $P$ are opposite. Wait, maybe the angles are equal? Wait, no, maybe the problem has a typo, or maybe I misread. Wait, maybe the angles are equal? Let's check: $4y + 7 = 10y - 37$
$7 + 37 = 10y - 4y$
$44 = 6y$? No, that's not integer. Wait, maybe consecutive angles: $ (4y + 7)+(10y - 37)=180$
$14y - 30 = 180$
$14y = 210$
$y = 15$. Wait, but let's check with $x = 4$, then $PS = 13*4 + 15 = 52 + 15 = 67$, $QR = 19*4 - 9 = 76 - 9 = 67$, which matches. Now for $y$: if $y = 15$, then $\angle Q=(4*15 + 7)=67^\circ$, $\angle R=(10*15 - 37)=150 - 37 = 113^\circ$, and $67 + 113 = 180$, which is correct for consecutive angles in parallelogram.
# Answer:
$x = 4$, $y = 15$

### Problem 7
# Explanation:
## Step1: Use parallelogram angle property
In parallelogram $ABCD$, $AB \parallel CD$, so $\angle B$ and $\angle D$ are same - side interior angles? Wait, no, $\angle B=(5x + 38)^\circ$ and $\angle D=(8x - 19)^\circ$. In a parallelogram, opposite angles are equal? Wait, no, $\angle B$ and $\angle D$: wait, $ABCD$ is a parallelogram, so $AB \parallel CD$ and $AD \parallel BC$. So $\angle B$ and $\angle D$: no, $\angle A=\angle C$, $\angle B=\angle D$? Wait, no, wait, $A - B - C - D - A$. So $\angle B$ and $\angle D$ are opposite angles? Wait, no, $\angle A$ and $\angle C$ are opposite, $\angle B$ and $\angle D$ are opposite. Wait, but the diagram shows $\angle B$ at $B$ (between $A$ and $C$) and $\angle D$ at $D$ (between $A$ and $C$). Wait, no, $AB \parallel CD$, so $\angle B$ and $\angle C$ are consecutive? Wait, no, the angles given: $\angle B=(5x + 38)^\circ$ and $\angle D=(8x - 19)^\circ$. In a parallelogram, $\angle B=\angle D$? Wait, no, that can't be. Wait, no, $AB \parallel CD$, so $\angle B$ and $\angle C$ are supplementary, and $\angle D$ and $\angle C$ are supplementary, so $\angle B=\angle D$. Wait, yes! In a parallelogram, opposite angles are equal. So $\angle B=\angle D$.
## Step2: Solve for $x$
$5x + 38 = 8x - 19$
$38 + 19 = 8x - 5x$
$57 = 3x$
$x = 19$.
## Step3: Find $m\angle B$
$m\angle B=5*19 + 38=95 + 38 = 133^\circ$.
## Step4: Find $m\angle C$
In parallelogram, consecutive angles are supplementary. So $\angle B + \angle C = 180^\circ$.
$m\angle C=180 - 133 = 47^\circ$. Or, since $\angle A=\angle C$, and $\angle A$ and $\angle B$ are supplementary. Wait, let's check with $x = 19$, $\angle D=8*19 - 19=152 - 19 = 133^\circ$, so $\angle B=\angle D = 133^\circ$, then $\angle C=180 - 133 = 47^\circ$.
# Answer:
$m\angle C = 47^\circ$

### Problem 8
# Brief Explanations:
- **Parallelograms**: In a general parallelogram, the diagonals bisect each other but do not necessarily bisect the opposite angles. For example, in a rectangle (a type of parallelogram), the diagonals are equal and bisect each other, but they do not bisect the angles (unless it is a square).
- **Rectangles**: The diagonals of a rectangle are equal and bisect each other, but they do not bisect the angles (except in the case of a square). For example, in a rectangle with length $l$ and width $w$ ($l
eq w$), the diagonal will not split the angles into two equal parts.
- **Rhombi**: In a rhombus, all sides are equal. The diagonals of a rhombus bisect the opposite angles. This is a property of a rhombus: the diagonals act as angle - bisectors.
- **Squares**: A square is a special type of rhombus (and also a special type of rectangle). Since it is a rhombus, its diagonals bisect the opposite angles. Also, since it is a rectangle, its diagonals are equal and bisect each other.
# Answer:
Rhombi, Squares (the options with checkmarks should be Rhombi and Squares)

### Problem 9
# Explanation:
## Part (a): Find $ZY$
In square $WXYZ$, all sides are equal. Given $WZ = 27$, and $ZY$ is a side of the square, so $ZY=WZ = 27$.
## Part (b): Find $WY$
The diagonal of a square with side length $s$ is given by $d = s\sqrt{2}$. Here, $s = 27$, so $WY=27\sqrt{2}$.
## Part (c): Find $RX$
In a square, the diagonals bisect each other. So the diagonals $WX$ and $ZY$ intersect at $R$, and $RX=\frac{1}{2}WX$. Since $WX$ is a diagonal, $WX = 27\sqrt{2}$, so $RX=\frac{27\sqrt{2}}{2}$ or we can also note that in a square, the diagonals are equal and bisect each other, and also, the length of the diagonal from $W$ to $Y$ is $27\sqrt{2}$, and the diagonals bisect each other, so $R$ is the mid - point. Also, the side length is $27$, and the distance from $R$ to $X$ can be found as follows: the diagonals of a square are perpendicular bisectors of each other. But more simply, since $WXYZ$ is a square, $WZ = ZY = 27$, and the diagonal $WX$ (wait, no, $WY$ is a diagonal). Wait, $WXYZ$: vertices $W$, $X$, $Y$, $Z$ in order. So $WZ$ and $XY$ are sides, $WX$ and $ZY$ are sides? Wait, no, square $WXYZ$: $W - X - Y - Z - W$. So $WZ$ is a side, $ZY$ is a side, so $WZ = ZY = 27$. The diagonal $WY$: using Pythagoras, $WY=\sqrt{WZ^{2}+ZY^{2}}=\sqrt{27^{2}+27^{2}}=\sqrt{2\times27^{2}} = 27\sqrt{2}$. The diagonals of a square bisect each other, so $R$ is the mid - point of $WY$ and also the mid - point of $XZ$. Wait, maybe I mixed up the diagonals. Let's correct: in square $WXYZ$, the diagonals are $WX$ and $YZ$? No, wait, the vertices are $W$, $X$, $Y$, $Z$ with $W$ connected to $X$ and $Z$, $X$ connected to $W$ and $Y$, $Y$ connected to $X$ and $Z$, $Z$ connected to $Y$ and $W$. So the diagonals are $WY$ and $XZ$. They intersect at $R$. In a square, diagonals are equal ($WY = XZ$) and bisect each other ($WR = RY=\frac{1}{2}WY$ and $XR = RZ=\frac{1}{2}XZ$). Since $WZ = 27$ (side length), $XZ$ (a diagonal) is $27\sqrt{2}$, so $RX=\frac{1}{2}XZ=\frac{27\sqrt{2}}{2}$. But also, we can think of $RX$ as half of the diagonal. Alternatively, since the side length is $27$, and in the square, the distance from $R$ to $X$ is half of the diagonal. The diagonal length is $27\sqrt{2}$, so $RX=\frac{27\sqrt{2}}{2}\approx19.09$ (but we can leave it in exact form).
## Part (d): Find $m\angle WRZ$
In a square, the diagonals are perpendicular to each other. So $\angle WRZ = 90^\circ$.
## Part (e): Find $m\angle XYZ$
In a square, each interior angle is $90^\circ$, so $m\angle XYZ = 90^\circ$.
## Part (f): Find $m\angle ZWY$
In a square, the diagonals bisect the angles. So $\angle ZWY$ is half of $\angle ZWX$. Since $\angle ZWX = 90^\circ$ (interior angle of square), $\angle ZWY = 45^\circ$.
# Answer:
a) $ZY = 27$
b) $WY = 27\sqrt{2}$
c) $RX=\frac{27\sqrt{2}}{2}$ (or approximately $19.09$)
d) $m\angle WRZ = 90^\circ$
e) $m\angle XYZ = 90^\circ$
f) $m\angle ZWY = 45^\circ$

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Problem 6

Step1: Use parallelogram side property

Step2: Solve for \(x\)

Step3: Use parallelogram angle property

Step1: Use parallelogram angle property

Step2: Solve for \(x\)

Step3: Find \(m\angle B\)

Step4: Find \(m\angle C\)

Answer:

Problem 7