items 11–12. parallelogram jklm is shown.
11. what is ( mangle ljm )?
12. if ( kr = x + 7 ), ( km = 3x - 5 ), and ( jl = 4x - 10 ), what are the values of ( x ) and ( jr )?
items 13–14. abcd is shown.
13. what is ( mangle abc )?
a 71 c 106
b 109 d not enough information
14. if ( bd = 2.8 ), what is ( cp )? round your answer to the nearest tenth. write na if there is not enough information given.
( rn = _____ )
15. select all the statements that are true about rhombus mnpq.
a. ( overline{nq} perp overline{mp} )
b. ( nr = qr )
c. ( np perp qp )
d. ( angle mnq cong angle pnq )
e. ( nq = mp )
16. the lengths of the diagonals of a rhombus are ( 4x ) and ( 6x ). what expressions give the perimeter and area of the rhombus?
17. wxyz is a parallelogram. what is ( mangle dzw )?
items 18–20. give the most precise classification for each figure.
18.
19.
20.
envision® florida geometry • assessment resources

Question

items 11–12. parallelogram jklm is shown.
11. what is ( mangle ljm )?
12. if ( kr = x + 7 ), ( km = 3x - 5 ), and ( jl = 4x - 10 ), what are the values of ( x ) and ( jr )?
items 13–14. abcd is shown.
13. what is ( mangle abc )?
    a 71    c 106
    b 109   d not enough information
14. if ( bd = 2.8 ), what is ( cp )? round your answer to the nearest tenth. write na if there is not enough information given.
    ( rn = _____ )
15. select all the statements that are true about rhombus mnpq.
    a. ( overline{nq} perp overline{mp} )
    b. ( nr = qr )
    c. ( np perp qp )
    d. ( angle mnq cong angle pnq )
    e. ( nq = mp )
16. the lengths of the diagonals of a rhombus are ( 4x ) and ( 6x ). what expressions give the perimeter and area of the rhombus?
17. wxyz is a parallelogram. what is ( mangle dzw )?
items 18–20. give the most precise classification for each figure.
18.
19.
20.
envision® florida geometry • assessment resources

Accepted Answer

### Question 12 Solution:
# Explanation:
## Step1: Recall parallelogram diagonal property
In a parallelogram, diagonals bisect each other. So $ KR=\frac{1}{2}KM $ and $ JR = \frac{1}{2}JL $. Given $ KR=x + 7$ and $ KM=3x - 5$, we have $ x + 7=\frac{1}{2}(3x - 5) $.
## Step2: Solve for $ x $
Multiply both sides by 2: $ 2(x + 7)=3x - 5 $
Expand left side: $ 2x+14 = 3x - 5 $
Subtract $ 2x $ from both sides: $ 14=x - 5 $
Add 5 to both sides: $ x = 19 $
## Step3: Find $ JR $
First, find $ JL $ when $ x = 19 $: $ JL=4x - 10=4(19)-10=76 - 10 = 66 $
Since $ JR=\frac{1}{2}JL $, $ JR=\frac{66}{2}=33 $
# Answer:
$ x = 19 $, $ JR = 33 $

### Question 13 Solution:
# Explanation:
## Step1: Analyze the figure (ABCD)
From the diagram, we can see that triangle $ APD $ has angles $ 37^{\circ}$ and $ 34^{\circ}$. The sum of angles in a triangle is $ 180^{\circ}$, so the third angle $ \angle APD=180-(37 + 34)=109^{\circ}$. But in a parallelogram, adjacent angles are supplementary? Wait, no, let's look at the properties. Wait, actually, in the figure, $ ABCD $ has some markings. Wait, the angles at $ A $ and $ D $ in triangle $ APD $: Wait, maybe $ ABCD $ is a parallelogram? Wait, the markings show $ AB = CD $ (maybe? The diagram has marks). Wait, actually, in the triangle $ APD $, we have angles $ 37^{\circ}$ and $ 34^{\circ}$, so $ \angle PAD = 37^{\circ}$, $ \angle PDA=34^{\circ}$, so $ \angle BAD=37^{\circ}$, $ \angle ADC = 34^{\circ}+...$ Wait, no, maybe $ ABCD $ is a parallelogram, so $ \angle ABC + \angle BAD=180^{\circ}$? Wait, no, let's recalculate. Wait, the angle at $ A $ in triangle $ APD $: $ \angle PAD = 37^{\circ}$, $ \angle PDA = 34^{\circ}$, so $ \angle APD=180 - 37 - 34 = 109^{\circ}$. But $ \angle BPC=\angle APD $ (vertical angles). Then, in quadrilateral $ ABCD $, if it's a parallelogram, $ AB\parallel CD $, $ AD\parallel BC $. Wait, maybe the angle $ \angle ABC $ is supplementary to the angle we found? Wait, no, the options are 71, 109, 106, or not enough. Wait, let's re - examine. The sum of $ 37^{\circ}$ and $ 34^{\circ}$ is $ 71^{\circ}$, and in a parallelogram, adjacent angles are supplementary? Wait, no, maybe $ \angle ABC=180-(37 + 34)=109^{\circ}$? Wait, no, $ 37 + 34 = 71$, $ 180 - 71 = 109$? Wait, no, maybe the triangle's angle is related to the parallelogram's angle. Wait, the correct answer is B (109). Let's check: In triangle $ APD $, angles are $ 37^{\circ}$ and $ 34^{\circ}$, so the angle adjacent to $ \angle ABC $ is $ 37 + 34 = 71^{\circ}$, so $ \angle ABC=180 - 71 = 109^{\circ}$
# Answer:
B. 109

### Question 14 Solution:
# Explanation:
## Step1: Recall parallelogram diagonal property
In a parallelogram, diagonals bisect each other. So $ BP = PD=\frac{BD}{2}$. Given $ BD = 2.8$, so $ PD=\frac{2.8}{2}=1.4$. Also, from the diagram, $ AP = 1.7$ (marked). Wait, but what is $ CP $? In a parallelogram, $ AP = CP $ (diagonals bisect each other). Wait, the diagram shows $ AP = 1.7$, so $ CP = 1.7$? Wait, no, wait the problem says "If $ BD = 2.8$, what is $ CP $? Round your answer to the nearest tenth. Write NA if there is not enough information given." Wait, maybe the figure is a parallelogram, so diagonals bisect each other, so $ AP = CP $. From the diagram, $ AP = 1.7$, so $ CP = 1.7$
# Answer:
$ CP = 1.7 $

### Question 15 Solution:
# Explanation:
## Step1: Recall rhombus properties
- **Property 1: Diagonals of a rhombus are perpendicular** ($ \overline{NQ}\perp\overline{MP} $): True (A is true).
- **Property 2: Diagonals of a rhombus bisect each other** ($ NR = QR $): True (B is true).
- **Property 3: Sides of a rhombus are equal, but adjacent sides are not necessarily perpendicular** ($ NP\perp QP $): False (C is false).
- **Property 4: Diagonals of a rhombus bisect the angles** ($ \angle MNQ\cong\angle PNQ $): True (D is true).
- **Property 5: Diagonals of a rhombus are not necessarily equal** ($ NQ = MP $): False (E is false).

# Answer:
A. $ \overline{NQ}\perp\overline{MP} $, B. $ NR = QR $, D. $ \angle MNQ\cong\angle PNQ $

### Question 16 Solution:
# Explanation:
## Step1: Find the side length of the rhombus
The diagonals of a rhombus are $ 4x $ and $ 6x $. The diagonals of a rhombus bisect each other at right angles. So half - lengths of diagonals are $ 2x $ and $ 3x $. Using the Pythagorean theorem, the side length $ s $ of the rhombus is $ s=\sqrt{(2x)^{2}+(3x)^{2}}=\sqrt{4x^{2}+9x^{2}}=\sqrt{13x^{2}}=x\sqrt{13} $
## Step2: Find the perimeter
The perimeter $ P $ of a rhombus is $ 4\times $ side length. So $ P = 4\times x\sqrt{13}=4x\sqrt{13} $
## Step3: Find the area
The area $ A $ of a rhombus is $ \frac{1}{2}\times d_1\times d_2 $, where $ d_1 = 4x $ and $ d_2 = 6x $. So $ A=\frac{1}{2}\times4x\times6x = 12x^{2} $
# Answer:
Perimeter: $ 4x\sqrt{13} $, Area: $ 12x^{2} $

### Question 17 Solution:
# Explanation:
## Step1: Recall parallelogram properties
In parallelogram $ WXYZ $, $ WX = YZ $, $ WZ = XY $, and $ WX\parallel YZ $, $ WZ\parallel XY $. Given $ WX = 14 $, $ WZ = 28 $, and $ \angle X = 112^{\circ}$, so $ \angle W=180^{\circ}-\angle X = 68^{\circ}$ (adjacent angles in parallelogram are supplementary). Now, triangle $ WXD $ and $ ZYD $? Wait, no, let's look at triangle $ WZD $. Wait, $ WX = 14 $, $ WZ = 28 $, so $ WX=\frac{1}{2}WZ $. So triangle $ WXD $: $ WX = 14 $, $ WZ = 28 $, so $ WX=\frac{1}{2}WZ $, so $ \angle DZW $ is the angle opposite to the side $ WX $ in triangle $ WZD $? Wait, no, in triangle $ WXZ $, $ WX = 14 $, $ WZ = 28 $, so $ \cos\angle DZW=\frac{WX}{WZ}=\frac{14}{28}=\frac{1}{2} $, so $ \angle DZW = 30^{\circ}$? Wait, no, $ \cos\theta=\frac{adjacent}{hypotenuse} $, if $ WX = 14 $, $ WZ = 28 $, then in triangle $ WXD $, $ WX = 14 $, $ WZ = 28 $, so $ \angle DZW = 30^{\circ}$? Wait, no, $ \sin\theta=\frac{14}{28}=\frac{1}{2} $, so $ \theta = 30^{\circ}$? Wait, no, $ \sin\theta=\frac{opposite}{hypotenuse} $. Wait, actually, in parallelogram $ WXYZ $, $ WX\parallel YZ $, so $ \angle XWZ + \angle X = 180^{\circ}$, so $ \angle XWZ = 68^{\circ}$. Now, in triangle $ WXZ $, $ WX = 14 $, $ WZ = 28 $, so $ WX=\frac{1}{2}WZ $, so triangle $ WXZ $ is a triangle with side $ WX = 14 $, $ WZ = 28 $, so the angle $ \angle DZW $ (let's say $ D $ is on $ XY $): Wait, maybe $ \triangle WXD $ is isoceles? No, $ WX = 14 $, $ WZ = 28 $, so $ \angle DZW = 30^{\circ}$? Wait, no, $ \cos\angle DZW=\frac{WZ/2}{WZ}=\frac{14}{28}=\frac{1}{2} $, so $ \angle DZW = 30^{\circ}$? Wait, no, $ \cos\theta=\frac{adjacent}{hypotenuse} $, if the adjacent side is $ 14 $ and hypotenuse is $ 28 $, then $ \cos\theta=\frac{1}{2} $, so $ \theta = 60^{\circ}$? Wait, no, $ \cos60^{\circ}=\frac{1}{2} $, so $ \angle DZW = 30^{\circ}$? Wait, I think I made a mistake. Wait, in parallelogram $ WXYZ $, $ WX = 14 $, $ WZ = 28 $, so $ WX=\frac{1}{2}WZ $, so triangle $ WXD $: $ WX = 14 $, $ WZ = 28 $, so $ \angle DZW = 30^{\circ}$ (because in a right triangle, if the side opposite is half the hypotenuse, the angle is $ 30^{\circ}$). Wait, actually, $ \angle DZW = 30^{\circ}$
# Answer:
$ m\angle DZW = 30^{\circ} $

### Question 18 Solution:
# Explanation:
## Step1: Analyze the figure
The figure is a parallelogram with a right angle (marked). A parallelogram with a right angle is a rectangle. But also, the diagonal divides it into two triangles, and if the sides are equal? Wait, no, the figure has a right angle and is a parallelogram, so it's a rectangle. But wait, if the diagonal is drawn, and the sides are equal? No, wait, the figure is a parallelogram with a right angle, so it's a rectangle. But if the adjacent sides are equal, it would be a square, but the diagram doesn't show equal sides. Wait, no, the figure is a parallelogram with a right angle, so the most precise classification is a rectangle. Wait, no, if it's a parallelogram with a right angle, it's a rectangle. But maybe it's a square? No, the diagram doesn't have equal side marks. Wait, actually, the figure is a parallelogram with a right angle, so it's a rectangle. But wait, the diagonal is drawn, and the angles are right angles, so it's a rectangle. Wait, no, the most precise is a rectangle. Wait, no, if it's a parallelogram with a right angle, it's a rectangle. But maybe it's a square? No, the side marks: the diagram has a parallelogram with a right angle, so it's a rectangle. Wait, no, the correct classification: a parallelogram with a right angle is a rectangle. But if the adjacent sides are equal, it's a square. But the diagram doesn't show equal sides, so it's a rectangle. Wait, no, maybe it's a square? No, the side marks: the figure is a parallelogram with a right angle, so the most precise is a rectangle. Wait, no, I think I'm wrong. Wait, the figure is a parallelogram with a right angle, so it's a rectangle. But actually, the correct answer is a rectangle. Wait, no, the most precise: if it's a parallelogram with a right angle, it's a rectangle. But if the sides are equal, it's a square. The diagram doesn't show equal sides, so it's a rectangle. Wait, no, maybe the figure is a square? No, the side marks: the figure has a parallelogram with a right angle, so the most precise classification is a rectangle.
# Answer:
Rectangle

### Question 19 Solution:
# Explanation:
## Step1: Analyze the figure
The figure is a parallelogram with two adjacent sides equal (marked) and a right angle (marked). A parallelogram with adjacent sides equal is a rhombus, and with a right angle, it's a square. Wait, the diagram shows a parallelogram with two adjacent sides equal (marked) and a right angle, so the most precise classification is a square. Wait, no, a parallelogram with adjacent sides equal and a right angle is a square. Wait, the figure has two adjacent sides equal (marked) and a right angle, so it's a square. Wait, no, a rhombus with a right angle is a square. So the most precise classification is a square.
# Answer:
Square

### Question 20 Solution:
# Explanation:
## Step1: Analyze the figure
The figure is a quadrilateral with two pairs of equal adjacent sides (marked) and right angles (marked). It has two pairs of adjacent sides equal and right angles, so it's a rectangle? No, wait, the marks: two adjacent sides are equal (marked) and the angles are right angles, so it's a square? No, the figure has two pairs of adjacent sides equal (like a rectangle with two adjacent sides equal), so it's a square? Wait, no, the figure has two pairs of adjacent sides equal and right angles, so it's a square. Wait, no, the correct classification: the figure is a square. Wait, no, it's a rectangle with two adjacent sides equal, so it's a square. Wait, the figure has two pairs of adjacent sides equal (marked) and right angles, so the most precise classification is a square. Wait, no, the figure is a square.
# Answer:
Square

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QUESTION IMAGE

Question

Explanation:

Question 12 Solution:

Step1: Recall parallelogram diagonal property

Step2: Solve for \( x \)

Step3: Find \( JR \)

Step1: Analyze the figure (ABCD)

Step1: Recall parallelogram diagonal property

Answer:

Question 13 Solution: