x =
db =
if ae= 6x - 55 & ec= 3x - 16, find db.
6x - 55 + 3x - 16
show your work for credit!
6.
if lo= 15x + 19 & qn= 10x + 2, find pn
x =
pn =
show your work for credit!
7.
find ( mangle gjk )
x =
( mangle gjk = )
show your work for credit!
8.
x =
( mangle ade = )
show your work for credit!

Question

Accepted Answer

### Problem 1 (Top - Rectangle Diagonals)
# Explanation:
## Step1: Recall Rectangle Diagonal Property
In a rectangle, the diagonals are equal and bisect each other. So $ AE = EC $ (since diagonals bisect each other).
Set $ 6x - 55=3x - 16 $
## Step2: Solve for $ x $
Subtract $ 3x $ from both sides: $ 6x - 3x-55 = 3x - 3x-16 $
$ 3x-55=-16 $
Add 55 to both sides: $ 3x-55 + 55=-16 + 55 $
$ 3x = 39 $
Divide by 3: $ x=\frac{39}{3}=13 $
## Step3: Find $ AC $ (then $ DB $ since diagonals are equal)
First, find $ AE $: $ AE = 6(13)-55=78 - 55 = 23 $
$ EC = 3(13)-16=39 - 16 = 23 $ (checks out)
$ AC=AE + EC=23 + 23 = 46 $
Since $ DB = AC $ (diagonals of rectangle are equal), $ DB = 46 $
# Answer:
$ x = 13 $
$ DB = 46 $

### Problem 6 (Middle - Rectangle Diagonals)
# Explanation:
## Step1: Recall Rectangle Diagonal Property
In a rectangle, diagonals are equal and bisect each other. So $ LO=PN $ (wait, actually, diagonals are equal, so $ LO = PN $? Wait, no, in rectangle $ LPNO $ (assuming it's a rectangle), diagonals $ LN $ and $ PO $ bisect each other at $ Q $, and $ LO $ is half of a diagonal? Wait, no, the problem says $ LO = 15x + 19 $ and $ QN=10x + 2 $. Wait, in a rectangle, diagonals are equal, so $ LN=PO $, and $ Q $ is the midpoint, so $ QN=\frac{1}{2}PO $ and $ LO=\frac{1}{2}LN $. Since $ LN = PO $, then $ LO = QN\times2 $? Wait, no, maybe $ LO $ is a diagonal? Wait, the figure is a rectangle $ LPNO $, so diagonals $ LN $ and $ PO $ intersect at $ Q $. So $ LO $ is a side? Wait, no, the labels are $ L, N, P, O $. Let's assume $ LO $ and $ PN $ are diagonals? Wait, no, the problem says "If $ LO = 15x + 19 $ & $ QN = 10x + 2 $, find $ PN $". In a rectangle, diagonals are equal, and the midpoint divides them into two equal parts. So $ PN $ is a diagonal, and $ QN $ is half of a diagonal? Wait, maybe $ LO $ is equal to $ PN $? No, let's re - examine. Wait, maybe the rectangle has diagonals $ LN $ and $ PO $, and $ LO $ is a side? No, the problem is likely that in the rectangle, $ LO $ and $ PN $ are related such that $ LO = PN $ (diagonals), but $ QN $ is half of a diagonal. Wait, no, let's correct: In a rectangle, diagonals are equal, so $ LN = PO $. The midpoint $ Q $ means $ QN=\frac{1}{2}LN $ and $ QO=\frac{1}{2}PO $. Since $ LN = PO $, then $ QN = QO $, and $ LO $ is a segment. Wait, maybe the problem has a typo, but assuming that $ LO $ is equal to $ PN $ (diagonals), but no, let's use the property that diagonals are equal, so $ LO = PN $ and $ QN=\frac{1}{2}PN $. So $ LO = 2\times QN $
So $ 15x + 19=2(10x + 2) $
## Step2: Solve for $ x $
Expand right side: $ 15x + 19 = 20x+4 $
Subtract $ 15x $: $ 19=5x + 4 $
Subtract 4: $ 15 = 5x $
Divide by 5: $ x = 3 $
## Step3: Find $ PN $
Since $ PN=LO $ (diagonals are equal), $ PN=15(3)+19=45 + 19 = 64 $
# Answer:
$ x = 3 $
$ PN = 64 $

### Problem 7 (Bottom - Rectangle Angles)
# Explanation:
## Step1: Recall Rectangle Properties
In a rectangle, opposite sides are parallel, so $ GJ\parallel II $ (wait, the figure has $ G, J, I, I $? Wait, the angle at $ I $ is $ (5x + 8)^\circ $ and at $ J $ is $ (7x - 16)^\circ $. Since $ GJ\parallel II $ (assuming $ GJ $ and $ II $ are parallel, and $ JI $ is a transversal), the alternate interior angles are equal? Wait, no, in a rectangle, the triangles formed by diagonals are isosceles. Wait, the angles $ \angle GIJ=(5x + 8)^\circ $ and $ \angle GJI=(7x - 16)^\circ $. In triangle $ GJI $, since $ GJ = II $ (sides of rectangle), triangle $ GJI $ is isosceles? Wait, no, maybe the angles are equal because of parallel sides. Wait, actually, in a rectangle, the diagonals are equal and bisect each other, so the triangles formed are isosceles. But maybe the angles $ (5x + 8) $ and $ (7x - 16) $ are equal? Let's set them equal:
$ 5x + 8=7x - 16 $
## Step2: Solve for $ x $
Subtract $ 5x $ from both sides: $ 8 = 2x-16 $
Add 16 to both sides: $ 24 = 2x $
Divide by 2: $ x = 12 $
## Step3: Find $ m\angle GJK $
First, find $ \angle GJI=(7x - 16)^\circ=7(12)-16 = 84 - 16 = 68^\circ $
Wait, $ \angle GJK $ is half of $ \angle GJI $? No, maybe $ \angle GJK $ is related. Wait, maybe the angle $ \angle GJK $ is equal to $ \angle GIJ $ or something. Wait, no, let's re - check. If $ x = 12 $, then $ 5x + 8=5(12)+8 = 68^\circ $, $ 7x - 16=7(12)-16 = 68^\circ $. So triangle $ GJI $ is isosceles with $ \angle GIJ=\angle GJI = 68^\circ $. Then $ \angle GJK $ is half of $ \angle GJI $? Wait, no, maybe $ K $ is the midpoint of the diagonals, so $ JK $ bisects $ \angle GJI $. So $ m\angle GJK=\frac{1}{2}\times68^\circ = 34^\circ $? Wait, no, maybe I made a mistake. Wait, the problem says "Find $ m\angle GJK $". If the two angles $ (5x + 8) $ and $ (7x - 16) $ are equal (since it's a rectangle, alternate interior angles), then $ x = 12 $, and then $ \angle GJK $ is equal to one of them? Wait, no, maybe the angle $ \angle GJK $ is equal to $ (5x + 8) $ or $ (7x - 16) $. Wait, when $ x = 12 $, $ 5x + 8=68 $, $ 7x - 16 = 68 $. So maybe $ m\angle GJK = 34^\circ $? Wait, no, let's think again. In a rectangle, diagonals bisect each other and are equal, so the triangles formed are isosceles. The angle at $ J $ and $ I $ are equal, so $ \angle GJI=\angle GIJ = 68^\circ $. Then $ \angle GJK $ is half of $ \angle GJI $ because $ K $ is the intersection of diagonals, so $ JK $ bisects $ \angle GJI $. So $ m\angle GJK=\frac{68^\circ}{2}=34^\circ $
# Answer:
$ x = 12 $
$ m\angle GJK = 34^\circ $ (Wait, but if the angles $ (5x + 8) $ and $ (7x - 16) $ are equal to $ 68^\circ $, and $ \angle GJK $ is half, then 34. But maybe I messed up. Alternatively, if $ \angle GJK $ is equal to $ (5x + 8) $ or $ (7x - 16) $, then $ m\angle GJK = 68^\circ $. Wait, the problem's figure: $ G, J, I, I $ (maybe a typo, should be $ H $ instead of second $ I $). So $ GJHI $ is a rectangle. Then $ GJ\parallel HI $, and $ JI $ is a transversal, so $ \angle GJI=\angle HIJ $, but the angles given are $ (5x + 8) $ at $ I $ and $ (7x - 16) $ at $ J $. So they are equal, so $ x = 12 $, and then $ \angle GJK $ is equal to one of them? Wait, maybe the answer is $ x = 12 $, $ m\angle GJK = 34^\circ $ or $ 68^\circ $. Wait, let's recalculate:
If $ x = 12 $, $ 5x + 8=68 $, $ 7x - 16 = 68 $. If $ K $ is the midpoint of diagonals, then $ \angle GJK=\frac{1}{2}\angle GJI=\frac{68}{2}=34 $. So I think that's it.
# Answer:
$ x = 12 $
$ m\angle GJK = 34^\circ $

### Problem 8 (Bottom - Rectangle Angles)
# Explanation:
## Step1: Recall Rectangle Properties
In a rectangle, $ AD\parallel BC $, so $ \angle ADE=\angle BEC $ (alternate interior angles). Also, $ \angle ABE=(4x + 15)^\circ $ and $ \angle EBC=(13x + 7)^\circ $, and $ \angle ABC = 90^\circ $ (angle of rectangle). So $ (4x + 15)+(13x + 7)=90 $
## Step2: Solve for $ x $
Combine like terms: $ 4x+13x + 15 + 7=90 $
$ 17x+22 = 90 $
Subtract 22: $ 17x=90 - 22 = 68 $
Divide by 17: $ x=\frac{68}{17}=4 $
## Step3: Find $ m\angle ADE $
First, find $ \angle EBC=(13x + 7)^\circ=13(4)+7 = 52 + 7 = 59^\circ $
Since $ AD\parallel BC $, $ \angle ADE=\angle EBC $ (alternate interior angles), so $ m\angle ADE = 59^\circ $
# Answer:
$ x = 4 $
$ m\angle ADE = 59^\circ $

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QUESTION IMAGE

Question

Explanation:

Problem 1 (Top - Rectangle Diagonals)

Step1: Recall Rectangle Diagonal Property

Step2: Solve for \( x \)

Step3: Find \( AC \) (then \( DB \) since diagonals are equal)

Step1: Recall Rectangle Diagonal Property

Step2: Solve for \( x \)

Step3: Find \( PN \)

Step1: Recall Rectangle Properties

Step2: Solve for \( x \)

Step3: Find \( m\angle GJK \)

Answer:

Problem 6 (Middle - Rectangle Diagonals)