geometry
review - quadrilaterals & their properties
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find the measurement indicated in each parallelogram,
1)
2)
3)
solve for x. each figure is a parallelogram.
4)
5)

Question

Accepted Answer

### Problem 1: Find the measurement indicated (angle at E) in parallelogram E F C D
# Explanation:
## Step 1: Recall parallelogram angle property
In a parallelogram, consecutive angles are supplementary (sum to $ 180^\circ $).
Given $ \angle F = 123^\circ $, $ \angle E $ and $ \angle F $ are consecutive angles.

## Step 2: Calculate $ \angle E $
$ \angle E + 123^\circ = 180^\circ $
$ \angle E = 180^\circ - 123^\circ = 57^\circ $

### Problem 2: Find the measurement indicated (angle at E) in rectangle (a type of parallelogram) E F C D
# Explanation:
## Step 1: Recall rectangle angle property
A rectangle is a parallelogram with all angles $ 90^\circ $ (right angles).

## Step 2: Determine $ \angle E $
Since $ \angle C $ is a right angle ($ 90^\circ $), $ \angle E = 90^\circ $ (all angles in a rectangle are right angles).

### Problem 3: Find the measurement indicated (angle at V) in parallelogram V W T U
# Explanation:
## Step 1: Recall parallelogram angle property
In a parallelogram, consecutive angles are supplementary (sum to $ 180^\circ $).
Given $ \angle W = 110^\circ $, $ \angle V $ and $ \angle W $ are consecutive angles.

## Step 2: Calculate $ \angle V $
$ \angle V + 110^\circ = 180^\circ $
$ \angle V = 180^\circ - 110^\circ = 70^\circ $

### Problem 4: Solve for $ x $ in parallelogram R S T U
# Explanation:
## Step 1: Recall triangle angle sum property
In $ \triangle R T S $, the sum of angles is $ 180^\circ $. Given angles: $ 70^\circ $, $ 60^\circ $, and $ \angle R S T $ (which is equal to $ 13x - 2 $ as opposite sides of a parallelogram are parallel, so alternate interior angles or triangle angle sum applies). Wait, actually, in the parallelogram, $ \angle U T R = 70^\circ $, $ \angle R T S = 60^\circ $, so $ \angle U T S = 70^\circ + 60^\circ = 130^\circ $. But in the triangle $ R T S $, angles sum to $ 180^\circ $: $ 70^\circ + 60^\circ + (13x - 2) = 180^\circ $? Wait, no—wait, the side $ R S $ is $ 13x - 2 $, but actually, the triangle $ R T S $: angles at $ T $ is $ 60^\circ $, at $ U $ (wait, no, the figure: $ U T R $ has $ 70^\circ $, $ R T S $ has $ 60^\circ $, so $ \angle S T R = 60^\circ $, $ \angle U T R = 70^\circ $, so $ \angle U T S = 130^\circ $. But in triangle $ R T S $, angles sum to $ 180^\circ $: $ 70^\circ + 60^\circ + \angle R S T = 180^\circ $? Wait, no, the angle at $ S $ in triangle $ R T S $ is $ 13x - 2 $. Wait, maybe the triangle is $ R T S $, with angles: $ \angle R T S = 60^\circ $, $ \angle T R S = 70^\circ $, so $ \angle R S T = 180^\circ - 60^\circ - 70^\circ = 50^\circ $? No, that doesn’t match. Wait, maybe the side $ R S = 13x - 2 $, but actually, the correct approach: in the parallelogram, $ U T \parallel R S $, so $ \angle U T R = \angle T R S = 70^\circ $ (alternate interior angles). Then in triangle $ R T S $, angles are $ 70^\circ $ (at $ R $), $ 60^\circ $ (at $ T $), so angle at $ S $ is $ 180 - 70 - 60 = 50^\circ $? No, this is confusing. Wait, the problem says “Solve for $ x $. Each figure is a parallelogram.” For problem 4, the triangle $ R T S $: angles at $ T $ is $ 60^\circ $, at $ U $ (wait, $ U T R $ is $ 70^\circ $), so $ \angle R T S = 60^\circ $, $ \angle U T R = 70^\circ $, so $ \angle U T S = 130^\circ $. But in the parallelogram, $ \angle U T S + \angle T S R = 180^\circ $ (consecutive angles), so $ \angle T S R = 50^\circ $. But the side $ R S = 13x - 2 $? No, maybe the angle at $ S $ is $ 13x - 2 $. Wait, perhaps the correct equation is $ 70 + 60 + (13x - 2) = 180 $? Let’s check:

$ 70 + 60 + 13x - 2 = 180 $
$ 128 + 13x = 180 $
$ 13x = 52 $
$ x = 4 $

### Problem 5: Solve for $ x $ in parallelogram S T U R
# Explanation:
## Step 1: Recall isosceles triangle property
In a parallelogram, $ S T \parallel U R $, so $ \angle S T R = \angle T R U = 70^\circ $ (alternate interior angles). Also, $ S T = U R $ and $ S R = T U $, but here, triangle $ S T R $: $ S T = x + 66 $, $ S R = x + 44 $, and $ \angle S T R = 70^\circ $, $ \angle T R S = x + 66 $? Wait, no—wait, in triangle $ S T R $, if $ \angle T R U = 70^\circ $, and $ T R $ is a diagonal, then $ \angle S T R = \angle T R U = 70^\circ $ (alternate interior angles). Also, in triangle $ S T R $, sides $ S T = x + 66 $, $ S R = x + 44 $, and angle at $ T $ is $ 70^\circ $. Wait, no—maybe it’s an isosceles triangle? Wait, the angles: $ \angle S T R = 70^\circ $, $ \angle S R T = x + 66 $, $ \angle S = x + 44 $. Wait, triangle angle sum: $ (x + 44) + (x + 66) + 70 = 180 $

## Step 2: Solve for $ x $
$ x + 44 + x + 66 + 70 = 180 $
$ 2x + 180 = 180 $
$ 2x = 0 $
$ x = 0 $? That can’t be right. Wait, maybe the angles are $ \angle S = x + 44 $, $ \angle S T R = 70^\circ $, and $ \angle S R T = x + 66 $, but actually, in the parallelogram, $ S T = U R $ and $ S R = T U $, but the diagonal $ T R $ splits the parallelogram into two triangles. Wait, maybe the correct approach is: in triangle $ S T R $, $ \angle S T R = 70^\circ $, $ \angle S R T = x + 66 $, and $ \angle S = x + 44 $. But triangle angle sum: $ (x + 44) + (x + 66) + 70 = 180 $
Simplify: $ 2x + 180 = 180 $ → $ 2x = 0 $ → $ x = 0 $. That seems odd, but maybe the problem has a typo, or I misinterpret the diagram. Alternatively, maybe $ \angle S T R = 70^\circ $, $ \angle S R T = x + 66 $, and $ \angle S = x + 44 $, but $ \angle S T R = \angle S R T $? No, that would be isosceles, but $ 70 = x + 66 $ → $ x = 4 $, then $ \angle S = 4 + 44 = 48 $, and $ 70 + 70 + 48 = 188 $, which is wrong. Alternatively, maybe $ \angle S = \angle S T R $? No. Wait, maybe the sides: $ S T = x + 66 $, $ S R = x + 44 $, and $ T R $ is a diagonal, but in a parallelogram, opposite sides are equal, but here it’s a triangle. Wait, perhaps the correct equation is $ x + 44 = x + 66 $? No, that’s impossible. I think I made a mistake. Let’s re-express:

Wait, the problem says “Solve for $ x $. Each figure is a parallelogram.” For problem 5, the triangle $ S T R $: angles at $ T $ is $ 70^\circ $, at $ S $ is $ x + 44 $, at $ R $ is $ x + 66 $. Then:

$ (x + 44) + (x + 66) + 70 = 180 $
$ 2x + 180 = 180 $
$ 2x = 0 $
$ x = 0 $

Even though it’s odd, mathematically, that’s the solution.

### Final Answers:
1) $ \boldsymbol{57^\circ} $
2) $ \boldsymbol{90^\circ} $
3) $ \boldsymbol{70^\circ} $
4) $ \boldsymbol{x = 4} $
5) $ \boldsymbol{x = 0} $ (Note: This may indicate a diagram misinterpretation, but following the angle sum gives $ x = 0 $.)

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QUESTION IMAGE

Question

Explanation:

Problem 1: Find the measurement indicated (angle at E) in parallelogram E F C D

Step 1: Recall parallelogram angle property

Step 2: Calculate \( \angle E \)

Problem 2: Find the measurement indicated (angle at E) in rectangle (a type of parallelogram) E F C D

Step 1: Recall rectangle angle property

Step 2: Determine \( \angle E \)

Problem 3: Find the measurement indicated (angle at V) in parallelogram V W T U

Step 1: Recall parallelogram angle property

Step 2: Calculate \( \angle V \)

Problem 4: Solve for \( x \) in parallelogram R S T U

Answer:

Step 1: Recall isosceles triangle property

Step 2: Solve for \( x \)

Final Answers: