If each quadrilateral below is a parallelogram, find the missing measures.

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If each quadrilateral below is a parallelogram, find the missing measures. Date: Bell: Homework 2: Parallelograms This is a 2-page document! Directions: If each quadrilateral below is a parallelogram; find the missing measures MN = KN = ZK = iZL = MM = 15 31 CF = 7E = Cz = GD = 45 10 1198 N = D = 22 Given PQ = 24, PS = 19, ∠2 = 42, TQ = 10, ∠ZPQR = 106, ∠EQS = 498, and m∠PRS = 35 QR = 8, PT = ∠ZQZS = ∠i2?QS = WL = PS = ∠ZPSQ = SQ = 4. Find KL 5. If AC = √14 and EC = 21 + 11, solve for : 7 – 2

The Correct Answer and Explanation is:

Correct Answers

1. Parallelogram KLMN

  • MN = 31
  • KN = 45
  • m∠K = 61°
  • m∠L = 119°
  • m∠M = 61°

2. Parallelogram CDEF

  • CF = 10
  • FE = 15
  • CE = 14
  • GD = 11

3. Parallelogram PQRS

  • QR = 19
  • SR = 24
  • PT = 21
  • SQ = 20
  • m∠QRS = 74°
  • m∠PQS = 49°
  • m∠RPS = 39°
  • m∠PSQ = 57°

Explanation

To solve these problems, we utilize the fundamental properties of a parallelogram. A parallelogram is a quadrilateral where opposite sides are parallel. This leads to several key geometric rules:

  1. Opposite sides are congruent (equal in length).
  2. Opposite angles are congruent (equal in measure).
  3. Consecutive angles (angles next to each other) are supplementary, adding up to 180°.
  4. The diagonals bisect each other, meaning they intersect at their midpoints.
  5. Alternate interior angles, formed by a transversal cutting across parallel sides, are congruent.

For Problem 1 (Parallelogram KLMN):

  • Side Lengths: Opposite sides are congruent. Therefore, side MN is congruent to side KL, making MN = 31. Side KN is congruent to side LM, making KN = 45.
  • Angles: Opposite angles are congruent, so m∠L is congruent to the given m∠N, making m∠L = 119°. Consecutive angles are supplementary. Thus, m∠K + m∠N = 180°, which means m∠K = 180° – 119° = 61°. Finally, m∠M is opposite m∠K, so m∠M = 61°.

For Problem 2 (Parallelogram CDEF):

  • Side Lengths: Following the rule of congruent opposite sides, CF is congruent to DE, making CF = 10, and FE is congruent to CD, making FE = 15.
  • Diagonals: The diagonals bisect each other at point G. This means G is the midpoint for both CE and FD. Since FD = 22, GD is half of that, so GD = 11. We are given that the segment CG = 7; because G is the midpoint, GE must also be 7. The full length of the diagonal is the sum of its parts: CE = CG + GE = 7 + 7 = 14.

For Problem 3 (Parallelogram PQRS):

  • Side Lengths & Diagonals: Opposite sides are congruent, so QR = PS = 19 and SR = PQ = 24. The diagonals bisect each other at T. Thus, PT is half the length of PR, making PT = 42 / 2 = 21. Since TQ = 10, the full diagonal SQ = 2 * TQ = 20.
  • Angles:
    • m∠QRS: It is consecutive to ∠PQR, so they are supplementary. m∠QRS = 180° – 106° = 74°.
    • m∠PQS: Since PQ || SR, the transversal SQ creates congruent alternate interior angles. Therefore, m∠PQS = m∠QSR = 49°.
    • m∠PSQ: Since PS || QR, m∠PSQ is the alternate interior angle to m∠RQS. We find m∠RQS by subtracting: m∠RQS = m∠PQR – m∠PQS = 106° – 49° = 57°. Thus, m∠PSQ = 57°.
    • m∠RPS: Since PS || QR, m∠RPS is the alternate interior angle to m∠PRQ. We find m∠PRQ by subtracting: m∠PRQ = m∠QRS – m∠PRS = 74° – 35° = 39°. Thus, m∠RPS = 39°.

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