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Prove that the opposite sides of a parallelogram are equal in length.

Mathematics

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Answer

x=8x = \text{8}

E^=84∘\hat{E} = \text{84}^\circ

Step-by-step solution

11 steps

Question 1

1.1 To prove that the opposite sides of parallelogram PQRS are equal, we can use the diagonal QS to divide the parallelogram into two triangles, △PQS\triangle PQS and △RSQ\triangle RSQ.

  1. Identify parallel lines and transversals.

    Since PQRS is a parallelogram, PQ∥RSPQ \parallel RS and PS∥QRPS \parallel QR.
    QS is a transversal for both pairs of parallel lines.

  2. Identify equal alternate interior angles.

    Because PQ∥RSPQ \parallel RS and QS is a transversal, the alternate interior angles are equal:

    ∠PQS=∠RSQ(or ∠Q1=∠S2 from the diagram)\angle PQS = \angle RSQ \quad (\text{or } \angle Q_1 = \angle S_2 \text{ from the diagram})

    Because PS∥QRPS \parallel QR and QS is a transversal, the alternate interior angles are equal:

    ∠PSQ=∠RQS(or ∠S1=∠Q2 from the diagram)\angle PSQ = \angle RQS \quad (\text{or } \angle S_1 = \angle Q_2 \text{ from the diagram})

  3. Identify the common side.

    The side QS is common to both △PQS\triangle PQS and △RSQ\triangle RSQ.

  4. Conclude triangle congruence.

    By the Angle-Side-Angle (ASA) congruence criterion, since two angles and the included side of △PQS\triangle PQS are equal to two angles and the included side of △RSQ\triangle RSQ:

    △PQS≅△RSQ\triangle PQS \cong \triangle RSQ

  5. Conclude equality of opposite sides.

    Since the triangles are congruent, their corresponding sides are equal:

    PQ=RSPQ = RS PS=QRPS = QR

    Therefore, the opposite sides of the parallelogram PQRS are equal.

    1.2 Assuming the parallelogram in the diagram is EFGH, as shown, and the given angles G^\hat{G} and F^\hat{F} refer to this parallelogram.

    1.2.1 Calculate the value of xx.

  6. In a parallelogram, consecutive angles are supplementary (they add up to 180∘180^\circ).

    Therefore, G^+F^=180∘\hat{G} + \hat{F} = 180^\circ.

  7. Substitute the given expressions for G^\hat{G} and F^\hat{F} into the equation.

    (9x+12∘)+(15x−24∘)=180∘(9x + 12^\circ) + (15x - 24^\circ) = 180^\circ

  8. Solve for xx.

    24x−12∘=180∘24x - 12^\circ = 180^\circ 24x=180∘+12∘24x = 180^\circ + 12^\circ 24x=192∘24x = 192^\circ x=19224x = \frac{192}{24} x=8x = \boxed{\text{8}}

    1.2.2 Determine the size of E^\hat{E}.

  9. In a parallelogram, opposite angles are equal.

    Therefore, E^=G^\hat{E} = \hat{G}.

  10. Substitute the value of xx into the expression for G^\hat{G}.

    G^=9x+12∘\hat{G} = 9x + 12^\circ G^=9(8)+12∘\hat{G} = 9(8) + 12^\circ G^=72∘+12∘\hat{G} = 72^\circ + 12^\circ G^=84∘\hat{G} = 84^\circ

  11. Since E^=G^\hat{E} = \hat{G}, the size of E^\hat{E} is

    E^=84∘\hat{E} = \boxed{\text{84}^\circ}
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