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In the trapezium ABCD, AB || DC and AB > DC. P and Q are the mid-points of the diagonals AC and BD. Then, PQ || AB and PQ = …..(AB - DC).

In the trapezium ABCD, AB || DC and AB and DC. P and Q are the mid-points of the diagonals AC and BD. Then, PQ || AB and PQ = .....(AB - DC).R.S. Aggarwal Mathematics Solutions ICSE Class 9.

  1. 2

  2. 3

  3. 12\dfrac{1}{2}

  4. 13\dfrac{1}{3}

Mid-point Theorem

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Answer

In △BQR and △CQD,

⇒ ∠BQR = ∠CQD (Vertically opposite angles are equal)

⇒ ∠BRQ = ∠QCD (Alternate angles are equal)

⇒ BQ = DQ (Q is the mid-point of BD)

∴ △BQR ≅ △CQD

⇒ BR = DC (Corresponding parts of congruent triangles are equal)

⇒ QR = CQ (Corresponding parts of congruent triangles are equal)

Given,

AB || DC and PQ || AB

∴ PQ || AB || DC

In △ARC,

Since, P and Q are the mid-points of AC and CR respectively.

By mid-point theorem,

⇒ PQ = 12\dfrac{1}{2} AR

⇒ PQ = 12\dfrac{1}{2} (AB - BR)

⇒ PQ = 12\dfrac{1}{2} (AB - DC) (∵ BR = DC)

Hence, option 3 is the correct option.

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