NCERT Class 9 Solutions: Quadrilaterals (Chapter 8) Exercise 8.1 – Part 5

Classification of quadrilaterals, key properties, shape , characteristic, and name

Classification of Quadrilaterals by Key Property

Classification of quadrilaterals, key properties, shape , characteristic, and name

Q-9 In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP=BQ (see Fig.). Show that:

  1. APDCQB

  2. AP=CQ

  3. AQBCPD

  4. AQ=CP

  5. APCQ is a parallelogram

Quadrilateral poly1 Quadrilateral poly1: Polygon A, D, C, B Segment a Segment a: Segment [A, D] of Quadrilateral poly1 Segment b Segment b: Segment [D, C] of Quadrilateral poly1 Segment c Segment c: Segment [C, B] of Quadrilateral poly1 Segment d Segment d: Segment [B, A] of Quadrilateral poly1 Segment f Segment f: Segment [D, B] Segment g Segment g: Segment [A, Q] Segment h Segment h: Segment [A, P] Segment i Segment i: Segment [P, C] Segment j Segment j: Segment [Q, C] Point A A = (-1.24, 3.66) Point A A = (-1.24, 3.66) Point A A = (-1.24, 3.66) Point D D = (2.6, 3.66) Point D D = (2.6, 3.66) Point D D = (2.6, 3.66) Point C C = (1.72, 0.8) Point C C = (1.72, 0.8) Point C C = (1.72, 0.8) Point B B = (-2.1, 0.78) Point B B = (-2.1, 0.78) Point B B = (-2.1, 0.78) Point Q Point Q: Point on f Point Q Point Q: Point on f Point Q Point Q: Point on f Point P Point P: Point on f Point P Point P: Point on f Point P Point P: Point on f

Triangle of APD and CQB

Triangle of APD and CQB, DP=BQ its diagonal of BD

Solution:

  1. In ΔAPDandΔCQB ,

    • DP=BQ (Given)

    • ADP=CBQ (Alternate interior angles)

    • AD=BC (Opposite sides of a parallelogram)

    Thus, ΔAPDΔCQB by Side-Angle-Side congruence condition.

  2. AP=CQ By Corresponding parts of congruent triangles as ΔAPDΔCQB .

  3. In ΔAQBandΔCPD ,

    • BQ=DP (Given)

    • ABQ=CDP (Alternate interior angles)

    • AB = CD (Opposite sides of a parallelogram)

    Thus, ΔAQB ≅ ΔCPD by Side-Angle-Side congruence condition.

  4. AQ=CP By Corresponding Parts of Congruent Triangles as ΔAQBΔCPD .

  5. The diagonal of a parallelogram bisect each other.

  • OB+OD

  • OBBQ=ODDPBQ=DP Given

  • OQ=OP ....equation (1)

Also, OA=OC ………….equation (2) (diagonal of a parallelogram bisect each other)

From equation (1) and (2), APCQ is parallelogram

Q-10 ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on on diagonal BD (see Fig.).Show that

  1. APBCQD

  2. AP=CQ

Segment f Segment f: Segment [D, C] Segment g Segment g: Segment [C, B] Segment j Segment j: Segment [D, A] Segment k Segment k: Segment [A, B] Segment l Segment l: Segment [D, B] Segment m Segment m: Segment [C, Q] Segment n Segment n: Segment [A, P] Point D D = (-1.54, 4.42) Point D D = (-1.54, 4.42) Point D D = (-1.54, 4.42) Point C C = (1.9, 4.42) Point C C = (1.9, 4.42) Point C C = (1.9, 4.42) Point B B = (1.1, 2) Point B B = (1.1, 2) Point B B = (1.1, 2) Point A Point A: Point on h Point A Point A: Point on h Point A Point A: Point on h Point Q Point Q: Point on l Point Q Point Q: Point on l Point Q Point Q: Point on l Point P Point P: Point on l Point P Point P: Point on l Point P Point P: Point on l

ABCD Parallelogram.

ABCD parallelogram, AP and CQ are perpendiculars from vertices A and C on diagonal BD

Solution:

Given,

  • ABCD is parallelogram

  • AP and CQ are perpendiculars from A and C on diagonal BD

Solution (i)

In ΔAPB and ΔCQD ,

  • AB=CD (Opposite side of parallelogram ABCD)

  • ABP=CDQ (Alternate interior angles)

  • ABDC

Now,

  • APB=CQD (Equal to right angles as AP and CQ are perpendiculars)

  • AB=CD (ABCD is a parallelogram)

  • Thus, ΔAPBΔCQD by Angle-Angle-Side congruence condition.

Solution (ii)

AP=CQ by Corresponding Parts of Congruent Triangles as ΔAPBΔCQD .

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