NCERT Class 9 Solutions: Quadrilaterals (Chapter 8) Exercise 8.2 – Part 1

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Quadiletrals and the properties of their diagonals

Diagonals of a Quadiletrals

Quadiletrals and the properties of their diagonals

Q-1 ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see Fig). AC is a diagonal. Show that:

  1. SRAC and SR=12AC

  2. PQ=SR

  3. PQRS is a parallelogram.

Quadrilateral poly1 Quadrilateral poly1: Polygon D, C, B, A Segment a Segment a: Segment [D, C] of Quadrilateral poly1 Segment b Segment b: Segment [C, B] of Quadrilateral poly1 Segment c Segment c: Segment [B, A] of Quadrilateral poly1 Segment d Segment d: Segment [A, D] of Quadrilateral poly1 Segment f Segment f: Segment [S, R] Segment g Segment g: Segment [R, Q] Segment h Segment h: Segment [Q, P] Segment i Segment i: Segment [P, S] Segment j Segment j: Segment [C, A] Point D D = (-1.56, 4.24) Point D D = (-1.56, 4.24) Point D D = (-1.56, 4.24) Point C C = (2.02, 3.78) Point C C = (2.02, 3.78) Point C C = (2.02, 3.78) Point B B = (2.02, 1.86) Point B B = (2.02, 1.86) Point B B = (2.02, 1.86) Point A A = (-2.08, 1.86) Point A A = (-2.08, 1.86) Point A A = (-2.08, 1.86) Point S Point S: Point on d Point S Point S: Point on d Point S Point S: Point on d Point R Point R: Point on a Point R Point R: Point on a Point R Point R: Point on a Point Q Point Q: Point on b Point Q Point Q: Point on b Point Q Point Q: Point on b Point P Point P: Point on c Point P Point P: Point on c Point P Point P: Point on c

ABCD Is a Quadrilateral

ABCD is a quadrilateral such that P, Q, R, S are mid-point of the sides AB, BC, CD and DA

Solution:

Give, ABCD is a quadrilateral

P, Q, R and S are mid-points of the sides AB, BC, CD and DA

  1. In ΔDAC ,

    • R is the midpoint of DC and S is the midpoint of DA.

    • Thus by mid-point theorem, SRAC and SR=12AC

  2. In ΔBAC ,

    • P is the midpoint of AB and Q is the midpoint of BC.

    • Thus by mid-point theorem, PQAC and PQ=12AC

    • Also, SR=12AC

    • Thus, PQ=SR

  3. SRAC - From (i)

  • And, PQAC - from (ii)

  • SRPQ - From (i) and (ii)

  • Also, PQ=SR

  • Thus, PQRS is a parallelogram.

Q-2 ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rectangle.

Solution:

Polygon poly1 Polygon poly1: Polygon[A, D, 4] Polygon poly1 Polygon poly1: Polygon[A, D, 4] Quadrilateral poly2 Quadrilateral poly2: Polygon P, Q, R, S Segment f Segment f: Segment [A, D] of Polygon poly1 Segment g Segment g: Segment [D, C] of Polygon poly1 Segment h Segment h: Segment [C, B] of Polygon poly1 Segment i Segment i: Segment [B, A] of Polygon poly1 Segment e Segment e: Segment [P, Q] of Quadrilateral poly2 Segment f_1 Segment f_1: Segment [Q, R] of Quadrilateral poly2 Segment g_1 Segment g_1: Segment [R, S] of Quadrilateral poly2 Segment h_1 Segment h_1: Segment [S, P] of Quadrilateral poly2 Segment j Segment j: Segment [B, D] Segment k Segment k: Segment [A, C] Point A A = (-0.44, 2.34) Point A A = (-0.44, 2.34) Point A A = (-0.44, 2.34) Point D D = (1.73, 0.24) Point D D = (1.73, 0.24) Point D D = (1.73, 0.24) Point C Point C: Polygon[A, D, 4] Point C Point C: Polygon[A, D, 4] Point C Point C: Polygon[A, D, 4] Point B Point B: Polygon[A, D, 4] Point B Point B: Polygon[A, D, 4] Point B Point B: Polygon[A, D, 4] Point P Point P: Point on i Point P Point P: Point on i Point P Point P: Point on i Point Q Point Q: Point on h Point Q Point Q: Point on h Point Q Point Q: Point on h Point R Point R: Point on g Point R Point R: Point on g Point R Point R: Point on g Point S Point S: Point on f Point S Point S: Point on f Point S Point S: Point on f Point O Point O: Point on j Point O Point O: Point on j Point O Point O: Point on j

ABCD Is Rhombus and PQRS Is Rectangle

ABCD is rhombus and PQRS is rectangle , P, Q, R, S are the mid-points of the side AB,BC,CD and DA respectively.

Given,

  • ABCD is a rhombus

  • P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively.

Construction, AC and BD are joined.

Proof,

In ΔDRS and ΔBPQ ,

  • DS=BQ (Halves of the opposite sides of the rhombus)

  • SDR=QBP (Opposite angles of the rhombus)

  • DR=BP (Halves of the opposite sides of the rhombus)

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

RS=PQ By Corresponding Parts of Congruent Triangles ……equation (1)

In ΔQCR and ΔSAP,

  • RC=PA (Halves of the opposite sides of the rhombus)

  • RCQ=PAS (Opposite angles of the rhombus)

  • CQ=AS (Halves of the opposite sides of the rhombus)

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

RQ=SP By Corresponding Parts of Congruent Triangles ……equation (2)

Now, on ΔCDB

  • R and Q are the mid points of CD and BC respectively

  • QRBD

Also,

  • P and S are the mid points of AD and AB respectively.

  • PSBD

  • QRPS

Thus, PQRS is a parallelogram.

Also, PQR=90°

Now, in PQRS,

  • RS=PQ and RQ=SP from equation (1) and (2) Q=90°

  • Thus, PQRS is a rectangle.

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