NCERT Class 9 Solutions: Quadrilaterals (Chapter 8) Exercise 8.1 – Part 6

An aritistic rendering of types of quadileterals which also dipicts the correct classification

Types of Quadileterals

An aritistic rendering of types of quadileterals which also dipicts the correct classification

A parallelogram is a quadrilateral with opposite sides parallel

Parallelogram of ABCD its opposite side are parallel

Parallelogram of ABCD

Parallelogram of ABCD its opposite side are parallel

Trapezium is a quadrilateral with one pair of sides parallel

Image of trapeziumImage of trapezium

Image of Trapezium

Image of trapeziumImage of trapezium

Q-11 In ABC and DEF , AB=DE,ABDE , BC=EF,BCEF . Vertices A, B, C are joined to vertices D, E, F respectively (see figure).show that

  1. Quadrilateral ABCD is a parallelogram

  2. Quadrilateral BEFC is a parallelogram

  3. ADCF and AD=CF

  4. Quadrilateral ACFD is a parallelogram

  5. AC=DF

  6. ABCDEF

Segment f Segment f: Segment [A, D] Segment j Segment j: Segment [A, C] Segment k Segment k: Segment [C, B] Segment l Segment l: Segment [B, A] Segment m Segment m: Segment [D, F] Segment n Segment n: Segment [F, E] Segment p Segment p: Segment [E, D] Segment g Segment g: Segment [B, E] Segment q Segment q: Segment [C, F] Point A A = (-0.32, 2.44) Point A A = (-0.32, 2.44) Point A A = (-0.32, 2.44) Point D D = (3.2, 1.34) Point D D = (3.2, 1.34) Point D D = (3.2, 1.34) Point B B = (-1.24, -0.18) Point B B = (-1.24, -0.18) Point B B = (-1.24, -0.18) Point C C = (1.58, 0.44) Point C C = (1.58, 0.44) Point C C = (1.58, 0.44) Point F Point F: Point on i Point F Point F: Point on i Point F Point F: Point on i Point E Point E: Point on h Point E Point E: Point on h Point E Point E: Point on h

Two Triangles of ABC and DEF

Two triangle of ABC and DEF, AB=DE, AB DE, BC=EF and BC EF

Solution:

Given,

  • ABC and DEF

  • AB=D

  • ABDE

  • BC=EF

  • BCEF .

  1. AB=DE and ABDE (Given)

    So, quadrilateral ABED is a parallelogram because: one of the two pairs of opposite sides of a quadrilateral are both equal and parallel to each other.

  2. Again BC=EF and BCEF .

    Therefore, quadrilateral BEFC is a parallelogram.

  3. ABED is a parallelogram therefore,

    • AD=BE And BE=CF ……equation (1) (Opposite sides of a parallelogram are equal)

    • Therefore, BEFC is a parallelogram.

    • Also, ADBE and BECF ……..equation (2) (Opposite sides of a parallelogram are parallel)

    • From equation (1) and (2), we obtains ADCF and AD=CF

  4. AD and CF are opposite sides of quadrilateral ACFD which are both equal and parallel to each other. Thus, it is a parallelogram.

  5. ACDF And AC=DF because ACFD is a parallelogram with opposite sides both parallel and equal length.

  6. In ΔABC and ΔDEF ,

  • AB=DE (Given)

  • BC=EF (Given)

  • AC=DF (Opposite sides of a parallelogram)

  • So, ΔABCΔDEF by SSS congruence condition.

Q-12 ABCD is a trapezium in which ABCD and AD=BC (see fig.). Show that

  1. A=B

  2. C=D

  3. ABCBAD

  4. Diagonal AC= diagonal BD

Segment f Segment f: Segment [A, B] Segment h Segment h: Segment [A, D] Segment i Segment i: Segment [B, C] Segment j Segment j: Segment [C, E] Segment k Segment k: Segment [D, C] Vector u Vector u: Vector[B, E] Vector u Vector u: Vector[B, E] Point A A = (-2, 3.7) Point A A = (-2, 3.7) Point A A = (-2, 3.7) Point B B = (2.2, 3.7) Point B B = (2.2, 3.7) Point B B = (2.2, 3.7) Point D D = (-3.32, 0.72) Point D D = (-3.32, 0.72) Point D D = (-3.32, 0.72) Point C Point C: Point on g Point C Point C: Point on g Point C Point C: Point on g Point E E = (4.96, 3.7) Point E E = (4.96, 3.7) Point E E = (4.96, 3.7)

Trapezium of ABCD

Trapezium of ABCD, also AB∥CD and AD=BC

Solution:

Given,

  • Trapezium ABCD

  • ABCD

  • AD=BC

Construction: Draw a line through C parallel to DA intersecting AB produced at E.

  1. ABCD is given and ADEC by construction therefore,

    AECD is a parallelogram therefore,

    • CE=AD (Opposite sides of a parallelogram)

    • AD=BC (Given)

    Therefore,

    • BC=CE

    • CBE=CEB …..equation (1) (Angle of opposite to equal side of a triangle are equal)

    • B+CBE=180° ………….equation (2) (Linear pair Axiom)

    • A+CEB=180° ……….equation (3) (The sum of consecutive interior angle on the sum side of the transversal is 180° )

    From Equation (2) and (3)

    • B+CBE=A+CEB

    • But CBE=CEB

    • B=A Or A=B

  2. ABCD

    • A+D=180° (Angles on the same side of transversal)

    • B+C=180°

    • A+D=B+D

    • But ( A=B )

    • ⇒ ∠D = ∠C

  3. In ΔABCandΔBAD ,

    • AB=AB (Common)

    • DBA=CBA

    • AD=BC (Given)

    • Thus, ΔABCΔBAD by SAS congruence condition.

  4. Diagonal AC= diagonal BD by Corresponding Parts of Congruent Triangles as ΔABCΔBA.

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