NCERT Class 9 Solutions: Quadrilaterals (Chapter 8) Exercise 8.1 – Part 4

Chart of Quadrilaterals

Chart of Quadrilaterals of various types

Chart of Quadrilaterals

Chart of Quadrilaterals of various types

Q-7 ABCD is a rhombus. Show that diagonal AC bisects A as well as C and

Diagonal BD bisects B as well as D .

Solution:

Quadrilateral poly1 Quadrilateral poly1: Polygon D, C, B, A Angle ? Angle ?: Angle between C, A, D Angle ? Angle ?: Angle between C, A, D Angle ? Angle ?: Angle between C, A, D Angle ? Angle ?: Angle between A, C, B Angle ? Angle ?: Angle between A, C, B Angle ? Angle ?: Angle between A, C, B Angle ? Angle ?: Angle between D, C, A Angle ? Angle ?: Angle between D, C, A Angle ? Angle ?: Angle between D, C, A Angle ? Angle ?: Angle between B, A, C Angle ? Angle ?: Angle between B, A, C Angle ? Angle ?: Angle between B, A, C Segment ?_1 Segment ?_1: 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 [D, B] Segment g Segment g: Segment [C, A] Point D D = (2.78, 2.98) Point D D = (2.78, 2.98) Point D D = (2.78, 2.98) Point C C = (7.44, 2.98) Point C C = (7.44, 2.98) Point C C = (7.44, 2.98) Point B B = (6, -0.28) Point B B = (6, -0.28) Point B B = (6, -0.28) Point A A = (1.16, -0.28) Point A A = (1.16, -0.28) Point A A = (1.16, -0.28)

ABCD Is Rhombus

ABCD is rhombus AC bisects ∠ A as well as ∠C

Given,

  • ABCD is a rhombus

  • AC and BD are its diagonals.

Proof,

  • AD=CD (Sides of a rhombus are equal)

  • DAC=DCA ………..equation (1) (Angles opposite of equal sides of a triangle are equal.)

  • α=γ

  • Also, ABCD , therefore, BAC=ACD ……..equation (2) (Alternate interior angles)

From equation (1) and (2),

DAC=BAC

α=β

Therefore, AC bisects A .

Similarly,

  • AB=BC (Sides of a rhombus are equal)

  • BAC=ACB …….equation (3) (Angles opposite of equal sides of a triangle are equal.)

  • β=δ

Also, ADBC and so BAC=ACD …..equation (4) (Alternate interior angles)

From Equation (3) and (4) ACD=ACB

Therefore, AC bisects C .

Similarly, diagonal BD bisects B as well as D .

Q-8 ABCD is a rectangle in which diagonal AC bisects A as well as C . Show that:

  1. ABCD is a square

  2. Diagonal BD bisects ∠B as well as ∠D.

Solution:

Quadrilateral poly1 Quadrilateral poly1: Polygon D, C, B, A Angle ? Angle ?: Angle between A, B, C Angle ? Angle ?: Angle between A, B, C Angle ? Angle ?: Angle between A, B, C Angle ? Angle ?: Angle between B, C, D Angle ? Angle ?: Angle between B, C, D Angle ? Angle ?: Angle between B, C, D Angle ? Angle ?: Angle between C, D, A Angle ? Angle ?: Angle between C, D, A Angle ? Angle ?: Angle between C, D, A Angle ? Angle ?: Angle between D, A, B Angle ? Angle ?: Angle between D, A, B Angle ? Angle ?: Angle between D, A, B Segment a Segment a: Segment [D, C] of Quadrilateral poly1 Segment b Segment b: Segment [C, B] of Quadrilateral poly1 Segment d Segment d: Segment [B, A] of Quadrilateral poly1 Segment c_1 Segment c_1: Segment [A, D] of Quadrilateral poly1 Segment j Segment j: Segment [A, C] Segment k Segment k: Segment [B, D] Point D D = (-1.54, 1.74) Point D D = (-1.54, 1.74) Point D D = (-1.54, 1.74) Point C C = (3.34, 1.74) Point C C = (3.34, 1.74) Point C C = (3.34, 1.74) Point A Point A: Intersection point of c, g Point A Point A: Intersection point of c, g Point A Point A: Intersection point of c, g Point B Point B: Intersection point of h, i Point B Point B: Intersection point of h, i Point B Point B: Intersection point of h, i

ABCD Is a Rectangle

ABCD is a rectangle its diagonal AC also bisects two angles ∠A as well as ∠C

  • Given ABCD is a rectangle, AC is bisects, A as well as C

Solution (i):

  • DAC=DCA (AC bisects A as well as C. )

  • AD=CD (Sides opposite to equal angles of a triangle are equal)

  • Also, CD=AB (Opposite sides of a rectangle)

  • Therefore, AB=BC=CD=AD

  • Thus, ABCD is a square.

Solution (ii):

  • BC=CD

  • CDB=CBD (Angles opposite to equal sides are equal)

  • Also, CDB=ABD (Alternate interior angles) therefore, CBD=ABD

  • Thus, BD bisects B

  • ADBC

Now,

  • CBD=ADB

  • CDB=ADB

Thus, BD bisects D

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