NCERT Class 9 Solutions: Quadrilaterals (Chapter 8) Exercise 8.1 – Part 3

Co-interior Angles

Give co-interior angles,also a+b=180

Give Co-Interior Angles

Give co-interior angles,also a+b=180

When two lines are cut by a third line (transversal) co-interior angles are between the pair of lines on the same side of the transversal. If the lines are parallel the co-interior angles are supplementary (add up to 180 degrees).

Q-5 Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.

Solution:

Quadrilateral poly1 Quadrilateral poly1: Polygon A, B, C, D Segment a Segment a: Segment [A, B] of Quadrilateral poly1 Segment b Segment b: Segment [B, C] of Quadrilateral poly1 Segment c Segment c: Segment [C, D] of Quadrilateral poly1 Segment d Segment d: Segment [D, A] of Quadrilateral poly1 Segment f Segment f: Segment [A, C] Segment g Segment g: Segment [B, D] Point A A = (-1.88, 3.64) Point A A = (-1.88, 3.64) Point A A = (-1.88, 3.64) Point B B = (1.88, 3.64) Point B B = (1.88, 3.64) Point B B = (1.88, 3.64) Point C C = (1.88, 0.48) Point C C = (1.88, 0.48) Point C C = (1.88, 0.48) Point D D = (-1.88, 0.48) Point D D = (-1.88, 0.48) Point D D = (-1.88, 0.48) Point O Point O: Intersection point of f, g Point O Point O: Intersection point of f, g Point O Point O: Intersection point of f, g

ABCD Is a Quadrilaterl

ABCD is a quadrilaterl diagonal AC and BD bisect each other right angle at O.

  • Given, ABCD is a quadrilateral in which diagonals AC and BD bisect each other at right angle at O.

  • To prove, Quadrilateral ABCD is a square.

Proof,

In Equation ,

  • Equation (Diagonals bisect each other)

  • Equation (Vertically opposite)

  • Equation (Diagonals bisect each other)

Therefore,

  • Equation by Side-Angle-Side congruence condition.

  • Thus, Equation by Corresponding Parts of Congruent Triangle…….equation(1)

    Also,

  • Equation (Alternate interior angles)

  • Equation

Now, In Equation ,

  • Equation (Diagonals bisect each other)

  • Equation (Vertically opposite)

  • Equation (Common)

Therefore, Equation by Side-Angle-Side congruence condition

Thus, Equation by Corresponding Parts of Congruent Triangle………equation(2)

Also,

  • Equation and Equation

  • Equation (From equation (1))

  • Equation by Corresponding Parts of Congruent Triangle

  • Equation (co-interior angles)

  • Equation ( Equation )

  • Equation ……..equation(3)

  • That is, one of the interior angle is right angle.

Thus, from (1), (2) and (3) given quadrilateral ABCD is a square.

Q-6 Diagonal AC of a parallelogram ABCD bisects Equation (see Fig.). Show that

  1. it bisects Equation also,

  2. ABCD is a rhombus.

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 [C, A] Point D D = (-1.58, 3.74) Point D D = (-1.58, 3.74) Point D D = (-1.58, 3.74) Point C C = (2.36, 3.74) Point C C = (2.36, 3.74) Point C C = (2.36, 3.74) Point B B = (1.26, 1.42) Point B B = (1.26, 1.42) Point B B = (1.26, 1.42) Point A A = (-2.84, 1.46) Point A A = (-2.84, 1.46) Point A A = (-2.84, 1.46)

Parallelogram ABCD With Diagonal Bisecting Angle

Parallelogram ABCD with diagonal AC which bisects ∠A

Solution:

Given, Parallelogram ABCD and its diagonal AC, it’s bisects Equation

In Equation ,

  • Equation (parallel sides are equal in a parallelogram)

  • Equation (parallel sides are equal in a parallelogram)

  • Equation (Common side)

Therefore, Equation by SSS congruence condition.

Thus,

  • Equation by Corresponding Parts of Congruent Triangle

  • And Equation (AC id diagonal)

  • Equation (AC id diagonal)

Therefore, AC bisects Equation also.

Since, Equation (Proved)

  • Equation (Opposite sides of equal angles of a triangle are equal)

  • Also, Equation (parallel sides are equal in a parallelogram)

Thus, ABCD is a rhombus.

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