NCERT Solutions for Class 9 Maths Chapter 8

In a parallelogram
Opposite sides are equal
Opposite angles are equal
Diagonals bisect each other.

The diagonals of following quadrilateral bisect each other:
Parallelogram
Rectangle
Square
Rhombus

Let the first angle = 3x
Therefore,
the second angle = 5x,
Third angle = 9x and
Fourth angle = 13x
Sum of all angles of a quadrilateral is 360°.
Therefore, 3x + 5x + 9x + 13x = 360°
⇒ 30x = 360°
⇒ x = (360°)/30 =12°
Hence,
The first angle = 3 × 12° =36°,
The second angle = 5 × 12° = 60°,
Third angle = 9 × 12° = 108°
The forth angle = 13 × 12° = 156°

Given: ABCD is a parallelogram with AC = BD.
To Prove: ABCD is a rectangle.
Solution: In ΔABC and ΔBAD,
BC = AD [∵ Opposite sides of a parallelogram are equal]
AC = BD [∵ Given] AB = AB [∵ Common]
Hence, ΔABC ≅ ΔBAD [∵SSS Congruency rule]
∠ABC = ∠BAD [∵ CPCT] But, ∠ABC + ∠BAD = 180° [∵ Co-interior angles]
⇒ 2∠BAD = 180° [∵ ∠ABC = ∠BAD]
⇒ ∠BAD = (180°)/2 = 90°
A parallelogram with one of its angle is 90° is a rectangle.
Hence, ABCD is a rectangle.

Given: ABCD is a quadrilateral in which AO = CO, BO = DO and ∠COD = 90°.
To prove: ABCD is a rhombus.
Solution: In ΔAOB and ΔAOD,
BO = DO [∵ Given]
∠AOB = ∠AOD [∵ Each 90°]
AO = AO [∵ Common]
Hence, ΔAOB ≅ ΔAOD [∵SAS Congruency rule]
AB = AD [∵ CPCT]
Similarly, AB = BC and BC = CD
Now, all the four sides of quadrilateral ABCD are equal.
Hence, ABCD is a rhombus.