Chapter 8: Quadrilaterals (Exercise 8.1)

Chapter 8: Quadrilaterals (Exercise 8.1)

Question 1:
The angles of a quadrilateral are in the ratio 3 : 5 : 9 : 13. Find all the angles of the quadrilateral.

Solution:
* Let the angles of the quadrilateral be 3x, 5x, 9x, and 13x.
* Sum of all interior angles of a quadrilateral = 360°:
3x + 5x + 9x + 13x = 360°
30x = 360°
x = 12°

* Calculating all four angles:
1. 1^st Angle = 3 × 12° = 36°
2. 2^nd Angle = 5 × 12° = 60°
3. 3^rd Angle = 9 × 12° = 108°
4. 4^th Angle = 13 × 12° = 156°

Question 2:
If the diagonals of a parallelogram are equal, then show that it is a rectangle.

Solution:
* Let ABCD be a parallelogram where diagonals AC = BD.
* In ΔABC and ΔBAD:
1. AB = BA (Common side)
2. BC = AD (Opposite sides of a parallelogram)
3. AC = BD (Given equal diagonals)

* By SSS Congruence Rule:
ΔABC ≅ ΔBAD

* By CPCT:
∠ABC = ∠BAD --- (Equation 1)

* Since AD ∥ BC, adjacent interior angles sum to 180°:
∠ABC + ∠BAD = 180°
∠ABC + ∠ABC = 180° ⇒ 2 ∠ABC = 180° ⇒ ∠ABC = 90°

* Since one angle of a parallelogram is 90°, ABCD is a rectangle.
(Hence proved.)

Question 3:
Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.

Solution:
* Let ABCD be a quadrilateral where diagonals AC and BD bisect each other at right angles at O.
* In ΔAOB and ΔAOD:
1. AO = AO (Common side)
2. ∠AOB = ∠AOD = 90° (Given perpendicular diagonals)
3. OB = OD (Given bisecting diagonals)

* By SAS Congruence Rule:
ΔAOB ≅ ΔAOD

* By CPCT:
AB = AD

* Similarly, we can prove AB = BC and BC = CD.
* Therefore, AB = BC = CD = DA.
* Since all four sides are equal, ABCD is a rhombus.
(Hence proved.)

Question 4:
Show that the diagonals of a square are equal and bisect each other at right angles.

Solution:
Let ABCD be a square.

1. Diagonals are equal (AC = BD):
In ΔABC and ΔBAD:
* AB = BA (Common side)
* ∠ABC = ∠BAD = 90°
* BC = AD (Sides of a square)
By SAS rule, ΔABC ≅ ΔBAD ⇒ AC = BD.

2. Diagonals bisect each other (AO = OC, BO = OD):
In ΔAOB and ΔCOD:
* ∠OAB = ∠OCD (Alternate interior angles)
* AB = CD (Sides of a square)
* ∠OBA = ∠ODC (Alternate interior angles)
By ASA rule, ΔAOB ≅ ΔCOD ⇒ AO = OC and BO = OD.

3. Diagonals bisect at right angles (∠AOB = 90°):
In ΔAOB and ΔAOD:
* AO = AO (Common side)
* OB = OD (Proved above)
* AB = AD (Sides of a square)
By SSS rule, ΔAOB ≅ ΔAOD ⇒ ∠AOB = ∠AOD.
Since ∠AOB + ∠AOD = 180° (Linear pair), ∠AOB = 90°.
(Hence proved.)