Class 9 Mathematics
Chapter 8: Quadrilaterals
Exercise 8.1 – Stepwise Solutions
Important Properties Used
1. In a parallelogram, opposite sides are equal and parallel.
2. Diagonals of a parallelogram bisect each other.
3. If diagonals are equal → parallelogram is a rectangle.
4. If all sides are equal → rhombus.
5. If rectangle + all sides equal → square.
6. CPCT (Corresponding Parts of Congruent Triangles are Equal).
Q1. If the diagonals of a parallelogram are equal, show that it is a rectangle.
Let ABCD be a parallelogram with AC = BD.
In ΔABC and ΔBAD:
AB = BA
BC = AD (opposite sides of parallelogram)
AC = BD
Therefore,
ΔABC ≅ ΔBAD by SSS
So, ∠ABC = ∠BAD
But adjacent angles in a parallelogram are supplementary:
∠ABC + ∠BAD = 180°
Thus,
2∠ABC = 180° ⇒ ∠ABC = 90°
Hence all angles are right angles.
ABCD is a rectangle
Q2. Show that the diagonals of a square are equal and bisect each other at right angles.
Let ABCD be a square.
Since it is a parallelogram, diagonals bisect each other.
Also,
AB = BC = CD = DA
In ΔABC and ΔBAD:
AB = AD
BC = BA
AC = AC
So diagonals are equal:
AC = BD
Also, diagonals intersect at right angles (property of square).
Hence proved
Q3. Diagonal AC of a parallelogram ABCD bisects ∠A. Show that:
(i) it bisects ∠C also
(ii) ABCD is a rhombus
Given: AC bisects ∠A ⇒ ∠DAC = ∠CAB
In ΔDAC and ΔCAB:
AD = BC
AC = AC
∠DAC = ∠CAB
Therefore,
ΔDAC ≅ ΔCAB by SAS
(i) By CPCT:
∠DCA = ∠ACB ⇒ AC bisects ∠C
(ii) Also, AD = AB
But in parallelogram, opposite sides are equal:
AB = CD and AD = BC
Thus, all sides are equal.
ABCD is a rhombus
Q4. ABCD is a rectangle in which diagonal AC bisects ∠A and ∠C. Show that:
(i) ABCD is a square
(ii) diagonal BD bisects ∠B and ∠D
Given: AC bisects ∠A and ∠C
In ΔDAC and ΔCAB:
AC = AC
∠DAC = ∠CAB
AD = BC
So,
ΔDAC ≅ ΔCAB
Hence, AD = AB
Thus rectangle has all sides equal ⇒ square
(ii) In square, diagonals bisect angles
Hence BD bisects ∠B and ∠D
Q5. In parallelogram ABCD, points P and Q on BD such that DP = BQ. Show that:
(i) ΔAPD ≅ ΔCQB
AD = BC
DP = BQ
∠ADP = ∠CBQ
Therefore,
ΔAPD ≅ ΔCQB
(ii) AP = CQ (CPCT)
(iii) ΔAQB ≅ ΔCPD
AQ = CP
AB = CD
∠AQB = ∠CPD
(iv) AQ = CP
(v) APCQ is a parallelogram (both pairs of opposite sides equal)
Q6. ABCD is a parallelogram and AP and CQ are perpendiculars from A and C to BD. Show that:
(i) ΔAPB ≅ ΔCQD
AP ⟂ BD and CQ ⟂ BD
∠APB = ∠CQD = 90°
AB = CD
BP = DQ
Therefore,
ΔAPB ≅ ΔCQD by RHS
(ii) AP = CQ (CPCT)
Q7. ABCD is a trapezium in which AB ∥ CD and AD = BC. Show that:
(i) ∠A = ∠B
Since AD = BC, triangle is isosceles
(ii) ∠C = ∠D
(iii) ΔABC ≅ ΔBAD
AB = AB
AD = BC
∠A = ∠B
(iv) AC = BD (CPCT)