Exercise 8.1 Practice
Properties of Quadrilaterals and Parallelograms
Q1: Angle Ratios
00:00
The angles of a quadrilateral are in the ratio $3 : 5 : 9 : 13$. Find all the angles of the quadrilateral.
Let the angles be $3x, 5x, 9x$ and $13x$.
Sum of angles of a quadrilateral is $360^\circ$.
$3x + 5x + 9x + 13x = 360^\circ$
$30x = 360^\circ$
$x = 12^\circ$.
$3x + 5x + 9x + 13x = 360^\circ$
$30x = 360^\circ$
$x = 12^\circ$.
Angles are:
$3x = 3(12) = 36^\circ$
$5x = 5(12) = 60^\circ$
$9x = 9(12) = 108^\circ$
$13x = 13(12) = 156^\circ$.
$3x = 3(12) = 36^\circ$
$5x = 5(12) = 60^\circ$
$9x = 9(12) = 108^\circ$
$13x = 13(12) = 156^\circ$.
Angles: $36^\circ, 60^\circ, 108^\circ, 156^\circ$
Q2: Parallelogram to Rectangle
00:00
If the diagonals of a parallelogram are equal, then show that it is a rectangle.
Let $ABCD$ be a parallelogram where $AC = BD$.
In $\triangle ABC$ and $\triangle DCB$:
1. $AB = DC$ (Opposite sides of parallelogram)
2. $BC = CB$ (Common)
3. $AC = DB$ (Given)
$\therefore \triangle ABC \cong \triangle DCB$ (SSS Rule).
1. $AB = DC$ (Opposite sides of parallelogram)
2. $BC = CB$ (Common)
3. $AC = DB$ (Given)
$\therefore \triangle ABC \cong \triangle DCB$ (SSS Rule).
By CPCT, $\angle ABC = \angle DCB$.
Since $AB \parallel DC$, consecutive interior angles sum to $180^\circ$:
$\angle ABC + \angle DCB = 180^\circ$
$2\angle ABC = 180^\circ \Rightarrow \angle ABC = 90^\circ$.
Since $AB \parallel DC$, consecutive interior angles sum to $180^\circ$:
$\angle ABC + \angle DCB = 180^\circ$
$2\angle ABC = 180^\circ \Rightarrow \angle ABC = 90^\circ$.
A parallelogram with one angle $90^\circ$ is a rectangle.
Proved.
Q3: Diagonals at 90°
00:00
Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.
Given: Diagonals bisect each other at $90^\circ$.
Since diagonals bisect each other, it is a parallelogram.
Since diagonals bisect each other, it is a parallelogram.
In $\triangle AOD$ and $\triangle COD$:
1. $OA = OC$ (Diagonals bisect)
2. $\angle AOD = \angle COD = 90^\circ$ (Given)
3. $OD = OD$ (Common)
$\therefore \triangle AOD \cong \triangle COD$ (SAS Rule).
1. $OA = OC$ (Diagonals bisect)
2. $\angle AOD = \angle COD = 90^\circ$ (Given)
3. $OD = OD$ (Common)
$\therefore \triangle AOD \cong \triangle COD$ (SAS Rule).
By CPCT, $AD = CD$.
A parallelogram with adjacent sides equal is a rhombus.
A parallelogram with adjacent sides equal is a rhombus.
Proved.
Q6: Diagonal Bisects Angles
00:00
Diagonal $AC$ of a parallelogram $ABCD$ bisects $\angle A$. Show that:
(i) it bisects $\angle C$ also,
(ii) $ABCD$ is a rhombus.
(i) it bisects $\angle C$ also,
(ii) $ABCD$ is a rhombus.
(i) Bisects $\angle C$:
Since $AB \parallel DC$, $\angle DAC = \angle BCA$ (Alt. Interior Angles).
Since $AD \parallel BC$, $\angle BAC = \angle DCA$ (Alt. Interior Angles).
Given $\angle DAC = \angle BAC$ (AC bisects A).
$\therefore \angle BCA = \angle DCA$. Thus, AC bisects $\angle C$.
Since $AB \parallel DC$, $\angle DAC = \angle BCA$ (Alt. Interior Angles).
Since $AD \parallel BC$, $\angle BAC = \angle DCA$ (Alt. Interior Angles).
Given $\angle DAC = \angle BAC$ (AC bisects A).
$\therefore \angle BCA = \angle DCA$. Thus, AC bisects $\angle C$.
(ii) ABCD is a rhombus:
From above, $\angle DAC = \angle DCA$.
In $\triangle ADC$, sides opposite to equal angles are equal.
$\therefore AD = CD$.
Since adjacent sides are equal in a parallelogram, it is a rhombus.
From above, $\angle DAC = \angle DCA$.
In $\triangle ADC$, sides opposite to equal angles are equal.
$\therefore AD = CD$.
Since adjacent sides are equal in a parallelogram, it is a rhombus.
Proved.
Q9: Points on Diagonal
00:00
In parallelogram $ABCD$, two points $P$ and $Q$ are taken on diagonal $BD$ such that $DP = BQ$. Show that:
(i) $\triangle APD \cong \triangle CQB$
(ii) $AP = CQ$
(iii) $\triangle AQB \cong \triangle CPD$
(iv) $AQ = CP$
(v) $APCQ$ is a parallelogram
(i) $\triangle APD \cong \triangle CQB$
(ii) $AP = CQ$
(iii) $\triangle AQB \cong \triangle CPD$
(iv) $AQ = CP$
(v) $APCQ$ is a parallelogram
(i) $\triangle APD \cong \triangle CQB$:
1. $AD = CB$ (Opposite sides of ||gm)
2. $\angle ADP = \angle CBQ$ (Alt. Interior Angles, $AD \parallel BC$)
3. $DP = BQ$ (Given)
$\therefore \triangle APD \cong \triangle CQB$ (SAS Rule).
1. $AD = CB$ (Opposite sides of ||gm)
2. $\angle ADP = \angle CBQ$ (Alt. Interior Angles, $AD \parallel BC$)
3. $DP = BQ$ (Given)
$\therefore \triangle APD \cong \triangle CQB$ (SAS Rule).
(ii) $AP = CQ$:
By CPCT from (i).
By CPCT from (i).
(iii) $\triangle AQB \cong \triangle CPD$:
1. $AB = CD$ (Opposite sides)
2. $\angle ABQ = \angle CDP$ (Alt. Interior Angles)
3. $BQ = DP$ (Given)
$\therefore \triangle AQB \cong \triangle CPD$ (SAS Rule).
1. $AB = CD$ (Opposite sides)
2. $\angle ABQ = \angle CDP$ (Alt. Interior Angles)
3. $BQ = DP$ (Given)
$\therefore \triangle AQB \cong \triangle CPD$ (SAS Rule).
(iv) $AQ = CP$:
By CPCT from (iii).
By CPCT from (iii).
(v) APCQ is a parallelogram:
In quadrilateral $APCQ$, opposite sides are equal ($AP = CQ$ and $AQ = CP$).
Therefore, $APCQ$ is a parallelogram.
In quadrilateral $APCQ$, opposite sides are equal ($AP = CQ$ and $AQ = CP$).
Therefore, $APCQ$ is a parallelogram.
Proved all parts.
Q12: Isosceles Trapezium
00:00
$ABCD$ is a trapezium in which $AB \parallel CD$ and $AD = BC$. Show that:
(i) $\angle A = \angle B$
(ii) $\angle C = \angle D$
(iii) $\triangle ABC \cong \triangle BAD$
(iv) diagonal $AC =$ diagonal $BD$
(i) $\angle A = \angle B$
(ii) $\angle C = \angle D$
(iii) $\triangle ABC \cong \triangle BAD$
(iv) diagonal $AC =$ diagonal $BD$
Extend $AB$ and draw a line through $C$ parallel to $DA$ intersecting $AB$ produced at $E$.
Now $ADCE$ is a parallelogram ($AD \parallel CE$ and $AE \parallel DC$).
$\therefore AD = CE$. But $AD = BC$ (Given).
So, $BC = CE$.
Now $ADCE$ is a parallelogram ($AD \parallel CE$ and $AE \parallel DC$).
$\therefore AD = CE$. But $AD = BC$ (Given).
So, $BC = CE$.
(i) $\angle A = \angle B$:
In $\triangle BCE$, $BC = CE \Rightarrow \angle CBE = \angle CEB$.
$\angle A + \angle CEB = 180^\circ$ (Consecutive interior angles, $AD \parallel CE$).
$\angle B + \angle CBE = 180^\circ$ (Linear Pair).
$\therefore \angle A = \angle B$.
In $\triangle BCE$, $BC = CE \Rightarrow \angle CBE = \angle CEB$.
$\angle A + \angle CEB = 180^\circ$ (Consecutive interior angles, $AD \parallel CE$).
$\angle B + \angle CBE = 180^\circ$ (Linear Pair).
$\therefore \angle A = \angle B$.
(ii) $\angle C = \angle D$:
$\angle A + \angle D = 180^\circ$ and $\angle B + \angle C = 180^\circ$ (Interior angles).
Since $\angle A = \angle B$, then $\angle C = \angle D$.
$\angle A + \angle D = 180^\circ$ and $\angle B + \angle C = 180^\circ$ (Interior angles).
Since $\angle A = \angle B$, then $\angle C = \angle D$.
(iii) $\triangle ABC \cong \triangle BAD$:
1. $AB = BA$ (Common)
2. $\angle B = \angle A$ (Proved)
3. $BC = AD$ (Given)
$\therefore \triangle ABC \cong \triangle BAD$ (SAS Rule).
1. $AB = BA$ (Common)
2. $\angle B = \angle A$ (Proved)
3. $BC = AD$ (Given)
$\therefore \triangle ABC \cong \triangle BAD$ (SAS Rule).
(iv) Diagonal AC = Diagonal BD:
By CPCT from (iii).
By CPCT from (iii).
Proved all parts.