# RD Sharma Solutions Class 9 Quadrilaterals Exercise 14.2

## RD Sharma Solutions Class 9 Chapter 14 Ex 14.2

Q1) Two opposite angles of a parallelogram are (3x-2)0 and (50-x)0. Find the measure of each angle of the parallelogram.

Solution:

We know that,

Opposite sides of a parallelogram are equal.

(3x-2)0 = (50-x)0

=>3x + x = 50 + 2

=>4x = 52

=>x = 130

Therefore, (3x-2)0 = (3*13-2) = 370

(50-x)0 = (50-13) = 370

Adjacent angles of a parallelogram are supplementary.

$∴ x+37=180^{0}$

$∴ x=180^{0}-37^{0}=143^{0}$

Hence, four angles are : 370, 1430, 370, 1430.

Q2) If an angle of a parallelogram is two-third of its adjacent angle, find the angles of the parallelogram.

Solution:

Let the measure of the angle be x.

Therefore, the measure of the angle adjacent is $\frac{2x}{3}$

We know that the adjacent angle of a parallelogram is supplementary.

Hence, $x+\frac{2x}{3}=180^{0}$

2x + 3x = 5400

=>5x = 5400

=>x = 1080

=>x + 1080 = 1800

=>x = 1800 – 1080 = 720

=>x = 720

Hence, four angles are 1800, 720, 1800, 720

Q3) Find the measure of all the angles of a parallelogram, if one angle is 240 less than twice the smallest angle.

Solution:

x + 2x – 24 = 1800

=>3x – 24 = 1800

=>3x = 1080 + 24

=>3x = 2040

=>x = $\frac{204}{3}=68^{0}$

=>x = 680

=>2x – 240 = 2*680 – 240 = 1120

Hence, four angles are 680, 1120, 680, 1120.

Q4) The perimeter of a parallelogram is 22cm. If the longer side measures 6.5cm what is the measure of the shorter side?

Solution:

Let the shorter side be ‘x’.

Therefore, perimeter = x + 6.5 + 6.5 + x                     [Sum of all sides]

22 = 2 ( x + 6.5 )

11 = x + 6.5

=>x = 11 – 6.5 = 4.5cm

Therefore, shorter side = 4.5cm

Q5) In a parallelogram ABCD, $\angle D=135^{0}$. Determine the measures of $\angle A\;and\;\angle B$.

Solution:

In a parallelogram ABCD

So, $\angle D+\angle C=180^{0}$

$\angle C=180^{0}-135^{0}$

$\angle C=45^{0}$

In a parallelogram opposite sides are equal.

$\angle A=\angle C=45^{0}$

$\angle B=\angle D=135^{0}$

Q6) ABCD is a parallelogram in which $\angle A=70^{0}$. Compute $\angle B,\;\angle C\;and\;\angle D$.

Solution:

In a parallelogram ABCD

$\angle A=70^{0}$

$\angle A+\angle B=180^{0}$                 [ Since, adjacent angles are supplementary ]

$70^{0}+\angle B=180^{0}$                      [ ∵ ∠ A=70°]

$\angle B=180^{0}-70^{0}$

$\angle B=110^{0}$

In a parallelogram opposite sides are equal.

$\angle A=\angle C=70^{0}$

$\angle B=\angle D=110^{0}$

Q7) In Figure 14.34, ABCD is a parallelogram in which $\angle A=60^{0}$. If the bisectors of $\angle A,\;and\;\angle B$ meet at P, prove that AD = DP, PC = BC and DC = 2AD.

Solution:

AP bisects $\angle A$

Then, $\angle DAP=\angle PAB=30^{0}$

Then, $\angle A+\angle B=180^{0}$

$\angle B+60^{0}=180^{0}$

$\angle B=180^{0}-60^{0}$

$\angle B=120^{0}$

BP bisects $\angle B$

Then, $\angle PBA=\angle PBC=30^{0}$

$\angle PAB=\angle APD=30^{0}$                     [Alternate interior angles]

Therefore, AD = DP                           [Sides opposite to equal angles are in equal length]

Similarly

$\angle PBA=\angle BPC=60^{0}$                     [Alternate interior angles]

Therefore, PC = BC

DC = DP + PC

DC = AD + BC                                  [ Since, DP = AD and PC = BC ]

DC = 2AD                              [ Since, AD = BC, opposite sides of a parallelogram are equal]

Q8) In figure 14.35, ABCD is a parallelogram in which $\angle DAB=75^{0}\;and\;\angle DBC=60^{0}$. Compute $\angle CDB,\;and\;\angle ADB$.

Solution:

To find $\angle CDB\;and\;\angle ADB$

$\angle CBD=\angle ABD=60^{0}$                    [Alternate interior angle. AD║BC and BD is the transversal]

In $\angle BDC$

$\angle CBD+\angle C+\angle CDB=180^{0}$               [Angle sum property]

$\Rightarrow 60^{0}+75^{0}+\angle CDB=180^{0}$

$\Rightarrow \angle CDB=180^{0}-(60^{0}+75^{0})$

$\Rightarrow \angle CDB=45^{0}$

Hence, $\angle CDB=45^{0},\;\angle ADB=60^{0}$

Q9) In figure 14.36, ABCD is a parallelogram and E is the mid-point of side BC. If DE and AB when produced meet at F, prove that AF = 2AB.

Solution:

In $\Delta BEF\;and\;\Delta CED$

$\angle BEF=\angle CED$                        [Verified opposite angle]

BE = CE                                                                     [Since, E is the mid-point of BC]

$\angle EBF=\angle ECD$                        [Since, Alternate interior angles are equal]

$∴ \Delta BEF\cong \Delta CED$                        [ASA congruence]

$∴ BF=CD\;\;\;\;\;\;[CPCT]$

AF = AB + AF

AF = AB + AB

AF = 2AB.

Hence proved.

Q10) Which of the following statements are true (T) and which are false (F)?

(i) In a parallelogram, the diagonals are equal.

(ii) In a parallelogram, the diagonals bisect each other.

(iii) In a parallelogram, the diagonals intersect each other at right angles.

(iv) In any quadrilateral, if a pair of opposite sides is equal, it is a parallelogram.

(v) If all the angles of a quadrilateral are equal, it is a parallelogram.

(vi) If three sides of a quadrilateral are equal, it is a parallelogram.

(vii) If three angles of a quadrilateral are equal, it is a parallelogram.

(viii) If all the sides of a quadrilateral are equal, it is a parallelogram.

Solution:

(i) False

(ii) True

(iii) False

(iv) False

(v) True

(vi) False

(vii) False

(viii) True

#### Practise This Question

Statement - 1: (pq) is equivalent to pq.
Statement - 2: (pq) is a tautology.