RD Sharma Solutions Class 9 Quadrilaterals Exercise 14.2

RD Sharma Solutions Class 9 Chapter 14 Exercise 14.2

RD Sharma Class 9 Solutions Chapter 14 Ex 14.2 Free Download

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

Adjacent angles are supplementary

=>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

Adjacent angles are supplementary

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.

1

 Solution:

AP bisects \(\angle A\)

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

Adjacent angles are supplementary

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 \(\angle CDB,\;and\;\angle ADB\).

2
2

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.

3

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

Preeti told Pavani that if the alternate angles are equal then lines need not be parallel. Is Preeti's statement true or false?