RD Sharma Solutions Class 9 Quadrilaterals Exercise 14.3

RD Sharma Class 9 Solutions Chapter 14 Ex 14.3 Free Download

RD Sharma Solutions Class 9 Chapter 14 Ex 14.3

Q1) In a parallelogram ABCD, determine the sum of angles \(\angle C\;and\;\angle D\).



\(\angle C\;and\;\angle D\) are consecutive interior angles on the same side of the transversal CD.

\(∴ \angle C+\angle D=180^{0}\)

Q2) In a parallelogram ABCD, if \(\angle B=135^{0}\), determine the measures of its other angles.


Given \(\angle B=135^{0}\)

ABCD is a parallelogram

\(∴ \angle A=\angle C,\;\angle B=\angle D\;and\;\angle A+\angle B=180^{0}\)

\(\Rightarrow \angle A+135^{0}=180^{0}\)

\(\Rightarrow \angle A=45^{0}\)

\(\Rightarrow \angle A=\angle C=45^{0}\;and\;\angle B=\angle C=135^{0}\)

Q3) ABCD is a square. AC and BD intersect at O. State the measure of \(\angle AOB\).



Since, diagonals of a square bisect each other at right angle.

\(∴ \angle AOB=90^{0}\)

Q4) ABCD is a rectangle with \(\angle ABD=40^{0}\). Determine \(\angle DBC\)



We have,

\(\angle ABC=90^{0}\)

\( \Rightarrow \angle ABD+\angle DBC=90^{0}\;\;\;\;\;\;\;\;[∵ \angle ABD=40^{0}]\)

\(\Rightarrow 40^{0}+\angle DBC=90^{0}\)

\(∴ \angle DBC=50^{0}\)

Q5) The sides AB and CD of a parallelogram ABCD are bisected at E and F. Prove that EBFD is a parallelogram.



Since ABCD is a parallelogram

\(∴ AB\parallel DC\;and\;AB=DC\)

\(\Rightarrow EB\parallel DF\;and\;\;\frac{1}{2}AB=\frac{1}{2}DC\)

\(\Rightarrow EB\parallel DF\;and\;EB=DF\)

EBFD is a parallelogram.

Q6) P and Q are the points of trisection of the diagonal BD of a parallelogram ABCD. Prove that CQ is parallel to AP. Prove also that AC bisects PQ.



We know that,

Diagonals of a parallelogram bisect each other.

Therefore, OA = OC and OB = OD

Since P and Q are point of intersection of BD.

Therefore, BP = PQ = QD

Now, OB = OD are BP = QD

=>OB – BP = OD – QD

=>OP = OQ

Thus in quadrilateral APCQ, we have

OA = OC and OP = OQ

Diagonals of Quadrilateral APCQ bisect each other.

Therefore APCQ is a parallelogram.

Hence \(AP\parallel CQ\).

Q7) ABCD is a square. E, F, G and H are points on AB, BC, CD and DA respectively, such that AE = BF = CG = DH. Prove that EFGH is a square.



We have,

AE = BF = CG = DH = x (say)

BE = CF = DG = AH = y (say)

\(In\;\Delta AEH\;and\Delta BEF,\;we\;have\)


\(\angle A=\angle B\)

And AH = BE

So, by SAS congruency criterion, we have

\(\Delta AEH\simeq \Delta BFE\)

\(\Rightarrow \angle 1=\angle 2\;and\;\angle 3=\angle 4\)

But \(\angle 1+\angle 3=90^{0}\;and\;\angle 2+\angle A=90^{0}\)

\(\Rightarrow \angle 1+\angle 3+\angle 2+\angle A=90^{0}+90^{0}\)

\(\Rightarrow \angle 1+\angle 4+\angle 1+\angle 4=180^{0}\)

\(\Rightarrow 2(\angle 1+\angle 4)=180^{0}\)

\(\Rightarrow \angle 1+\angle 4=90^{0}\)


Similarly we have \(\angle F=\angle G=\angle H=90^{0}\)

Hence, EFGH is a Square.


Q8) ABCD is a rhombus, EAFB is a straight line such that EA = AB = BF. Prove that ED and FC when produced meet at right angles.



We know that the diagonals of a rhombus are perpendicular bisector of each other.

\(∴ OA=OC,\;OB=OD,\;and\;\angle AOD=\angle COD=90^{0}\)

And \(\angle AOB=\angle COB=90^{0}\)

In \(\Delta BDE\), A and O are mid-points of BE and BD respectively.

\(OA\parallel DE\)

\(OC\parallel DG\)

In \(\Delta CFA\), B and O are mid-points of AF and AC respectively.

\(OB\parallel CF\)

\(OD\parallel GC\)

Thus, in quadrilateral DOGC, we have

\(OC\parallel DG\;and\;OD\parallel GC\)

=>DOCG is a parallelogram

\(\angle DGC=\angle DOC\)

\(\angle DGC=90^{0}\)

Q9) ABCD is a parallelogram, AD is produced to E so that DE = DC and EC produced meets AB produced in F. Prove that BF = BC.


Draw a parallelogram ABCD with AC and BD intersecting at O.

Produce AD to E such that DE = DC

Join EC and produce it to meet AB produced at F.


In \(\Delta DCE\),

\(\angle DCE=\angle DEC…..(i)\)              [In a triangle, equal sides have equal angles]

\(AB\parallel CD\)                   [Opposite sides of the parallelogram are parallel]

\( ∴ AE\parallel CD\)                      [AB lies on AF]

\(AF\parallel CD\) and EF is the Transversal.

\(\angle DCE=\angle BFC…..(ii)\)              [Pair of corresponding angles]

From (i) and (ii) we get

\(\angle DEC=\angle BFC\)

In \(\Delta AFE\),

\(\angle AFE=\angle AEF\)                         [∠ DEC = ∠ BFC]

Therefore, AE = AF                [In a triangle, equal angles have equal sides opposite to them]

=>AD + DE = AB + BF

=>BC + AB = AB + BF                     [Since, AD = BC, DE = CD and CD = AB, AB = DE]

=> BC = BF

Hence proved.

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