RD Sharma Solutions Class 9 Quadrilaterals Exercise 14.3

RD Sharma Solutions Class 9 Chapter 14 Ex 14.3

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

Solution:

$\angle C\;and\;\angle D$ 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.

Solution:

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 $\angle AOB$.

Solution:

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 $\angle DBC$

Solution:

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.

Solution:

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.

Solution:

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.

Solution:

We have,

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

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

$In\;\Delta AEH\;and\Delta BEF,\;we\;have$

AE = BF

$\angle A=\angle B$

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 $\angle 1+\angle 3=90^{0}\;and\;\angle 2+\angle A=90^{0}$

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

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

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

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

HEF=900

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.

Solution:

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.

Solution:

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.