Q1) In a parallelogram ABCD, determine the sum of angles \(\angle C\;and\;\angle D\)
Solution:
\(\angle C\;and\;\angle D\)
\(∴ \angle C+\angle D=180^{0}\)
Q2) In a parallelogram ABCD, if \(\angle B=135^{0}\)
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 \(\angle ABD=40^{0}\)
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 \(\Delta BDE\)
\(OA\parallel DE\)
\(OC\parallel DG\)
In \(\Delta CFA\)
\(OB\parallel CF\)
\(OD\parallel GC\)
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 \(\Delta DCE\)
\(\angle DCE=\angle DEC…..(i)\)
\(AB\parallel CD\)
\( ∴ AE\parallel CD\)
\(AF\parallel CD\)
\(\angle DCE=\angle BFC…..(ii)\)
From (i) and (ii) we get
\(\angle DEC=\angle BFC\)
In \(\Delta AFE\)
\(\angle AFE=\angle AEF\)
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.