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Q.1. In a parallelogram ABCD, determine the sum of angles ∠C and ∠D.
Solution:
∠C and ∠D are consecutive interior angles on the same side of the transversal CD.
∴∠C+∠D = 180^{0}
Q.2. In a parallelogram ABCD, if ∠B = 135^{0}, determine the measures of its other angles.
Solution:
Given ∠B = 135^{0}
ABCD is a parallelogram
∴∠A = ∠C,∠B =∠Dand∠A+∠B = 180^{0}
⇒∠A+135^{0 }= 180^{0}
⇒∠A = 45^{0}
⇒∠A = ∠C = 45^{0} and ∠B = ∠C = 135^{0}
Q.3. ABCD is a square. AC and BD intersect at O. State the measure of ∠AOB.
Solution:
Since, diagonals of a square bisect each other at right angle.
∴ ∠AOB = 90^{0}
Q.4. ABCD is a rectangle with ∠ABD = 40^{0}. Determine ∠DBC
Solution:
We have,
∠ABC = 90^{0}
⇒∠ABD+∠DBC = 90^{0} [∵∠ABD = 40^{0}]
⇒40^{0}+∠DBC = 90^{0}
∴∠DBC = 50^{0}
Q.5. 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
EBFD is a parallelogram.
Q.6. 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∥CQ.
Q.7. 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 ΔAEH and ΔBEF,we have
AE = BF
∠A=∠B
And AH = BE
So, by SAS congruency criterion, we have
ΔAEH≃ΔBFE
⇒∠1 = ∠2 and ∠3 = ∠4
But ∠1+∠3 = 90^{0} and ∠2+∠A = 90^{0}
⇒ ∠1+∠3+∠2+∠A = 90^{0}+90^{0}
⇒ ∠1+∠4+∠1+∠4 = 180^{0}
⇒ 2(∠1+∠4) = 180^{0}
⇒ ∠1+∠4 = 90^{0}
HEF = 90^{0}
Similarly we have ∠F = ∠G = ∠H = 90^{0}
Hence, EFGH is a Square.
Q.8. 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∠AOD = ∠COD = 90^{0}
And ∠AOB = ∠COB = 90^{0}
In ΔBDE, A and O are midpoints of BE and BD respectively.
OA∥DE
OC∥DG
In ΔCFA, B and O are midpoints of AF and AC respectively.
OB∥CF
OD∥GC
Thus, in quadrilateral DOGC, we have
OC∥DGandOD∥GC
⇒ DOCG is a parallelogram
∠DGC = ∠DOC
∠DGC = 90^{0}
Q.9. 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 ΔDCE,
∠DCE =∠DEC…..(i) [In a triangle, equal sides have equal angles]
AB∥CD [Opposite sides of the parallelogram are parallel]
∴AE∥CD [AB lies on AF]
AF∥CD and EF is the Transversal.
∠DCE = ∠BFC…..(ii) [Pair of corresponding angles]
From (i) and (ii) we get
∠DEC = ∠BFC
In ΔAFE,
∠AFE = ∠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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