Chapter 14: Quadrilaterals Exercise – 14.3
Question: 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°
Question: 2
In a parallelogram ABCD, if ∠B = 135°, determine the measures of its other angles.
Solution:
Given ∠B = 135°
ABCD is a parallelogram
∴ ∠A = ∠C, ∠B = ∠D and ∠A + ∠B = 180°
⇒ ∠A + 135° = 180°
⇒ ∠A = 45°
⇒ ∠A = ∠C = 45° and ∠B = ∠C = 135°
Question: 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°
Question: 4
ABCD is a rectangle with ∠ABD = 40°. Determine ∠DBC
Solution:
We have,
∠ABC = 90°
⇒ ∠ABD + ∠DBC = 90° [∵ ∠ABD = 40°]
⇒ 400 + ∠DBC = 90°
∴ ∠DBC = 50°
Question: 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
∴ AB ∥ DC and AB = DC
⇒ EB ∥ DF and (1/2) AB = (1/2) DC
⇒ EB ∥ DF and EB = DF
EBFD is a parallelogram.
Question: 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.
Question: 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° and ∠2 + ∠A = 90°
⇒ ∠1 + ∠3 + ∠2 + ∠A = 90° + 90°
⇒ ∠1 + ∠4 + ∠1 + ∠4 = 180°
⇒ 2(∠1 + ∠4) = 180°
⇒ ∠1 + ∠4 = 90°
HEF = 90°
Similarly we have ∠F = ∠G = ∠H = 90°
Hence, EFGH is a Square.
Question: 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°
And ∠AOB = ∠COB = 90°
In ΔBDE, A and O are mid-points of BE and BD respectively.
OA ∥ DE
OC ∥ DG
In ΔCFA, B and O are mid-points of AF and AC respectively.
OB ∥ CF
OD ∥ GC
Thus, in quadrilateral DOGC, we have
OC ∥ DG and OD ∥ GC
⟹ DOCG is a parallelogram
∠DGC = ∠DOC
∠DGC = 90°
Question: 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.