Click to Chat
1800-1023-196
+91-120-4616500
CART 0
Use Coupon: CART20 and get 20% off on all online Study Material
Welcome User
OR
LOGIN
Complete Your Registration (Step 2 of 2 )
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.
In a parallelogram ABCD, determine the sum of angles ∠C and ∠D.
∠C and ∠D are consecutive interior angles on the same side of the transversal CD.
∴ ∠C + ∠D = 180°
In a parallelogram ABCD, if ∠B = 135°, determine the measures of its other angles.
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°
ABCD is a square. AC and BD intersect at O. State the measure of ∠AOB.
Since, diagonals of a square bisect each other at right angle.
∴ ∠AOB = 90°
ABCD is a rectangle with ∠ABD = 40°. Determine ∠DBC
We have,
∠ABC = 90°
⇒ ∠ABD + ∠DBC = 90° [∵ ∠ABD = 40°]
⇒ 400 + ∠DBC = 90°
∴ ∠DBC = 50°
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 ∥ DC and AB = DC
⇒ EB ∥ DF and (1/2) AB = (1/2) DC
⇒ EB ∥ DF and EB = DF
EBFD is a parallelogram.
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 ∥ CQ.
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.
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.
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 ∠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°
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 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.
Signing up with Facebook allows you to connect with friends and classmates already using askIItians. It’s an easier way as well. “Relax, we won’t flood your facebook news feed!”
Post Question
Dear , Preparing for entrance exams? Register yourself for the free demo class from askiitians.
Chapter 14: Quadrilaterals Exercise – 14.2...
Chapter 14: Quadrilaterals Exercise – 14.4...
Chapter 14: Quadrilaterals Exercise – 14.1...