Telanagana SCERT Solution Class IX (9) Math Chapter 8 Quadrilaterals Exercise 8.3
(1) The opposite angles of a parallelogram are (3x − 2)0 and (x + 48)0 . Find the measure of each angle of the parallelogram.
Solution: Given,
Opposite angles of a parallelogram are (3x – 2)o, (x + 4x)o
We know,
Opposite angles of a parallelogram are equal
∴ 3x – 2 = x + 48
Or, 3x – x = 48 + 2
Or, x = 50/2
Or, x = 25o
∴ The two angles of the parallelogram is
x + 48o
= 25o + 48o
= 73o
Now, the other two angles are also opposite to each other hence equal.
Let, the other two angles be y
We know,
Sum of angles of a parallelogram is 360o
∴ y + y + 73o + 73o = 360o
Or, 2y = 360o – 146o
Or, y = 214/2
= 107o
∴ The angles of the parallelogram are 73o, 107o, 73o, 107o
(2) Find the measure of all the angles of a parallelogram, if one angle is 24o less than the twice of the smallest angle
Solution: Given,
One angle of a parallelogram is 24o less than twice the smallest angles.
Let,
The smallest angle be x
∴ We know,
Opposite angles of a parallelogram are equal
∴ The opposite angle to x is also equal to x
Now, the other angle is 2x – 24o
∴ The 4th angle is also 2x – 24o sides opposite angles are equal
We know,
Sum of angles of a parallelogram is 360o
∴ x+ x + 2x – 24 + 2x – 24 = 360o
Or, 6x – 48 = 360o
Or, 6x = 360 + 48
Or, x = 408/6
Or, x = 68o
∴ The other angle is
2x – 24
= 2 x 68 – 24
= 136 – 24
= 112o
The angles of the parallelogram are 68o, 112o, 68o, 112o
(3) In the adjacent figure ABCD is a parallelogram and E is the midpoint of the side BC. If DE and AB are produced to meet at F, show that AF = 2AB
Solution: Given,
ABCD is a parallelogram
E is the mid point ABC
We know,
Opposite sides of a parallelogram are equal and parallel
∴ AB = DC
∴ AB ∥ DC also BE ∥ DC [∵ AF is straight line when AB is produce to F]
Now, In △DCE & △BEF
(i) ∠DCE = ∠EBF [alternate angles, BF ∥ BC]
(ii) ∠CED = ∠BEF [vertically opposite angles]
(iii) CE = BE [∵ E is the raid point of BC]
∴ DCE ≅ BEF by ASA
Now,
DC = BF [∵ Corresponding sides of congruent triangle are equal]
Or, AB = BF [∴DC = AB opposite sides of parallelogram ABCD]
∴ AF = AB + BF
Or, AF = AB + AB [∴ AB + BF]
Or, AF = 2AB hence proved
(4) In the adjacent figure ABCD is a parallelogram P and Q are the midpoints of sides AB and DC respectively. Show that PBCQ is also a parallelogram.
Solution: Given,
ABCD is a parallelogram. Q & P are mid point of DC and AB
Q & P are mid points of DC and AB respectively
Join point PQ and QB as shown in the adjacent figure
(i) AB = DC [opposite sides of parallelogram ABCD]
∴ AB ∥ DC [opposite sides of parallelogram ABCD]
∴ QC ∥ PB [∵ Q and pare the mid point of DC and AB respectively]
∴ QC = PB [∵ Q and P are the mid points of DC and AB respectively]
∴ In △PQB & △QBC
(i) ∠PBQ = ∠BQC [alternate angles, QC ∥ PB]
(ii) QC = PB [Q and P are the midpoints of DC and AB respectively]
(iii) QB is the common side
∴ △PQB ≅ △QBC by SAS
∴ ∠PQB = ∠QBC [a corresponding angles of congruent triangles are equal]
Hence we can say that PQ ∥ BC as ∠PQB = ∠QBC are alternate angles of sides PQ and BC
Now,
PQ = BC [∵ corresponding sides of congruent triangles are equal]
From the calculations on previous page we see that
∴ QC = PB & QC ∥ PB
∴ PQ = BC & PQ ∥ BC
∴ PBCQ is a parallelogram since the opposite sides of the quadrilateral are equal and parallel to each other.
(5) ABC is an isosceles triangle in which AB = AC. AD bisects exterior angle QAC and CD || BA as shown in the figure. Show that (i) ∠DAC = ∠BCA (ii) ABCD is a parallelogram
Solution: Given,
ABC is an isosceles triangle
AB = AC
AD bisects ∠QAC
CD ∥ BA
AB is produce to point Q
Now, In △ABC
∠ABC = ∠BCA [opposite angles of equal sides of a triangle are equal]
Or, ∠QAC = ∠ABC + ∠BCA [∵ ∠QAC is the exterior angles of ABC]
Or, ∠QAC = ∠BCA + ∠BCA [∵ ∠BCA = ∠ABC]
Or, ∠QAC = 2∠BCA – (i)
Now,
AD bisects QAC
∴ ∠DAC = ½ QAC
Or, ∠DAC = ½ x 2 BCA [putting the value of QAC from equation (i)]
(i) Or, ∠DAC = ∠BCA shown
(ii) Now,
∠DAC + ∠BAC = ∠A – (ii)
∠BCA + ∠DCA = ∠C – (iii)
∠DCA = ∠BAC [alternate angle, BA CD]
& ∠DAC = ∠BAC
Subtracting equation (ii) from equation (iii) we get
∠DAC + ∠BAC – ∠BCA – ∠DCA = ∠A – ∠C
Or, ∠DAC + ∠BCA – ∠DAC – ∠BAC = ∠A – ∠C
Or, 0 = ∠A – ∠C
Or, ∠A = ∠C
Now,
BA ∥ CD
∴ BQ ∥ CD [∵ BA is produced to point Q]
∴ ∠D = ∠DAQ [alternate angles] – (iv)
∠B = ∠BCA [∵ isosceles ABC where AB = AC]
Now,
∠BCA = ∠DAC [Found before]
And ∠DAC = ∠DAQ [∵ AD is the bisector of QAC]
∴ ∠BCA = ∠DAQ
∠B = ∠DAQ
Also, ∠DAQ = ∠D [from equation (iv)]
∴ ∠B = ∠D
∴ In quadrilateral ABCD opposite angles ∠A = ∠C and ∠B = ∠D therefore ABCD is a parallelogram
(6) ABCD is a parallelogram AP and CQ are perpendiculars drawn from vertices A and C on diagonal BD (see figure) show that (i) ΔAPB ≅ ΔCQD (ii) AP = CQ
Solution: Given,
ABCD is a parallelogram
BD is and diagonal
AP and CQ are ⊥ to BD
∴ ∠APB = ∠CQD = 90o
∴ In △AOB & △CQD
(i) ∠CQD = ∠APB = 90o [given]
(ii) AB = DC [opposite sides of a parallelogram are equal]
AB and DC are hypotenuse of △APB & △CQD respectively because they are the side opposite to the right angle.
(iii) ∠ABP = ∠CDQ [alternate angle, since opposite sides of parallelogram ABCD are parallel, AB ∥ DC]
(i) ∴△APB ≅ △CQD by AAS
(ii) ∴AP = CQ [∵corresponding sides of congruent triangles of equal]
(7) In Δs ABC and DEF, AB || DE; BC = EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F respectively (see figure). Show that (i) ABED is a parallelogram (ii) BCFE is a parallelogram (iii) AC = DF (iv) ΔABC ≅ ΔDEF
Solution: Given,
AB ∥ DE
AB = DE
BC ∥ EF
BC = EF
(i) ABDE is a parallelogram since opposite sides AB and DE are equal and parallel to each others.
(ii) BCEF is a parallelogram since opposite sides BC and EF are equal and parallel to each other.
(iii) AD = BE [opposite sides of parallelogram ABED are equal and parallel to each other]
AD ∥ BE
BE = CF [opposite sides of parallelogram BCFE are equal and parallel to each other]
BE ∥ CF
∴ AD = CF
AD ∥ CF [∵ both parallel and equal to BE]
∴ ACFD is a parallelogram since opposite sides AD and CF are parallel to each other and equal in length.
∴ AC = DF [opposite side of parallelogram ACED are equal] – (i)
(iv) In ABC & DEF
(i) AB = DE [given us]
(ii) BC = EF [given us]
(iii) AC = DF [from equation (i)]
∴ △ABC ≅ △DEF by SSS
(8) ABCD is a parallelogram. AC and BD are the diagonals intersect at O. P and Q are the points of tri section of the diagonal BD. Prove that CQ || AP and also AC bisects PQ (see figure)
Solution: Given,
ABCD is a parallelogram
AB & BD are diagonals intersecting at O
P & Q are point of trisection of diagonal BD
∴ BP = PQ = QD
In △APB & △CQD
(i) AB = DC [opposite sides of parallelogram ABCD]
(ii) ∠ABP = ∠CDQ [alternate angles, DC ∥ AB]
(iii) DQ = BP [∵ P and Q are bisector of BD]
△APB ≅ △CQD by SAS
∠DQC = ∠APB [∵ corresponding angles of congruent △s are equal]
∠APQ + ∠APB = 180o [collinear angles]
Or, ∠APQ = 180o – ∠APB – (i)
Now, ∠DQC + ∠CQD = 180o [collinear angles]
Or, ∠CQD = 180o – ∠DQC – (ii)
On comparing equation (i) and (ii) we get
∠APD = ∠CQD = 180o – ∠APB since ∠APB = ∠DQC
CQ ∥ AP since alternate angles APD = ∠CQD
Also CQ = AP [∵△CQD ≅ △APB, corresponding sides of congruent triangles]
CQ ∥ AP & CQ = AP
∴ Quadrilateral APCQ is a parallelogram since opposite sides CQ & AP are equal and parallel to each other.
∴ AC and PQ are the diagonals of parallelogram AP CQ
∴ AC bisects PQ since diagonals of parallelogram APCQ bisects each other (proved)
(9) ABCD is a square. E, F, G and H are the mid points of AB, BC, CD and DA respectively. Such that AE = BF = CG = DH. Prove that EFGH is a square
Solution: Given,
ABCD is a square
E, F, G, H are mid points of AB, BC, CD, DA respectively
∴ AB = BC = CD = DA
AE = EB = BF = FC = CG = GD = DH = HA
[∵ E, F, G, H are mid points of corresponding sides]
In △HAE & △EBF
(i) ∠HAE = ∠EBF = 90o [angles of square ABCD]
(ii) AE = EB [shown above]
(iii) AH = BF [shown above]
△HAE ≅ △EBF by SAS
∴ EH = EF [corresponding sides of congruent △s are equal]
Similarly
△EBF ≅ △FCG by SAS
∴ EF = FG [corresponding sides of congruent △s are equal]
Again,
△FCG ≅ △GDH by SAS
∴ FG = GH [corresponding sides of congruent △s are equal]
Similarly,
△GDH ≅ △HAF by SAS
∴ GH = HF [corresponding sides of congruent △s are equal]
∴ EF = FG = GH = HE
In △HAE
∠HAE + ∠AEH + ∠EHA = 180o [sum of angles of is 180o]
Or, 90o + ∠AEH + ∠EHA = 180o
Or, ∠AFH + ∠EHA = 90o
Now,
∠AEH = ∠EHA [∵ AF = HA]
2∠EHA = 90o
Or, ∠EHA = 45o
Similarly, In △GDH
∠DHG = 45o
Now, ∠DHG + ∠GHE + ∠EHA = 180o [supplementary angles]
Or, GHE = 180o – (45o + 45o)
= 180o – 90o
= 90o
Similarly,
∠HEF = ∠EFG = ∠FGH = 90o
∴ In quadrilateral EGFH, all sides are equal and all angles are 90o
∴ The quadrilateral EFGH is a square.
nice
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