# Telangana SCERT Class 9 Math Solution Chapter 8 Quadrilaterals Exercise 8.3

## 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

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,

∴ ∠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]

BE = CF [opposite sides of parallelogram BCFE are equal and parallel to each other]

BE ∥ 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.

Updated: October 21, 2020 — 1:55 pm