Maths NCERT Solutions for Class 9 Chapter 6 Exercise 6.3

Class 8 Maths Chapter 6 Exercise 6.3


The solutions to exercise 6.3 (from class 9th's maths NCERT book) are given below. The exercise has 6 questions which have been explained thoroughly with proper reasoning. 


∠PQT + ∠PQR = 1800 (Linear Pair)

1100 + ∠PQR = 1800 

∠PQR = 1800 – 1100

∠PQR = 700 … (1)


∠SPR = ∠PQR + ∠PRQ (Exterior angle is equal to the sum of the opposite interior angles)

1350 = 700 + ∠PRQ       [Using (1)]

650 = ∠PRQ       




 In triangle XYZ

∠X + ∠Y +∠Z = 1800

620 + 540 + ∠Z = 1800

∠Z = 1800 - 1160

∠Z = 1800 - 1160

∠Z = 640 … (1)


Since YO and ZO are bisectors of ∠XYZ and ∠XZY

∠XYO = ∠OYZ = 54/2 = 270

∠XZO = ∠OZY =>  64/2 = 320 (Using (1))

In triangle OYZ

∠OZY +∠OYZ +∠YOZ = 1800 (Angle sum property of a triangle)

270 +320 +∠YOZ = 1800

∠YOZ = 1800 – 590

∠YOZ = 1210

AB//DE  and AE is the transversal

∠BAE = ∠CED (Alternate angles)

∠CED = 350

In triangle DCE 

∠DCE + ∠CDE +∠CED = 1800 (Angle sum property of a triangle)

∠DCE + 530 + 350 = 1800

∠DCE = 1800 - 880

∠DCE = 920

In triangle PRT

∠TPR + ∠PRT + ∠RTP = 1800

950 + 400 +∠RTP = 1800

∠RTP = 1800 - 1350

∠RTP = 450


∠RTP = 450 = ∠STQ (Vertically opposite angles)

In triangle STQ

∠STQ + ∠TQS + ∠QST = 1800 (Angle sum property of a triangle)

450 + ∠TQS + 750 = 1800

∠TQS + 1200 = 1800

∠TQS = 1800 - 1200

∠TQS = 600


In triangle QSR

∠QSR + ∠RQS = ∠QRT (Exterior angle is equal to the sum of opposite interior angles)

∠QSR + 280 = 650 

∠QSR = 370

PQ//SR and QS is the transversal

x= ∠QSR = 370 (Alternate angles)

PS is not // to QR, therefore the same can be done in the case of Y, so we will use the angle sum property of a triangle.

In triangle PQS

∠P + x + y = 1800 (Angle sum property of a triangle)

90 + 370 + y = 1800

y= 1800 – 1270

y= 530


In this question we will use the exterior angles

Let ∠PRT = ∠TRS = x

∠PQT = ∠TQR = y


In triangle PQR

∠PRS = ∠P + ∠PQT (Exterior angle is equal to the sum of opposite interior angles)

2x =∠P + 2y … (1)


In triangle QTR

∠TRS = ∠T+ ∠TQR (Exterior angle is equal to the sum of opposite interior angles)

x= ∠T+ y … (2)


Using (2) in (1)

2(∠T+ y) =∠P + 2y

2∠T+ 2y =∠P + 2y

2∠T=∠P + 2y - 2y

2∠T=∠P 

∠T= 1/2 ∠P  

Or

 ∠QTR = 1/2 ∠QPR