NCERT Solutions Class 9 Chapter 6 Ex 6.2
The solutions to exercise 6.2 (from class 9th's maths NCERT book) are given below. The exercise has 6 questions which have been explained thoroughly with proper reasoning.
50+ x = 1800 (Linear Pair)
x= 1300
y= 1300 (vertically opposite angles)
The measure of x and y is equal, and that is only possible if AB//CD because only then they will be treated as alternate angles.
Given: AB//CD and CD//EF
Let y = 3a and z =7a
AB//CD and CD//EF, therefore AB//EF
x= z = 7a (Alternate angles) … (1)
AB//CD
x + y = 1800 (interior angles are supplementary)
Using (1)
7a + 3a = 1800
10a = 1800
a= 180
Therefore x = 7a => 7 x 18 => 1260
AB//CD
∠GED = ∠AGE= 1260 (alternate angles)
∠FEC = 900
∠GED can be written as:
∠GED = ∠GEF +∠FED
1260 = ∠GEF + 900
∠GEF = 360
∠AGE + ∠FGE = 1800 (Linear Pair)
1260 + ∠FGE = 1800
∠FGE = 1800 – 1260
∠FGE = 540
Extend PQ to form PU and let PQ intersect RS at O
∠SOQ = ∠TSQ (Alternate angles)
∠SOQ= 1300
∠SOQ + ∠ROQ = 1800(Linear Pair)
130 +∠ROQ = = 1800
∠ROQ = = 500
∠PQR + ∠RQO = 1800 (Linear Pair)
1100 +∠RQO = 1800
∠RQO = 700
In triangle QRO
∠RQO + ∠ROQ + ∠QRO = 1800 (Angle sum property of a triangle)
700 + 500 + ∠QRO = 1800
∠QRO = 600
∠QRO can also be written as ∠QRS. Therefore ∠QRS = 600
Alternate method
As per the hint provided by NCERT
Draw a line XY through R, such that XY //PQ, and XY//ST
∠XRQ + ∠RQP = 1800 (Interior angles are supplementary)
∠XRQ + 1100 = 1800
∠XRQ = 700
Similarly ∠YRS + ∠RST = 1800 (Interior angles are supplementary)
∠YRS + 1300 = 1800
∠YRS = 500
Now ∠XRQ + ∠YRS + ∠QRS = 1800
700 + 500 + ∠QRS = 1800
∠QRS = 600
AB//CD, therefore x= ∠QPA = 500 (Alternate angles)
∠PRQ + ∠PRD = 1800 (Linear Pair)
∠PRQ + ∠PRD = 1800
∠PRQ + 1270 = 1800
∠PRQ = 530
In triangle PQR
∠PRQ + x+ y = 1800 (Sum angle property of a triangle)
530 + 500 + y = 1800
1030 + y= 1800
y= 770
∠ABN1 = ∠N1BC (Angle of incidence equals angle of reflection)
Similarly ∠BCN2 = ∠N2 CD (Angle of incidence equals angle of reflection)
But BN1// CN2
Therefore, ∠N1BC = ∠BCN2 (Alternate angles)
Multiplying 2 on both sides
2∠N1BC = 2∠BCN2
∠N1BC+ ∠N1BC = ∠BCN2 + ∠BCN2
Using (1) & (2)
∠ABN1 + ∠N1BC = ∠BCN2 + ∠N2 CD
∠ABC = ∠B CD
But this is only possible if AB//CD (As only in that case they will be treated as alternate equal angles)
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