Construction Of Parallel Lines
Question 1. With a scale and pencil let’s join two points C and Q and produce the line obtained on both sides. Thus, I got a straight-line PR.
Solution:
Now I prove logically step-by-step that PR||AB.
I joined D,Q and O,Q.
In Δ s CDQ and DOQ.
DC = OQ, CQ= DO and DQ is their common side.
∴ ΔCDQ = ΔDOQ (by s-s-s criteria of congruency)
∴ ∠CQD = ∠QDO, but they are alternate angles.
∴ CQ || DO
Therefore, PR || AB
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∴ I got straight line PR through the point C, which is parallel to the straight line AB, i.e., I draw a line PR through the point C which was not on the given line AB and parallel to the line PR AB.
Parallel Lines Exercise 22.1
Question 1. Let’s see how many line segment parallel to the line XY are possible to draw through the point Z which is not on XY.
Solution:
On straight line XY a point Q is taken. Points Z and Q are joined, Consequently ∠ZQY is formed. Now on point Z on the opposite side of ∠ZQY ∠QZP is drawn equal to angle ∠ZQY.
Joining P and Z with scale and extending on both sides AB straight line is got.
∵ ∠PZQ = ∠ZQY, are alternate angles, AB // XY
Now outside the line XY from a point Z, no other parallel line can be drawn because the alternate angle at Z will not be equal in measurement.
∴ From point Z parallel to XY only one straight line can be drawn.
“WBBSE Class 8 Maths Chapter 22 solutions, Construction of Parallel Lines”
Question 2. Habib has drawn a line segment PQ in his exercise book and he has also considered a point R outside the line segment PQ. Let’s draw a line parallel to the line segment PQ that passes through R by a scale and a compass.
Solution:
Given
Habib has drawn a line segment PQ in his exercise book and he has also considered a point R outside the line segment PQ.
On straight line PQ a point S is taken Both points. R and S are joined with scale, consequently, ∠RSQ is drawn.
Now on point R on opposite side of ∠RSQ ∠SRT is drawn equal to ∠RSQ.
∵ ∠SRT = ∠RSQ, alternate angles
∴ PQ and TU straight lines are parallel, i.e., PQ // TU
“Class 8 WBBSE Maths Chapter 22 solutions, Construction of Parallel Lines study material”
Question 3. By a scale and a compass, Megha draws an angle ∠ABC = 60° between rays BA and BC respectively. Let’s take two points P and Q. Let’s draw a straight line through point P parallel to the ray BC and also draw a straight line through the point Q parallel to the ray BA. Let D be the intersecting point. Let’s write the type, of which quadrilateral PBQD is.
Solution:
Given
By a scale and a compass, Megha draws an angle ∠ABC = 60° between rays BA and BC respectively. Let’s take two points P and Q.
Let’s draw a straight line through point P parallel to the ray BC and also draw a straight line through the point Q parallel to the ray BA. Let D be the intersecting point.
∠ABC = 60°
“WBBSE Class 8 Maths Chapter 22, Construction of Parallel Lines solved examples”
At AB on point P ∠APE is drawn equal to ∠ABC. At BC on point Q ∠FQC is drawn equal to ∠ABC. PE and QF are joined which cut each other at D.
∠ABC = ∠APE (Corresponding angles)
∴ PE//BC
∠ABC = ∠FQC (Corresponding angles)
∴ AB // FQ
Now in PBQD,
BP // QD and BQ // PD
∴ PBQD is a parallelogram.