
Let us draw BM \(⊥\) PQ and CN \(⊥\) RS.
As PQ \(||\) RS,
Therefore, BM \(||\) CN
Thus, BM and CN are two parallel lines and a transversal line BC cut them at B and C respectively.
\(∴ \) \(∠\)2 = \(∠\)3 (Alternate interior angles)
However, \(∠\)1 = \(∠\)2 and \(∠\)3 = \(∠\)4 (By laws of reflection)
∴ \(∠\)1 = \(∠\)2 = \(∠\)3 = \(∠\)4
Also, \(∠\)1 + \(∠\)2 = \(∠\)3 + \(∠\)4
\(∠\)ABC = \(∠\)DCB
However, these are alternate interior angles.
∴ AB \(|| \)CD





Section A | Section B | ||
|---|---|---|---|
Marks | Frequency | Marks | Frequency |
0 − 10 | 3 | 0 − 10 | 5 |
10 − 20 | 9 | 10 − 20 | 19 |
20 − 30 | 17 | 20 − 30 | 15 |
30 − 40 | 12 | 30 − 40 | 10 |
40 − 50 | 9 | 40 − 50 | 1 |
Represent the marks of the students of both the sections on the same graph by two frequency polygons. From the two polygons compare the performance of the two sections.