# Class 9 RD Sharma Solutions – Chapter 8 Introduction to Lines and Angles- Exercise 8.4 | Set 2

**Question 11. In the given figure, lines AB and CD are parallel and P is any point as shown in the figure. Show that âˆ ABP+ âˆ CDP= âˆ DPB.**

**Solution:**

Here in the given figure:

Given:

AB || CD

Now draw a line XY passing through point P and parallel to AB and CD.

Here,XY || CD, thus, âˆ CDP and âˆ 1are alternate interior opposite angles. Therefore,

âˆ 1 = âˆ CDP ……(i)

Similarly, we have XY || AB, thus, âˆ ABP andâˆ 2are alternate interior opposite angles. Therefore,

âˆ 2 = âˆ ABP ….(ii)

On adding (i) and (ii)

âˆ 1 + âˆ 2 = âˆ CDP + âˆ ABP

âˆ DPB = âˆ CDP + âˆ ABP

Thus proved.

**Question 12. In the given figure, AB||CD and P is any point shown in the figure. Prove that:**

**âˆ ABP+ âˆ BPD+ âˆ CDP= 360Â°**

**Solution:**

The given figure is as follows:

It is given that AB || CD

Let us draw a line XY passing through point P and parallel to AB and CD.

We have XY || CD, thus, âˆ CDP andâˆ 2are consecutive interior angles. Therefore,

âˆ 2 + âˆ CDP = 180Â° ……(i)

Similarly, we have XY || AB, thus,âˆ CDP andâˆ 2 are consecutive interior angles. Therefore,

âˆ 1 + âˆ ABP = 180Â° ……(ii)

On adding equation (i) and (ii), we get:

âˆ 2 + âˆ CDP + âˆ 1 + âˆ ABP = 180Â° + 180Â°

(âˆ 2 + âˆ 1) + âˆ CDP + âˆ ABP = 360Â°

âˆ ABP + âˆ BPD + âˆ CDP = 360Â°

Hence proved.

**Question 13. **Two unequal angles of a parallelogram are in the ratio 2 : 3. Find all its angles in degrees.

**Solution:**

The parallelogram can be drawn as follows:

It is given that

âˆ A : âˆ C = 2 : 3

Therefore, let:

âˆ A = 2x and âˆ C = 3x

We know that opposite angles of a parallelogram are equal.

Therefore,

âˆ A = âˆ D

âˆ D = 2x

Similarly

âˆ B = 3x

Also, if AB || CD, then sum of consecutive interior angles is equal to180Â°. .

Therefore,

âˆ A + âˆ C = 180Â°

2x + 3x = 180Â°

5x = 180Â°

x =

x = 36Â°

We have

âˆ A = 2x

âˆ A = 2(36Â°)

âˆ A = 72Â°

Also,

âˆ C = 3x

âˆ C = 3(36Â°)

âˆ C = 108Â°

Similarly,

âˆ D = 72Â°

And

âˆ B = 108Â°

Hence, the four angles of the parallelogram are as follows:

âˆ A = 72Â°, âˆ B = 108Â°, âˆ C = 72Â° and âˆ D = 108Â°

**Question 14. In each of the two lines is perpendicular to the same line, what kind of lines are they to each other?**

**Solution:**

The figure can be drawn as follows:

Here, l âŠ¥ n and m âŠ¥n.

We need to find the relation between lines l and m

It is given that l âŠ¥n, therefore,

âˆ 1 = 90Â° …….(i)

Similarly, we have m âŠ¥n, therefore,

âˆ 2 = 90Â° …….(ii)

From (i) and (ii), we get:

âˆ 1 = âˆ 2

But these are the pair of corresponding angles.

Theorem states: If a transversal intersects two lines in such a way that a pair of corresponding angles is equal, then the two lines are parallel.

Thus, we can say that l || m

Hence, the lines are parallel to each other.

**Question 15. **In the given figure, âˆ 1 = 60Â° and âˆ 2 =(23)rd23rdof a right angle. Prove that l||m.

**Solution:**

The figure is given as follows:

It is given that âˆ 1 = 60Â°

Also,

âˆ 2 =(90Â°)

âˆ 2 = 2(30Â°)

âˆ 2 = 60Â°

Thus, we have âˆ 1 = âˆ 2

But these are the pair of corresponding angles.

Thus, l || m

Hence, proved.

**Question 16. **In the given figure, if l||m||n and âˆ 1 = 60Â°, find âˆ 2.

**Solution:**

The given figure is as follows:

We have l || m || n and âˆ 60Â°

Thus, we get âˆ 1 and âˆ 3 as corresponding angles.

Therefore,

âˆ 3 = âˆ 1

âˆ 3 = 60Â° ……….(i)

We have âˆ 3 and âˆ 4 forming a linear pair.

Therefore, they must be supplementary. That is;

âˆ 3 + âˆ 4 = 180Â°

From equation (i)

60Â° + âˆ 4 = 180Â°

âˆ 4 = 180Â° – 60Â°

âˆ 4 = 120Â° ……(ii)

We have m || n

Thus, we get âˆ 2 and âˆ 4 as alternate interior opposite angles.

Therefore, these must be equal. That is,

âˆ 2 = âˆ 4

From equation (ii), we get :

âˆ 2 = 120Â°

Hence, the required value for âˆ 2 is 120Â°

**Question 17. **Prove that the straight lines perpendicular to the same straight line are parallel to one another.

**Solution:**

The figure can be drawn as follows:

Here, l âŠ¥ m and m âŠ¥ n.

We need to prove that l || m

It is given that l âŠ¥n, therefore,

âˆ 1 = 90Â° …..(i)

Similarly, we have m âŠ¥ n, therefore,

âˆ 2 = 90Â° ……(ii)

From (i) and (ii), we get

âˆ 1 = âˆ 2

But these are the pair of corresponding angles.

Theorem states: If a transversal intersects two lines in such a way that a pair of corresponding angles is equal, then the two lines are parallel.

Thus, we can say that l || m.

**Question 18. **The opposite sides of a quadrilateral are parallel. If one angle of the quadrilateral is 60Â°, find the other angles.

**Solution:**

The quadrilateral can be drawn as follows:

Here, we have AB || CD and AC || BD.

Also, âˆ ACD = 60Â°.

Thus, AB || CD.

Thus, âˆ ACD and âˆ BAC are consecutive interior angles.

Therefore, these two must be supplementary. i.e.

âˆ ACD + âˆ BAC = 180Â°

60Â° + âˆ BAC = 180Â°

âˆ BAC = 180Â° – 60Â°

âˆ BAC = 120Â°

Similarly, AC || BD .

Hence, âˆ ACD and âˆ CDB are consecutive interior angles.

Therefore, these two must be supplementary. i.e.

âˆ ACD + âˆ CDB = 180Â°

60Â° + âˆ CDB = 180Â°

âˆ CDB = 180Â° + 60Â°

âˆ CDB = 120Â°

Similarly, AB || CD.Thus, âˆ ABD and âˆ CDB are consecutive interior angles.

Therefore, these two must be supplementary. i.e.

âˆ ABD + âˆ CDB = 180Â°

âˆ ABD + 120Â° = 180Â°

âˆ ABD = 180Â° – 120Â°

âˆ ABD = 60Â°

Thus, the other angles are as follows:

âˆ BAC = 120Â°

âˆ CDB = 120Â°

âˆ ABD = 60Â°

**Question 19. **Two lines AB and CD intersect at O. If âˆ AOC+ âˆ COB+ âˆ BOD= 270Â°, find the measures of âˆ AOC, âˆ COB, âˆ BOD and âˆ DOA.

**Solution:**

Since, lines AB and CD intersect each other at point O.

Thus, âˆ AOC and âˆ BOD are vertically opposite angles.

Therefore,

âˆ AOC = âˆ BOD â€¦â€¦ (i)

Similarly,

âˆ COB = âˆ AOD â€¦… (ii)

Also, we have âˆ AOC, âˆ BOD, and âˆ AOD forming a complete angle.

Therefore, âˆ AOC + âˆ BOD + âˆ COB + âˆ AOD = 360Â°

Given:

âˆ AOC + âˆ COB + âˆ BOD = 270Â°

Hence, we get

(âˆ AOC + âˆ BOD + âˆ COB) + âˆ AOD = 360Â°

270Â° + âˆ AOD = 360Â°

âˆ AOD = 360Â° – 270Â°

âˆ AOD = 90Â°

From (ii), we will get;

âˆ COB = 90Â°

As we know that âˆ AOC and âˆ COB form a linear pair. Therefore, these must be supplementary.

âˆ AOC + âˆ COB = 180Â°

âˆ AOC + 90Â° = 180Â°

âˆ AOC = 180Â° – 90Â°

âˆ AOC = 90Â°

From (i), we get;

âˆ BOD = 90Â°

**Question 20. **In the given figure, p is a transversal to lines m and n, âˆ 2 = 120Â° and âˆ 5 = 60Â°. Prove that m||n.

**Solution:**

The given figure is;

Here we have that p is a transversal to lines m and n.

Also,

âˆ 2 = 120Â° and âˆ 5 = 60Â°.

To prove: m || n

Here we have âˆ 2 = 120Â°.

Also, âˆ 2 and âˆ 4 are vertically opposite angles, therefore, these two must be equal. i.e.

âˆ 4 = 120Â° ……(i)

Also, âˆ 5 = 60Â°

Adding this equation to (i), we will get :

âˆ 4 + âˆ 5 = 120Â° + 60Â°

âˆ 4 + âˆ 5 = 60Â°

But these are the consecutive interior angles.

Theorem states: If a transversal intersects two lines in such a way that a pair of consecutive interior angles is supplementary, then the two lines are parallel.

Thus, m || n.

Therefore, the lines are parallel to each other.

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