The DAV Maths Class 8 Solutions and **DAV Class 8 Maths Chapter 10 Worksheet 2** Solutions of Parallel Lines offer comprehensive answers to textbook questions.

## DAV Class 8 Maths Ch 10 WS 2 Solutions

Question 1.

In the given figure l ∥ m and p ⊥ l and q ⊥ m. Show that p ∥ q. Give reasons for your answer.

Answer:

p ⊥1 (given)

q ⊥ m (given)

l ∥ m (given)

∴ p ∥ q

(The lines perpendicular to the parallel lines are also parallel to each other).

Hence proved.

Question 2.

What can you say about the quadrilateral ABCD given in figure. Is it a rectangle? Justify your answer.

Answer:

In Fig, p ⊥ l, q ⊥ m and l ∥ m

∴ the quadrilateral ABCD may be square or a rectangle.

Reason: ∠ABC = 90° (given)

∠CDA = 180° – 90° = 90°

Hence the quadrilateral ABCD is a square or rectangle.

Question 3.

In the given figure, ABC is a triangle and AD is an altitude, show that:

(i) \(\overrightarrow{\mathrm{BP}} \| \overrightarrow{\mathrm{AD}}\)

Answer:

∠PBA = 40° (Given)

∠BAD = 40° (Given)

∴ ∠PBA = ∠BAD

But they are alternate angles.

∴ \(\overrightarrow{\mathrm{BP}} \| \overrightarrow{\mathrm{AD}}\)

(ii) \(\overrightarrow{\mathrm{CQ}} \| \overrightarrow{\mathrm{AD}}\)

Answer:

∠ADC = 90° (Given)

∠QCD = 90° (Given)

∴ ∠ADC ∥∠QCD

But they are alternate angles.

(iii) \(\overrightarrow{\mathrm{BP}} \| \overrightarrow{\mathrm{CQ}}\)

Answer:

\(\overrightarrow{\mathrm{BP}} \| \overrightarrow{\mathrm{AD}}\) [From (i)]

\(\overrightarrow{\mathrm{CQ}} \| \overrightarrow{\mathrm{AD}}\) [from (ii)]

∴ \(\overrightarrow{\mathrm{BP}} \| \overrightarrow{\mathrm{CQ}}\)

Hence Proved.

Question 4.

Draw a line segment AB of length 5 cm. At A and B, construct lines perpendicular to AB. Also, draw the perpendicular bisector of AB. Are these three lines parallel to each other? Justify your answer.

Answer:

AD ⊥ AB (Given)

CE ⊥ AB (By construction)

AD ∥ CE …(i) [The lines perpendicular to the same line are parallel to each other.]

Similarly, CE ∥ BF …(ii)

From (i) and (ii), we get

AD ∥ CE ∥ BF

Hence proved.

Question 5.

If ∠D AC = 30°, find the angles of ΔABC in the figure.

Answer:

∠DAC = 30° (given)

∴ ∠BAC = ∠DAC + ∠BAD

= 30° + 40° = 70°

∠PBD = 90° (given)

∴ ∠ABC = ∠PBD – 40°

= 90° – 40° = 50°

∠ACB = 180° – (70° + 50°)

= 180° – 120° = 60°.