Selina solutions

Selina solutions

Grade 6

Properties of Angles and Lines (Including Parallel Lines) | Revision Exercise

Question 1

Q1)

In the following figures, AB is parallel to CD; find the values of angles x, y and z :

Solution:

AB II CD

And LM is its transversal

∠ALM = ∠LMN (Alternate angles)

x = 105°

Similarly AB II CD and LN is its transversal

∠BLN = ∠LNM (Alternate angles)

∠z = 60°

z = 60°

But x + y + z = 180° (Straight line angles)

> 105° + y + 60° = 180°

> y + 165° = 180°

> y = 180° - 165° = 15°

Hence x = 105°, y = 15° and z = 60°

AB II CD

MN is its transversal

∠LNM = ∠NMD (Alternate angles)

= y = 45°

And AB II CD and LM is its transversal

∠ALM = ∠CMP (Corresponding angles)

> 75° = x

x = 75°

And ∠ALM = ∠LMD (Alternate angles)

75° = z + 45°

> z = 75° - 45° = 30°

Hence x = 75°, y = 45° and z = 30°.

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Question 1

Q1)

In the following figures, AB is parallel to CD; find the values of angles x, y and z :

Solution:

AB II CD

And LM is its transversal

∠ALM = ∠LMN (Alternate angles)

x = 105°

Similarly AB II CD and LN is its transversal

∠BLN = ∠LNM (Alternate angles)

∠z = 60°

z = 60°

But x + y + z = 180° (Straight line angles)

> 105° + y + 60° = 180°

> y + 165° = 180°

> y = 180° - 165° = 15°

Hence x = 105°, y = 15° and z = 60°

AB II CD

MN is its transversal

∠LNM = ∠NMD (Alternate angles)

= y = 45°

And AB II CD and LM is its transversal

∠ALM = ∠CMP (Corresponding angles)

> 75° = x

x = 75°

And ∠ALM = ∠LMD (Alternate angles)

75° = z + 45°

> z = 75° - 45° = 30°

Hence x = 75°, y = 45° and z = 30°.

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