Parallel and Intersecting Lines Class 7 Short Question Answer

Parallel and Intersecting Lines Class 7 Short Question Answer

Parallel and Intersecting Lines Class 7 Short Question Answer

Question 1.

Two angles are making a linear pair. If one of them is one-third of the other, then find the angles.

Solution:

Let one angle be 3x. So, the other angle is x.

So, 3x + x = 180° or 4x = 180° or x = [latex]\frac{180^{\circ}}{4}[/latex] = 45°

So, one angle is 45° and the other angle is 3 x 45° = 135°.

Question 2.

In the given figure, find the values of angles x, y and z.

Solution:

x + 40° + 25° = 180°.

So, x + 65° = 180°.

Hence, x = 180°- 65° = 115°.

Also,y + 40° = 180°.

So,y = 180° - 40° = 140°.

Again, y + z = 180°. So, 140° + z = 180°.

That is, z = 180° - 140° = 40°.

Question 3.

In the given figure, AB is parallel to DE and BC is parallel to EF. Find :

(i) ∠DGC

(ii) ∠DEF.

Solution:

(i) Since AB is parallel to ED and BC is a transversal, so, ∠DGC = ∠ABC (Corresponding angles)

Hence, ∠DGC = 70°.

(ii) Since BC is parallel to EF and DE is a transversal, so, ∠DEF = ∠DGC (Corresponding angles)

Hence, ∠DEF = 70°.

Question 4.

In the given figure, if lines l and m are parallel, then find ∠BAC.

Solution:

∠EAC + ∠ACF = 180°

(Interior angles on the same side of the transversal)

So, ∠EAC + 120° = 180°

or ∠EAC = 180° - 120° = 60° ...(i)

Now, ∠DAB + ∠BAC + ∠EAC = 180°

So, 40° + ∠BAC + 60° = 180°

or ∠BAC = 180° - 40° - 60° = 80°.

Question 5.

In the given figure, if l is parallel to m, then find the value of p + q + r.

Solution:

p = 130° (Vertically opposite angles)

q = p = 130° (Alternate angles)

r = q = 130° (Vertically opposite angles)

So, p + q + r = 130° + 130° + 130° = 390°.