**NCERT Solutions for Class 7 Maths**

Chapter 5 Lines and Angles

Exercise 5.2

Chapter 5 Lines and Angles

Exercise 5.2

**1. State the property that is used in each of the following statements?**

**(i) If a || b, then ∠1 = ∠5.**

**(ii) If ∠4 = ∠6, then a || b.**

**(iii) If ∠4 + ∠5 = 180°, then a || b.**

**Answer**

(i) We have, if $a \| b$, then $\angle 1=\angle 5$.

If a transversal intersects two parallel lines, then the corresponding angles are equal.

$\therefore$ By corresponding angle property, it is true.

$\therefore$ By alternate interior angle property, it is true.(iii) We have, if $\angle 4+\angle 5=180^{\circ}$, then $a \| b$.

If a transversal intersects two parallel lines, then the interior angles on the same side of the transversal are supplementary.

**2. In the adjoining figure, identify**

**(i) the pairs of corresponding angles.**

**(ii) the pairs of alternate interior angles.**

**(iii) the pairs of interior angles on the same side of the transversal.**

**(iv) the vertically opposite angles.**

Answer

(i) Pairs of corresponding angles are

$\angle 1, \angle 5 ; \angle 2, \angle 6 ; \angle 3, \angle 7 ; \angle 4, \angle 8$

(ii) Pairs of alternate interior angles are $\angle 2, \angle 8 ; \angle 3, \angle 5$.

(iii) Pairs of interior angles on the same sides of the transversal are

$\angle 2, \angle 5 ; \angle 3, \angle 8$.

(iv) Vertically opposite angles are

$\angle 1, \angle 3 ; \angle 2, \angle 4 ; \angle 5, \angle 7 ; \angle 6, \angle 8$.

**3. In the adjoining figure, p || q. Find the unknown angles.**

**Answer**

Given, $p \| q$

∠e + 125° = 180° (by linear pair)

⇒ ∠e = 180° – 125 ⇒ ∠e = 55°

∴ ∠f = ∠e = 55° (vertically opposite angles)

Since, $p \| q$ and t is a transversal.

∴ ∠a = ∠f = 55° (alternate interior angles)

∠d = 125° (corresponding angles)

∠c = ∠a = 55° (vertically opposite angles)

and ∠b = ∠d = 125° (vertically opposite angle)

Hence, ∠a = 55°, ∠b = 125°, ∠c = 55°, ∠d = 125° and ∠f = 55°.

**4. Find the value of x in each of the following figures if l || m.**

Answer

(i) We have, $l \| m$ and $t$ is a transversal. (alternate interior angles)

$\therefore \quad \angle x=\angle 1$

Now, $\angle 1+110^{\circ}=180^{\circ} \quad$ [by linear pair]

$\therefore \quad \angle 1=180^{\circ}-110^{\circ}=70^{

$\therefore \quad \angle x=70^{\circ}$

Hence, the required value of $x$ is $70^{\circ}$.

(ii) Since, $l \| m$ and $a$ is a transversal.

Hence, the required value of x is 100°.

**5. In the given figure, the arms of two angles are parallel.**

**If ∠ABC = 70º, then find**

**(i) ∠DGC**

**(ii) ∠DEF**

**Answer**

(i) Given, ∠ABC = 70°

Since, $AB \| DE$ and BC is a transversal.

∴ ∠DGC = ∠ABC = 70° (corresponding angles)

Hence, the value of ∠DGC is 70°

(ii) Since, $BC \| EF$ and DE is a transversal.

∴ ∠DEF = ∠DGC = 70° (corresponding angles)

Hence, the value of ∠DEF is 70°

**6. In the given figures below, decide whether l is parallel to m.**

**Answer**

(i) Since, $44^{\circ}+126^{\circ}=170^{\

But $\quad 170^{\circ} \neq 180^{\circ}$

i.e. the sum of the interior angles on the same side of the transversal is not $180^{\circ} .$

So, $l$ and $m$ are not parallel.

(ii) Since, $n$ is a transversal to $l$ and $m .$

$\therefore \quad \angle 1=75^{\circ}$ (vertically opposite angles)

Also, $\quad \angle 1=75^{\circ}=75^{\circ}+75^{\

$\therefore \quad 150^{\circ} \neq 180^{\circ}$

So, $l$ and $m$ are not parallel.

So, $l$ and $m$ are parallel.

(iv) Since, $\angle 1+98^{\circ}=180^{\circ} \quad$ (by linear pair)

$\therefore $ $\angle 1=180^{\circ}-98^{\circ}=82^{\

Let $\angle 2=98^{\circ}$ and $\angle 3=72^{\circ}$

$\therefore $ $\angle 1 \neq \angle 3$

i.e., the corresponding angles are not equal.

So, l and m are not parallel.

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