NCERT Solutions for Class 7 Maths
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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