Telangana SCERT Class 7 Math Solution Chapter 4 Lines and Angles

Telanagana SCERT Solution Class VII (7) Math Chapter 4 Lines and Angles .

Exercise – 1

(1) Name the figures drawn below.

Solution: (i) Line Segment AB

(ii) Ray CD

(iii) Line XY

(iv) Point ‘P’

(2.) Draw the figures for the following

(4.) Write any five examples of angles that you have observed arround. Example : The angle formed when a scissor is opened.

Solution:

(i) Angle between the door and the wall.

(ii) Angle between the corners of a room.

(iii) Angle between the table and ground.

(iv) Angle between the two bags in a bag.

(v) Angle between the tyres and the spokes of a bike.

(5.) Identify the following given angles as acute, right or obtuse.

Ans. (i) acute

(ii) Obtuse

(iii) Right

(iv) acute

(v) Obtuse

(6.) Name all the possible angles you can find in the following figure. Which are acute, right, obtuse and straight angles?

Solution:

<AOF, <FOE, <EOD, <DOC, <COB, <FOD, <EOC, <DOB – Acute angles.

<AOE, <EOB, <FOC – Right angles.

<AOD, <AOC, <FOB – Obtuse angles.

<AOB – Straight angle.

(7.) Which of the following pairs of lines are parallel? Why?

Solution:

(i) and (iv) Parallel

(ii) and (iii) non-parallel

(8.) Which of the following lines are intersecting?

Ans. (i), (ii) and (iv) are intersecting lines and (iii) non-intersecting lines.

Exercise – 2

(1.) Which of the following pairs of angles are complementary?

Solution: (1) (iii) is complementary.

(2.) Find the complementary angles of the following. (i) 25^o (ii) 40^o (iii) 89^o (iv) 55^o

Solution:

90^o – 25^o = 65^o

90^o – 40^o = 50^o

90^o – 89^o = 1^o

90^o – 55^o = 35^o

(3.) Two angles are complement to each other and are also equal. Find them.

Solution: let, the angles be x.

Therefore, x + x = 180^o

=> 2x = 180^o

=> x = 180^o/2

= 90^o

The sum of two acute angles is always less than 180^o

Exercise – 3

(1.) Which of the following pairs of angles are supplementary?

Solution:

110 degree + 70 degree = 180 degree.

These are a pair of supplementary angles as their sum is 110 degree + 70 degree = 180 degree.

(ii) There are a pair of supplementary angles as their sum is 90 degree + 90 degree = 180 degree.

(iii) These are not a pair of supplementary angles because their sum is 50 degree + 140 degree = 190 degree.

(2.) Find the supplementary angles of the given angles. (i) 105^o (ii) 95^o (iii) 150^o (iv) 20^o

Solution:

180^o – 105^o = 75^o

180^o – 95^o = 85^o

180^o – 150^o = 30^o

180^o – 20^o = 160^o

(3.) Two acute angles cannot form a pair of supplementary angles. Justify.

Solution: We know acute angles are angles less than 90 degree. and supplementary angles are the two angles whose sum would be 180 degree. Since the acute angles are less than 90 degree, their sum would never add to 180 degree and hence could never be supplementary.

( 4.) Two angles are equal and supplementary to each other. Find them

Solution: Let the one angle be x, therefore other will be x.

Therefore, x + x = 180^o

=> 2x = 180^o

=> x = 180^o/2

= x = 90^o

Exercise – 4

(1) The angles (i) ab and (ii) cd are adjacent angles, as they have a common vertex and common arm.
(2) (i) <AOD, <DOB
(ii) <DOB, <BOC
(iii) <BOC, <COA
(iv) <COA, <AOD
Those angles are called adjant angles because they have a common vertex and common arm.

(1) Which of the following are adjacent angles?

Solution: The angles (i) ab and (ii) cd are adjacent angles, as they have a common vertex and common arm.

(2) Name all pairs of adjacent angles in the figure. How many pairs of adjacent angles are formed? Why these angles are called adjacent angles?

Solution:  (i) <AOD, <DOB

(ii) <DOB, <BOC

(iii) <BOC, <COA

(iv) <COA, <AOD

Those angles are called adjacent angles because they have a common vertex and common arm.

(3.) Can two adjacent angles be supplementary? Draw figure.

(4.) Can two adjacent angles be complementary? Draw figure.

(5.) Give four examples of adjacent angles in daily life.

Example : Angles between the spokes at the centre of a cycle wheel.

(i) ______ (ii) __________

(iii) _______ (iv) _________

Solution:

Exercise – 5

(iii) 140^o + 35^o = 175^o is not a linear pair.

(2) Solution: In the given figure, angles are not adjacent to each other to means they don’t have any common arm.

Exercise – 6

(1.) Name two pairs of vertically opposite angles in the figure.

Solution: (i) <AOD, <BOC

(ii) <AOC, <BOD

(2.) Find the measure of x, y and z without actually measuring them

Solution:

In given figure,

<y = 160^o and <x = <z

We can conclude that vertically opposite angles are equal.

Therefore, y = 160^o

x + 160^o = 180^o (in linear pair)

=> x = 180^o  – 160^o

=> x = 20^o

Therefore <x = <z = 20^o

(3.) Give some examples of vertically opposite angles in your surroundings.

Solution: (a) a open scissor

(b) The points where two roads intersect each other.

(c) Railroad crossing signs.

(d) An hourglass.

Exercise 7

(1) (i) Transversal

(ii) Parallel

(iii) Parallel

(iv) One

(2.) In the adjacent figure, the lines ‘l’ and ‘m’ are parallel and ‘n’ is a transversal.

Fill in the blanks for all the situations given below-

(i) If ∠ 1 = 800 then ∠ 2 = _______

(ii) If ∠ 3 = 450 then ∠ 7 = _____

(iii) If ∠ 2 = 900 then ∠ 8 = ____

(iv) If ∠ 4 = 1000 then ∠ 8 = ______

Solution:

(i) If <1 = 80 then <2 = 180 – 80 = 100 (linear pair)

(ii) If <3 = 45 then <7 = 45 (interior alternate angles)

(iii) If <2 = 90 then <8 = 90 (Vertically opposite angles)

(iv) If <4 = 100 then <8 = 100 (exterior alternate angles)

(3.) Find the measures of x,y and z in the figure, where l ll BC

Solution: <y = 75^o (interior alternate angles)

<z = 45^o (interior alternate angles)

<x = 180^o – (75^o + 45^o) (angles on straight line)

-> ,x = 180^o – 120^o – 60^o

(4.) ABCD is a quadrilateral in which AB ll DC and AD ll BC. Find ∠ b, ∠ c and ∠ d

Solution:

AB || CD and AD ||BC

Therefore.

ABCD is a parallelogram

So, (i) <c = 50^o (Opposite angles)

(ii) <b = 180^o – 50^o

= 130^o (Supplementary)

(iii) <d = <b = 130^o (Opposite angles)

(5) In a given figure, ‘l’ and ‘m’ are intersected by a transversal ‘n’. Is l || m?

Solution: l || m

<PQA = 100^o

<PQA + <AQR = 180^o (Linear Pair)

100^o + <AQR = 180^o

=> <AQR = 180^o – 100^o = 80^o

We are given <CRS = 80^o

So, <CRS = 80^o = <AQR

<CRS and <AQR are corresponding angles

Since, <CRS = <AQR corresponding angles are equal

So, l||m

(6.) Find ∠a, ∠b, ∠c, ∠d and ∠e in the figure? Give reasons

Solution: All the angles will be 50 degree because they all are alternate interior angles.