Selina Solutions Concise Maths Class 7 Chapter 14 Lines and Angles (Including Construction of Angles) are designed by highly knowledgeable faculty at BYJU’S, keeping in mind the understanding capacity of students. The PDF of solutions prepare students to solve problems without any difficulty. It also improves confidence among students to appear for the annual exam. Selina Solutions Concise Maths Class 7 Chapter 14 Lines and Angles (Including Construction of Angles), PDF links are given here, with a free download option.
Chapter 14 deals with basic terms, types of lines, types of angles and the construction of angles. In the solutions, students are provided with the steps to be followed in the construction of various structures.
Selina Solutions Concise Maths Class 7 Chapter 14: Lines and Angles (Including Construction of Angles) Download PDF
Exercises of Selina Solutions Concise Maths Class 7 Chapter 14 – Lines and Angles (Including Construction of Angles)
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Exercise 14A page: 162
1. State, true or false:
(i) A line segment 4 cm long can have only 2000 points in it.
(ii) A ray has one end point and a line segment has two end-points.
(iii) A line segment is the shortest distance between any two given points.
(iv) An infinite number of straight lines can be drawn through a given point.
(v) Write the number of end points in
(a) a line segment AB (b) a ray AB (c) a line AB
(vi) Out of 
which one has a fixed length?
(vii) How many rays can be drawn through a fixed point O?
(viii) How many lines can be drawn through three
(a) collinear points?
(b) non-collinear points?
(ix) Is 40° the complement of 60°?
(x) Is 45° the supplement of 45°?
Solution:
(i) False.
It contains infinite number of points.
(ii) True.
(iii) True.
(iv) True.
(v) (a) 2 (b) 1 (c) 0
(vi) 
has fixed length.
(vii) Infinite rays can be drawn through a fixed point O.
(viii) (a) 1 line can be drawn through three collinear points.
(b) 3 lines can be drawn through three non-collinear points.
(ix) False.
40o is the complement of 50o as 40o + 50o = 90o
(x) False.
45o is the supplement of 135o not 45o.
2. In which of the following figures, are ∠AOB and ∠AOC adjacent angles? Give, in each case, reason for your answer.
Solution:
If ∠AOB and ∠AOC are adjacent angle, they have OA as their common arm.
(i) From the figure
OB is the common arm
∠AOB and ∠AOC are not adjacent angles.
(ii) From the figure
OC is the common arm
∠AOB and ∠AOC are not adjacent angles.
(iii) From the figure
OA is the common arm
∠AOB and ∠AOC are adjacent angles.
(iv) From the figure
OB is the common arm
∠AOB and ∠AOC are not adjacent angles.
3. In the given figure, AOC is a straight line.
Find: (i) x (ii) ∠AOB (iii) ∠BOC
Solution:
We know that
∠AOB and ∠COB are linear pairs
It can be written as
∠AOB + ∠COB = 180o
Substituting the values
x + 25o + 3x + 15o = 180o
By further calculation
4x + 40o = 180o
So we get
4x = 180 – 40 = 140o
(i) x = 140/4 = 35o
(ii) ∠AOB = x + 25
Substituting the value of x
∠AOB = 25 + 35 = 60o
(iii) ∠BOC = 3x + 15o
Substituting the value of x
∠BOC = (3 × 35) + 15
∠BOC = 120o
4. Find y in the given figure.
Solution:
Here AOC is a straight line
We can write it as
∠AOB + ∠BOD + ∠DOC = 180o
Substituting the values
y + 150 – x + x = 180
By further calculation
y + 150 = 180
So we get
y = 180 – 150 = 30o
5. In the given figure, find ∠PQR.
Solution:
Here SQR is a straight line
We can write it as
∠SQT + ∠TQP + ∠PQR = 180o
Substituting the values
x + 70 + 20 – x + ∠PQR = 180o
By further calculation
90o + ∠PQR = 180o
So we get
∠PQR = 180o – 90o = 90o
6. In the given figure, po = qo = ro, find each.
Solution:
We know that
po + qo + ro = 180o is a straight angle
It is given that
po = qo = ro
We can write it as
po + po + po = 180o
3p = 180
p = 180/3 = 60o
Therefore, po = qo = ro = 60o
7. In the given figure, if x = 2y, find x and y.
Solution:
It is given that
x = 2y
For a straight angle
xo + yo = 180o
Substituting the values
2y + y = 180
By further calculation
3y = 180
y = 180/3 = 60o
x = 2y = 2 × 60o = 120o
8. In the adjoining figure, if bo = ao + co, find b.
Solution:
It is given that
bo = ao + co
For a straight angle
ao + bo + co = 180o
Substituting the values
bo + bo = 180o
2bo = 180o
bo = 180/2 = 90o
9. In the given figure, AB is perpendicular to BC at B.
Find : (i) the value of x.
(ii) the complement of angle x.
Solution:
(i) From the figure
AB || BC at B
Here ∠ABC = 90o
Substituting the values
x + 20 + 2x + 1 + 7x – 11 = 90
By further calculation
10x + 10 = 90
10x = 90 – 10 = 80
x = 80/10 = 8o
(ii) The complement of angle x = 90 – x
So we get
= 90 – 8 = 82o
10. Write the complement of:
(i) 25o
(ii) 90o
(iii) ao
(iv) (x + 5)o
(v) (30 – a)o
(vi) ½ of a right angle
(vii) 1/3 of 180o
(viii) 21o 17’
Solution:
(i) The complement of 25o = 90o – 25o = 65o
(ii) The complement of 90o = 90o – 90o = 0
(iii) The complement of ao = 90o – ao
(iv) The complement of (x + 5)o = 90o – (x + 5)o
By further calculation
= 90o – x – 5o
= 85o – x
(v) The complement of (30 – a)o = 90o – (30 – a)o
By further calculation
= 90o – 30o + ao
= 60o + ao
(vi) The complement of ½ of a right angle = 90o – ½ of a right angle
So we get
= 90o – ½ × 90o
= 90o – 45o
= 45o
(vii) The complement of 1/3 of 180o = 90o – 1/3 of 180o
By further calculation
= 90o – 60o
= 30o
(viii) The complement of 21o 17’ = 90o – 21o 17’
So we get
= 68o 43’
11. Write the supplement of:
(i) 100°
(ii) 0°
(iii) x°
(iv) (x + 35)°
(v) (90 +a + b)°
(vi) (110 – x – 2y)°
(vii) 1/5 of a right angle
(viii) 80° 49′ 25″
Solution:
(i) The supplement of 100° = 180 – 100 = 800
(ii) The supplement of 0° = 180 – 0 = 180o
(iii) The supplement of x° = 1800 – x0
(iv) The supplement of (x + 35)° = 1800 – (x + 35)0
We can write it as
= 180 – x – 35
= 1450 – x0
(v) The supplement of (90 + a + b)° = 1800 – (90 + a + b)0
We can write it as
= 180 – 90 – a – b
So we get
= 900 – a0 – b0
= (90 – a – b)0
(vi) The supplement of (110 – x – 2y)° = 1800 – (110 – x – 2y)°
We can write it as
= 180 – 110 + x + 2y
= 700 + x0 + 2y0
(vii) The supplement of 1/5 of a right angle = 1800 – 1/5 of a right angle
We can write it as
= 1800 – 1/5 × 900
So we get
= 1800 – 180
= 1620
(viii) The supplement of 80° 49′ 25″ = 1800 – 80° 49′ 25″
We know that 10 = 60’ and 1’ = 60”
So we get
= 990 10’ 35”
12. Are the following pairs of angles complementary?
(i) 10° and 80°
(ii) 37° 28′ and 52° 33′
(iii) (x+ 16)°and (74 – x)°
(iv) 54° and 2/5 of a right angle.
Solution:
(i) 10° and 80°
Yes, they are complementary angles as their sum = 100 + 800 = 900
(ii) 37° 28′ and 52° 33′
No, they are not complementary angles as their sum is not equal to 900
37° 28′ + 52° 33′ = 9001’
(iii) (x+ 16)° and (74 – x)°
Yes, they are complementary angles as their sum = x + 16 + 74 – x = 900
(iv) 54° and 2/5 of a right angle
We can write it as
= 540 and 2/5 × 900
= 540 and 360
Yes, they are complementary angles as their sum = 54 + 36 = 900
13. Are the following pairs of angles supplementary?
(i) 139o and 39o
(ii) 26o59’ and 153o1’
(iii) 3/10 of a right angle and 4/15 of two right angles
(iv) 2xo + 65o and 115o – 2xo
Solution:
(i) 139o and 39o
No, they are not supplementary angles as their sum is not equal to 1800
1390 + 390 = 1780
(ii) 26o59’ and 153o1’
Yes, they are supplementary angles as their sum = 26o59’ + 153o1’ = 1800
(iii) 3/10 of a right angle and 4/15 of two right angles
We can write it as
= 3/10 of 900 and 4/15 of 1800
= 270 and 480
No, they are not supplementary angles as their sum is not equal to 1800
270 + 480 = 750
(iv) 2xo + 65o and 115o – 2xo
Yes they are supplementary angles as their sum = 2x + 65 + 115 – 2x = 1800
14. If 3x + 18° and 2x + 25° are supplementary, find the value of x.
Solution:
It is given that 3x + 18° and 2x + 25° are supplementary
We can write it as
3x + 18° + 2x + 25° = 1800
By further calculation
5x + 430 = 1800
So we get
5x = 180 – 43 = 1370
x = 137/5 = 27.40 or 270 24’
15. If two complementary angles are in the ratio 1:5, find them.
Solution:
It is given that two complementary angles are in the ratio 1:5
Consider x and 5x as the angles
We can write it as
x + 5x = 900
6x = 900
So we get
x = 90/6 = 150
Here the angles will be 150 and 15 × 5 = 750
Exercise 14B page: 166
1. In questions 1 and 2, given below, identify the given pairs of angles as corresponding angles, interior alternate angles, exterior alternate angles, adjacent angles, vertically opposite angles or allied angles:
(i) ∠3 and ∠6
(ii) ∠2 and ∠4
(iii) ∠3 and ∠7
(iv) ∠2 and ∠7
(v) ∠4 and∠6
(vi) ∠1 and ∠8
(vii) ∠1 and ∠5
(viii) ∠1 and ∠4
(ix) ∠5 and ∠7
Solution:
(i) ∠3 and ∠6 are interior alternate angles.
(ii) ∠2 and ∠4 are adjacent angles.
(iii) ∠3 and ∠7 are corresponding angles.
(iv) ∠2 and ∠7 are exterior alternate angles.
(v) ∠4 and∠6 are allied or co-interior angles.
(vi) ∠1 and ∠8 are exterior alternate angles.
(vii) ∠1 and ∠5 are corresponding angles.
(viii) ∠1 and ∠4 are vertically opposite angles.
(ix) ∠5 and ∠7 are adjacent angles.
2. (i) ∠1 and ∠4
(ii) ∠4 and ∠7
(iii) ∠10 and ∠12
(iv) ∠7 and ∠13
(v) ∠6 and ∠8
(vi) ∠11 and ∠8
(vii) ∠7 and ∠9
(viii) ∠4 and ∠5
(ix) ∠4 and ∠6
(x) ∠6 and ∠7
(xi) ∠2 and ∠13
Solution:
(i) ∠1 and ∠4 are vertically opposite angles.
(ii) ∠4 and ∠7 are interior alternate angles.
(iii) ∠10 and ∠12 are vertically opposite angles.
(iv) ∠7 and ∠13 are corresponding angles.
(v) ∠6 and ∠8 are vertically opposite angles.
(vi) ∠11 and ∠8 are allied or co-interior angles.
(vii) ∠7 and ∠9 are vertically opposite angles.
(viii) ∠4 and ∠5 are adjacent angles.
(ix) ∠4 and ∠6 are allied or co-interior angles.
(x) ∠6 and ∠7 are adjacent angles.
(xi) ∠2 and ∠13 are allied or co-interior angles.
3. In the following figures, the arrows indicate parallel lines. State which angles are equal. Give reasons.
Solution:
(i) From the figure (i)
a = b are corresponding angles
b = c are vertically opposite angles
a = c are alternate angles
So we get
a = b = c
(ii) From the figure (ii)
x = y are vertically opposite angles
y = l are alternate angles
x = l are corresponding angles
1 = n are vertically opposite angles
n = r are corresponding angles
So we get
x = y = l = n = r
Similarly
m = k are vertically opposite angles
k = q are corresponding angles
Hence, m = k = q.
4. In the given figure, find the measure of the unknown angles:
Solution:
From the figure
a = d are vertically opposite angles
d = f are corresponding angles
f = 1100 are vertically opposite angles
So we get
a = d = f = 1100
We know that
e + 1100 = 1800 are linear pair of angles
e = 180 – 110 = 700
b = c are vertically opposite angles
b = e are corresponding angles
e = g are vertically opposite angles
So we get
b = c = e = g = 700
Therefore, a = 1100, b = 700, c = 700, d = 1100, e = 700, f = 1100 and g = 700.
5. Which pair of the dotted line, segments, in the following figures, are parallel. Give reason:
Solution:
(i) From the figure (i)
If the lines are parallel we get 1200 + 500 = 1800
There are co-interior angles where 1700 ≠ 1800
Therefore, they are not parallel lines.
(ii) From the figure (ii)
∠1 = 450 are vertically opposite angles
We know that the lines are parallel if
∠1 + 1350 = 1800 are co-interior angles
Substituting the values
450 + 1350 = 1800
1800 = 1800 which is true
Therefore, the lines are parallel.
(iii) From the figure (iii)
The lines are parallel if corresponding angles are equal
Here 1200 ≠ 1300
Hence, lines are not parallel.
(iv) ∠1 = 1100 are vertically opposite angles
We know that if lines are parallel
∠1 + 700 = 1800 are co-interior angles
Substituting the values
1100 + 700 = 1800
1800 = 1800 which is correct
Therefore, the lines are parallel.
(v) ∠1 + 1000 = 1800
So we get
∠1 = 1800 – 1000 = 800 which is a linear pair
Here the lines l1 and l2 are parallel if ∠1 = 700
But here ∠1 = 800 which is not equal to 700
So the lines l1 and l2 are not parallel
Now, l3 and l5 will be parallel if corresponding angles are equal
But here, the corresponding angles 800 ≠ 700
Hence, l3 and l5 are not parallel.
We know that, in l2 and l4
∠1 = 800 are alternate angles
800 = 800 which is true
Hence, l2 and l4 are parallel.
(vi) Two lines are parallel if alternate angles are equal
Here, 500 ≠ 400 which is not true
Hence, the lines are not parallel.
6. In the given figures, the directed lines are parallel to each other. Find the unknown angles.
Solution:
(i) If the lines are parallel
a = b are corresponding angles
a = c are vertically opposite angles
a = b = c
Here b = 600 are vertically opposite angles
Therefore, a = b = c = 600
(ii) If the lines are parallel
x = z are corresponding angles
z + y = 1800 is a linear pair
y = 550 are vertically opposite angles
Substituting the values
z + 550 = 1800
z = 180 – 55 = 1250
If x = z we get x = 1250
Therefore, x = 1250, y = 550 and z = 1250.
(iii) If the lines are parallel
c = 1200 (corresponding angles)
a + 1200 = 1800 are co-interior angles
a = 180 – 120 = 600
We know that a = b are vertically opposite angles
So b = 600
Therefore, a = b = 600 and c = 1200.
(iv) If the lines are parallel
x = 500 are alternate angles
y + 1200 = 1800 are co-interior angles
y = 180 – 120 = 600
We know that
x + y + z = 3600 are angles at a point
Substituting the values
50 + 60 + z = 360
By further calculation
110 + z = 360
z = 360 – 110 = 2500
Therefore, x = 500, y = 600 and z = 2500.
(v) If the lines are parallel
x + 900 = 1800 are co-interior angles
x = 1800 – 900 = 900
∠2 = x
∠2 = 900
We know that the sum of angles of a triangle
∠1 + ∠2 + 300 = 1800
Substituting the values
∠1 + 900 + 300 = 1800
By further calculation
∠1 + 1200 = 1800
∠1 = 180 – 120 = 600
Here ∠1 = k are vertically opposite angles
k = 600
Here ∠1 = z are corresponding angles
z = 600
Here k + y = 1800 are co-interior angles
Substituting the values
600 + y = 1800
y = 180 – 60 = 1200
Therefore, x = 900, y = 1200, z = 600, k = 600.
7. Find x, y and p is the given figures:
Solution:
(i) From the figure (i)
The lines are parallel
x = z are corresponding angles
y = 400 are corresponding angles
We know that
x + 400 + 2700 = 3600 are the angles at a point
So we get
x + 3100 = 3600
x = 360 – 310 = 500
So z = x = 500
Here p + z = 1800 is a linear pair
By substituting the values
p + 500 = 1800
p = 180 – 50 = 1300
Therefore, x = 500, y = 400, z = 500 and p = 1300.
(ii) From the figure (ii)
The lines are parallel
y = 1100 are corresponding angles
We know that
250 + p + 1100 = 1800 are angles on a line
p + 1350 = 1800
p = 180 – 135 = 450
We know that the sum of angles of a triangle
x + y + 250 = 1800
x + 1100 + 250 = 1800
By further calculation
x + 1350 = 1800
x = 180 – 135 = 450
Therefore, x = 450, y = 1100 and p = 450.
8. Find x in the following cases:
Solution:
(i) From the figure (i)
The lines are parallel
2x + x = 1800 are co-interior angles
3x = 1800
x = 180/3 = 600
(ii) From the figure (ii)
The lines are parallel
4x + ∠1 = 1800 are co-interior angles
∠1 = 5x are vertically opposite angles
Substituting the values
4x + 5x = 1800
So we get
9x = 1800
x = 180/9 = 200
(iii) From the figure (iii)
The lines are parallel
∠1 + 4x = 1800 are co-interior angles
∠1 = x are vertically opposite angles
Substituting the values
x + 4x = 1800
5x = 1800
So we get
x = 180/5 = 360
(iv) From the figure (iv)
The lines are parallel
2x + 5 + 3x + 55 = 1800 are co-interior angles
5x + 600 = 1800
By further calculation
5x = 180 – 60 = 1200
So we get
x = 120/5 = 240
(v) From the figure (v)
The lines are parallel
∠1 = 2x + 200 are alternate angles
∠1 + 3x + 250 = 1800 is a linear pair
Substituting the values
2x + 200 + 3x + 250 = 1800
5x + 450 = 1800
So we get
5x = 180 – 45 = 1350
x = 135/5 = 270
(vi) From the figure (vi)
Construct a line parallel to the given parallel lines
∠1 = 4x and ∠2 = 6x are corresponding angles
∠1 + ∠2 = 1300
Substituting the values
4x + 6x = 1300
10x = 1300
So we get
x = 130
e + 1100 = 1800 are linear pair of angles
e = 180 – 110 = 700
b = c are vertically opposite angles
b = e are corresponding angles
e = g are vertically opposite angles
So we get
b = c = e = g = 700
Therefore, a = 1100, b = 700, c = 700, d = 1100, e = 700, f = 1100 and g = 700.
5. Which pair of the dotted line, segments, in the following figures, are parallel. Give reason:
Solution:
(i) From the figure (i)
If the lines are parallel we get 1200 + 500 = 1800
There are co-interior angles where 1700 ≠ 1800
Therefore, they are not parallel lines.
(ii) From the figure (ii)
∠1 = 450 are vertically opposite angles
We know that the lines are parallel if
∠1 + 1350 = 1800 are co-interior angles
Substituting the values
450 + 1350 = 1800
1800 = 1800 which is true
Therefore, the lines are parallel.
(iii) From the figure (iii)
The lines are parallel if corresponding angles are equal
Here 1200 ≠ 1300
Hence, lines are not parallel.
(iv) ∠1 = 1100 are vertically opposite angles
We know that if lines are parallel
∠1 + 700 = 1800 are co-interior angles
Substituting the values
1100 + 700 = 1800
1800 = 1800 which is correct
Therefore, the lines are parallel.
(v) ∠1 + 1000 = 1800
So we get
∠1 = 1800 – 1000 = 800 which is a linear pair
Here the lines l1 and l2 are parallel if ∠1 = 700
But here ∠1 = 800 which is not equal to 700
So the lines l1 and l2 are not parallel
Now, l3 and l5 will be parallel if corresponding angles are equal
But here, the corresponding angles 800 ≠ 700
Hence, l3 and l5 are not parallel.
We know that, in l2 and l4
∠1 = 800 are alternate angles
800 = 800 which is true
Hence, l2 and l4 are parallel.
(vi) Two lines are parallel if alternate angles are equal
Here, 500 ≠ 400 which is not true
Hence, the lines are not parallel.
6. In the given figures, the directed lines are parallel to each other. Find the unknown angles.
Solution:
(i) If the lines are parallel
a = b are corresponding angles
a = c are vertically opposite angles
a = b = c
Here b = 600 are vertically opposite angles
Therefore, a = b = c = 600
(ii) If the lines are parallel
x = z are corresponding angles
z + y = 1800 is a linear pair
y = 550 are vertically opposite angles
Substituting the values
z + 550 = 1800
z = 180 – 55 = 1250
If x = z we get x = 1250
Therefore, x = 1250, y = 550 and z = 1250.
(iii) If the lines are parallel
c = 1200 (corresponding angles)
a + 1200 = 1800 are co-interior angles
a = 180 – 120 = 600
We know that a = b are vertically opposite angles
So b = 600
Therefore, a = b = 600 and c = 1200.
(iv) If the lines are parallel
x = 500 are alternate angles
y + 1200 = 1800 are co-interior angles
y = 180 – 120 = 600
We know that
x + y + z = 3600 are angles at a point
Substituting the values
50 + 60 + z = 360
By further calculation
110 + z = 360
z = 360 – 110 = 2500
Therefore, x = 500, y = 600 and z = 2500.
(v) If the lines are parallel
x + 900 = 1800 are co-interior angles
x = 1800 – 900 = 900
∠2 = x
∠2 = 900
We know that the sum of angles of a triangle
∠1 + ∠2 + 300 = 1800
Substituting the values
∠1 + 900 + 300 = 1800
By further calculation
∠1 + 1200 = 1800
∠1 = 180 – 120 = 600
Here ∠1 = k are vertically opposite angles
k = 600
Here ∠1 = z are corresponding angles
z = 600
Here k + y = 1800 are co-interior angles
Substituting the values
600 + y = 1800
y = 180 – 60 = 1200
Therefore, x = 900, y = 1200, z = 600, k = 600.
7. Find x, y and p is the given figures:
Solution:
(i) From the figure (i)
The lines are parallel
x = z are corresponding angles
y = 400 are corresponding angles
We know that
x + 400 + 2700 = 3600 are the angles at a point
So we get
x + 3100 = 3600
x = 360 – 310 = 500
So z = x = 500
Here p + z = 1800 is a linear pair
By substituting the values
p + 500 = 1800
p = 180 – 50 = 1300
Therefore, x = 500, y = 400, z = 500 and p = 1300.
(ii) From the figure (ii)
The lines are parallel
y = 1100 are corresponding angles
We know that
250 + p + 1100 = 1800 are angles on a line
p + 1350 = 1800
p = 180 – 135 = 450
We know that the sum of angles of a triangle
x + y + 250 = 1800
x + 1100 + 250 = 1800
By further calculation
x + 1350 = 1800
x = 180 – 135 = 450
Therefore, x = 450, y = 1100 and p = 450.
8. Find x in the following cases:
Solution:
(i) From the figure (i)
The lines are parallel
2x + x = 1800 are co-interior angles
3x = 1800
x = 180/3 = 600
(ii) From the figure (ii)
The lines are parallel
4x + ∠1 = 1800 are co-interior angles
∠1 = 5x are vertically opposite angles
Substituting the values
4x + 5x = 1800
So we get
9x = 1800
x = 180/9 = 200
(iii) From the figure (iii)
The lines are parallel
∠1 + 4x = 1800 are co-interior angles
∠1 = x are vertically opposite angles
Substituting the values
x + 4x = 1800
5x = 1800
So we get
x = 180/5 = 360
(iv) From the figure (iv)
The lines are parallel
2x + 5 + 3x + 55 = 1800 are co-interior angles
5x + 600 = 1800
By further calculation
5x = 180 – 60 = 1200
So we get
x = 120/5 = 240
(v) From the figure (v)
The lines are parallel
∠1 = 2x + 200 are alternate angles
∠1 + 3x + 250 = 1800 is a linear pair
Substituting the values
2x + 200 + 3x + 250 = 1800
5x + 450 = 1800
So we get
5x = 180 – 45 = 1350
x = 135/5 = 270
(vi) From the figure (vi)
Construct a line parallel to the given parallel lines
∠1 = 4x and ∠2 = 6x are corresponding angles
∠1 + ∠2 = 1300
Substituting the values
4x + 6x = 1300
10x = 1300
So we get
x = 130/10 = 130
Exercise 14C PAGE: 172
1. Using ruler and compasses, construct the following angles:
(i)30°
(ii)15°
(iii) 75°
(iv) 180°
(v) 165°
Solution:
(i) 30°
Steps of Construction:
1. Construct a line segment BC.
2. Taking B as centre and a suitable radius construct an arc which meets BC at the point P.
3. Taking P as centre and same radius cut off the arc at the point Q.
4. Consider P and Q as centre construct two arcs which intersect each other at the point R.
5. Now join BR and produce it to point A forming ∠ABC = 300
(ii) 150
Steps of Construction:
1. Construct a line segment BC.
2. Taking B as centre and a suitable radius construct an arc which meets BC at the point P.
3. Taking P as centre and same radius cut off the arc at the point Q.
4. Consider P and Q as centres, construct two arcs which intersect each other at the point D and join BD.
5. Taking P and R as centre construct two more arcs which intersect each other at the point S.
6. Now join BS and produce it to point A.
∠ABC = 150
(iii) 75°
Steps of Construction:
1. Construct a line segment BC.
2. Taking B as centre and suitable radius construct an arc and cut off PQ then QR of the same radius.
3. Taking Q and R as centre, construct two arcs which intersect each other at the point S.
4. Now join SB.
5. Taking Q and D as centre construct two arcs which intersect each other at the point T.
6. Now join BT and produce it to point A.
∠ABC = 75°
(iv) 180°
Steps of Construction:
1. Construct a line segment BC.
2. Taking B as centre and some suitable radius construct an arc which meets BC at the point P.
3. Taking P as centre and with same radius cut off the arcs PQ, QR and RS.
4. Now join BS and produce it to point A.
∠ABC = 180°
(v) 165°
Steps of Construction:
1. Construct a line segment BC.
2. Taking B as centre and some suitable radius construct an arc which meets BC at the point P.
3. Taking P as centre and same radius cut off arcs at PQ, QR and then RS.
4. Now join SB and produce it to point E.
5. Taking R and S as centres construct two arcs which intersect each other at the point M.
6. Taking T and S as centres construct two arcs which intersect each other at the point L.
7. Now join BL and produce it to point A.
∠ABC = 165°
2. Draw ∠ABC = 120°. Bisect the angle using ruler and compasses only. Measure each angle so obtained and check whether the angles obtained on bisecting ∠ABC are equal or not.
Solution:
Steps of Construction:
1. Construct a line segment BC.
2. Taking B as centre and some suitable radius construct an arc which meets BC at the point P.
3. Taking P as centre and with same radius cut off the arcs PQ and QR.
4. Now join BR and produce it to point A
∠ABC = 120°
5. Taking P and R as centres construct two arcs which intersect each other at the point S.
6. Now join BS and produce it to point D.
Here BD is the bisector of ∠ABC
By measuring each angle we get to know that is it 600
Yes, both the angles are of equal measure.
3. Draw a line segment PQ = 6 cm. Mark a point A in PQ so that AP = 2 cm. At point A, construct angle QAR = 60°.
Solution:
Steps of Construction:
1. Construct a line segment PQ = 6cm.
2. Now mark point A on PQ so that AP = 2 cm.
3. Taking A as centre and some suitable radius construct an arc which meets AQ at the point C.
4. Taking C as centre and with same radius cut the arc CB.
5. Now join AB and produce it to point R.
∠QAR = 600.
4. Draw a line segment AB = 8 cm. Mark a point P in AB so that AP = 5 cm. At P, construct angle APQ = 30°.
Solution:
Steps of Construction:
1. Construct a line segment AB = 8 cm.
2. Now mark the point P on AB such that AP = 5 cm.
3. Taking P as centre and some suitable radius construct an arc which meets AB at L.
4. Taking L as centre and same radius cut the arc LM.
5. Now bisect the arc LM at the point N.
6. Join PN and produce it to point Q.
∠APQ = 300
5. Construct an angle of 75° and then bisect it.
Solution:
Steps of Construction:
1. Construct a line segment BC.
2. At the point B construct an angle ABC which is equal to 750.
3. Taking P and T as centres construct arcs which intersect each other at the point L.
4. Now join BL and produce it to point D.
Here BD bisects ∠ABC.
6. Draw a line segment of length 6 .4 cm. Draw its perpendicular bisector.
Solution:
Steps of Construction:
1. Construct a line segment AB = 6.4 cm.
2. Taking A and B as centres and with some suitable radius construct arcs which intersect each other at the points S and R.
3. Now join SR which intersects AB at the point Q.
Here PQR is the perpendicular bisector of the line segment AB.
7. Draw a line segment AB = 5.8 cm. Mark a point P in AB such that PB = 3.6 cm. At P, draw perpendicular to AB.
Solution:
Steps of Construction:
1. Construct a line segment AB = 5.8 cm.
2. Now mark a point P on the line segment AB such that PB = 3.6 cm.
3. Taking P as centre and some suitable radius construct an arc which meets AB at the point L.
4. Taking L as centre and same radius cut off the arcs LM and MN.
5. Now bisect the arc MN at the point S.
6. Join PS and produce it to point Q.
Here PQ is perpendicular to the line segment AB at point P.
8. In each case, given below, draw a line through point P and parallel to AB:
Solution:
Steps of Construction:
1. From the point P construct a line segment meeting AB
2. Taking Q as centre and some suitable radius construct an arc CD.
3. Taking P as centre and same radius construct another arc which meets PQ at the point E.
4. Taking E as centre and radius equal to CD cut this arc at the point F.
5. Now join PF and produce it to both sides to L and M.
Here the line LM is parallel to the given line AB.
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