14
FUNDAMENTAL GEOMETRICAL
CONCEPTS, LINE AND ANGLES,
CIRELES
1. At 7 o’clock the hands of clock make an obtuse angle and a reflex angle. Calculate
the size, in degrees, of each of these angles.
Ans. At 6 o’clock the hands of clock make a straight line i.e. the angle of 180°
12
∴
180°
× 7 = 210°
At 7 o’clock, the angle made =
6
Reflex angle = 360° – 210° = 150°.
REFLEX
ANGLE
Z
B
L
A
A
N
L IO
U AT
D
N
© E ER
T
IN
180°
6
7
2. At 5 o’clock the hands of a clock make an obtuse angle and a reflex angle. Find the
measure of each of these angles.
12 1
11
Ans. At 5 o’clock, the two hands of a clock make on obtuse angle
2
10
3
9
5
= 150°
= 360° ×
12
4
8
7 6
5
Hence, measure of reflex angle will be = 360° – 150° = 210°
3. At 10 o’clock the hands of a clock make an acute angle and a reflex angle. Find the
measure of each of these angles:
12 1
11
Ans. At 10 o’clock, the two hands of the clock make a reflex angle
2
10
(i) x (ii) ∠ AOB (iii) ∠ BOC
Ans (i) ∠ AOB and ∠ COB are linear pairs
3
9
360° × 10
=
= 300°
12
Then the measure of acute angle = 360° – 300° = 60°
Hence, angles are 60°, 300°.
4. In the given figure, AC is a straight line.
Find :
4
8
7 6
5
B
A
x + 25°
3x + 15°
O
C
∴ ∠ AOB + ∠ COB = 180°
⇒ x + 25° + 3x + 15° = 180° ⇒ 4x + 40° = 180°
⇒ 4x = 180° – 40° = 140°
ICSE Math Class VII
∴ x=
1
140°
= 35°
4
Question Bank
(ii) ∠ AOB = x + 25° = 35° + 25° = 60°
(iii) ∠ BOC = 3x + 15° = 3 × 35° + 15° = 105° + 15° = 120°.
P
T
20 – x°
5. Find y in the given figure.
Ans. ∵ AOC is a straight line.
∴ ∠ AOB + ∠ BOD + ∠ DOC = 180°
⇒ y + 150° – x + x = 180° ⇒ y + 150° = 180°
⇒ y = 180° – 150° = 30°, Hence, y = 30°
6. In the given figure, find ∠ PQR.
Ans. SQR is a straight line.
S
x + 70°
∴
∠ SQT + ∠ TQP + ∠ PQR = 180°
⇒
x + 70° + 20° – x + ∠ PQR = 180°
⇒
90° + ∠ PQR = 180° ⇒ ∠ PQR = 180° – 90° = 90°
Z
B
L
A
A
N
L IO
U AT
D
N
© E ER
T
IN
Hence, ∠ PQR = 90°
7. In the given figure p° = q° = r°, find each.
Ans.
p° = q° = r°
∴ p + q + r = p° + p° + p° = 180°
⇒ 3p° = 180° ⇒ p° =
q°
p°
180°
= 60°
3
(given)
(straight angle)
r°
Hence, p° = q° = r° = 60°
8. In the given figure, if x = 2y, find x and y.
Ans.
x° + y° = 180
2y + y = 180° ⇒ 3y = 180°
But
x = 2y ∴
180°
= 60°
3
Hence, y = 60° and x = 2y = 2 × 60° = 120°
9. In the adjoining figure, if b° = a° + c°, find b.
Ans. a° + b° + c° = 180°
But b° = a° + c°
∴ a° + c° + b° = 180°
⇒ b° + b° = 180° ⇒ 2b° = 180°
⇒
R
Q
(straight angle)
(given)
y=
x°
(straight angle)
(given)
c°
b°
a°
180°
= 90°
2
10. In the given figure AB is perpendicular to BC at B.
Find : (i) the value of x.
(ii) the complement of angle x.
A
2
x+
20
°
⇒ b° =
ICSE Math Class VII
y°
B
+
1°
1°
2x
–1
x
7
C
Question Bank
Ans. (i) In the given figure.
AB ⊥ BC at B. ∴ ∠ABC = 90°
⇒ x + 20° + 2x + 1° + 7x – 11° = 90° ⇒ 10x + 10° = 90°
80°
= 8°
10
(ii) Complement of angle x = 90° – x = 90° – 8° = 82°
11. Find each angle shown in the diagram.
⇒ 10x = 90° – 10° = 80° ∴ x =
1
1
Ans. In the figure, 3 y° + 2 y ° + 2 y° + 2 y° = 360°
2
2
(Angles at a point)
⇒
3
2y°
2y°
Z
B
7
5
y ° + y ° + 4 y ° = 360°
2
2
1 y°
2
2
1 y°
2
12
y° + 4 y ° = 360° ⇒ 6y° + 4y° = 360° ⇒ 10y° = 360°
2
360°
= 36°
⇒ y=
10°
1
7
7
∠ AOB = 3 y ° = y ° = × 36° = 126°
∴
2
2
2
∠ BOC = 2y° = 2 × 36 = 72°
⇒
L
A
A
N
L IO
U AT
D
N
© E ER
T
IN
∠ COD = 2y°= 72°
1
5
5
∠ DOA = 2 y° = y° = y°×36° = 90°
2
2
2
12. In the given figure BC is produced on both sides to points D and E respectively.
Find the values of x and y.
Ans. ∠ DCA + ∠ ABC = 180°
13.
Ans.
⇒
⇒
[Linear pair]
A
⇒ x + 115° = 180°
⇒ x = 180° – 115° = 65°
⇒ ∠ BCA + ∠ ACE = 180°
⇒ y + 132° = 180°
D
⇒ y = 180° – 132° = 48°
In the given figure, find the value of x.
We know that sum of all the angles at a point = 360°
∠ AOB + ∠ BOC + ∠ COD + ∠ DOA = 360°
x + 100° + 30° + 150° = 360°
x + 280° = 360° ⇒ x = 360° – 280° = 80°
ICSE Math Class VII
3
x° 115°y° 132°
C
B
E
B
C
D
100°
30°
x°
O
150°
A
Question Bank
14. In the given figure, find the measure of each of the angles ∠ AOB, ∠ BOC , ∠ COD
and ∠ DOA
Ans. We know that the sum of all the angles at a point = 360°
x + 2x+ 3x + 4x = 360
C
B
2x°
360
⇒ 10x = 360 ⇒ x =
= 36
10
Thus, ∠ AOB = 36°, ∠BOC = 2x° = (2 × 36)° = 72°
3x°
x°
A
4x°
O
D
∠COD = 3x° = (3 × 36)° = 108°
∠DOA = 4x° = (4 × 36)° = 144°.
15. In the given figure, find the measure of each of the angles
∠DOE and ∠ EOA.
Z
B
B
Ans. In the given figure, ∠ AOB = 90°, ∠ BOC = 112°, ∠COD
= 86°, ∠ DOE = x and ∠ EOA = 3x
90°
O 3x°
86° x°
112°
L
A
A
N
L IO
U AT
D
N
© E ER
T
IN
But ∠ AOB + ∠ BOC + ∠ COD + ∠ DOE + ∠ EOA =
360° (Angles at a point)
⇒ 90° + 112° + 86° + x + 3x = 360°
⇒ 288° + 4x = 360° ⇒ 4x = 360° – 288° = 72°
C
D
A
E
72
= 18°. Thus, ∠ DOE = 18° and ∠ EOA = 3x = 3 × 18° = 54°
4
C
16. In the given figure, find the value of x.
∴
x=
What is the measure of ∠ COD ?
Ans. In the figure, ∠ AOB = (x – 5)°, ∠ BOC = (2x – 5)°,
∠ COD = (3x + 20)° and ∠ DOA = 4x°
But ∠ AOB + ∠ BOC + ∠ COD + ∠ DOA = 360°
°
5)
B
–
x
(2 – 5)°
(x
(3x + 20)°
A
O 4x°
D
(Angles at a point)
x – 5 + 2x – 5 + 3x + 20 + 4x = 360°
∴
⇒ 10x + 10 = 360° ⇒ 10x = 360° – 10 = 350
350
= 35
∴ x=
10
Now, ∠ COD = 3x + 20° = 3 × 35° + 20° = 105 + 20° = 125°
17. From the adjoining diagram, find the value of x and hence complete the following:
(i) ∠ BOC = ....
ICSE Math Class VII
(ii) ∠ DOE = ....
4
Question Bank
As the sum of ∠ s at a point,
Ans.
∠ AOE + ∠ AOB + ∠ BOC + ∠ COD + ∠ DOE = 360°
C
⇒ 90° + 55° + 2x + 36° + 3x + 4° = 360°
5x + 185 = 360°
⇒ 5x = 360° – 185°
B
D
36°
2x
3x + 4°
175°
= 35°
x=
5
(i) ∠ BOC = 2x = 2 × 35° = 70°
55°
A
O
E
(ii) ∠ DOE = 3x + 4° = 3 × 35° + 4° = 105° + 4° = 109°.
18. In the adjoining diagram, angles x and y form a linear pair. If y – x = 52°, find the
values of x and y.
Ans. From the given figure,
y + x = 180°
(Linear pair)
But y – x = 52°
(Given)
y x
A
C
B
Adding, we have
2y = 232°
Z
B
L
A
A
N
L IO
U AT
D
N
© E ER
T
IN
232
= 116°
2
Now, y + x = 180° ⇒ 116° + x = 180° ⇒ x = 180° – 116° = 64°
19. Find k in the given figures
k+
15°
42°
(i)
y=
30°
⇒
(ii)
K
3K
2K
k + 150°
Ans.(i) In the fig. (i) :
k + 150° + k – 15° + 30° – 90° = 360°
⇒ 2k + 150° + 90° + 30° – 15° = 360°
⇒ 2k + 270° – 15° = 360°
⇒ 2k = 360° – 270° + 15° = 105°
105°
= 52.5° or 52° 30′
2
Hence, k = 52.5° or 52° 30′
(ii) In the fig. (ii)
⇒ k + 2k + 3k + 42° = 360°
(Angles at a point)
⇒ k=
ICSE Math Class VII
(Angles at a point)
5
Question Bank
⇒ 6k + 42° = 360°
⇒ 6k = 360° – 42° = 318°
318°
= 53°
6
20. If two complementary angles are in the ratio 1 : 5, find them.
Ans. Two complementary angles are in the ratio = 1 : 5
Let these angles be x and 5x
∴ x + 5x + 90° ⇒ 6x = 90°
⇒ k=
90°
= 15°
6
Thus, angles will be 15° and 15° × 5 = 75°
21. Three angles which add upto 180° are in the ratio 2 : 3 : 7. Find them.
Ans. Ratio of three angles = 2 : 3 : 7
Let first angle be = 2x, second angles = 3x, and third angle = 7x
∴ 2x + 3x + 7x = 180°
⇒ x=
Z
B
L
A
A
N
L IO
U AT
D
N
© E ER
T
IN
180°
= 15°
12
∴ First angle = 2x = 2 × 15° = 30°, second angle = 3x = 3 × 15° = 45° and third
angle = 7x = 7 × 15° = 105°,
Hence, angles are 30°, 45° and 105°
22. 20% of an angle is the supplement of 60°. Find the angle.
Ans. Let the given angle be x.
Then 20% of x + 60° = 180°
⇒ 20% of x = 180° – 60° = 120°
⇒ 12x = 180° ⇒ x =
20
120 × 100
× x = 120° ⇒ x =
= 600
100
20
Hence, x = 600°.
23. How many degrees are there is an angle which is one-fifth of its supplement?
Ans. Let the angle be x°, then its supplement is 180° – x°.
As per condition,
⇒
1
(180° − x°) ⇒ 5x° = 180° – x°
5
⇒ 5x° + x° = 180° ⇒ 6x° = 180°
x° =
∴
x° =
ICSE Math Class VII
180
= 30°
6
6
Question Bank
24. An angle is 30° more than one-half of its complement. Find the angle in degree.
Ans. Let the angle be x° then its complement is 90° – x°.
As per given condition,
x°
1
x° = 30° + (90° − x°) = 30° + 45° −
2
2
x°
⇒ x° +
= 30° + 45°
2
2 x° + x°
= 75° ⇒ 3x° = 2 × 75
⇒
2
∴ x=
150°
= 50°
3
Z
B
25. 25% of an angle is the complement of 50°. Find the angle.
Ans. Let angle be x and complement of 50° = 90° – 50° = 40°
L
A
A
N
L IO
U AT
D
N
© E ER
T
IN
∴ 25% of x = 40° ⇒
40 × 100
= 160°
25
Hence, required angle = 160°.
80% of an angle is the supplement of 140°. Find the angle.
Let the required angle be x and supplement of 140° = 180° – 140° = 40°
∴ As per condition 80% of x
80 × x
80
⇒ =
= 40 ⇒ x = 40 ×
⇒ x = 50
100
100
Hence, the required angle = 50°
An angle is double its complement. Find the angle.
Let the angle be x°
∴ x = 2(90 – x) = 180 – 2x
(Given)
∴x=
26.
Ans.
27.
Ans.
25
x = 40°
100
180
⇒ = 60
3
Hence, the angle = 60°.
28. Find an angle which is two-thirds of its complement.
Ans. Let the required angle be x and complement be 90° – x
x + 2x = 180 ⇒ x =
2
(90° – x) = x
3
⇒ 2(90° – x) = 3x ⇒ 180° – 2x = 3x ⇒ 3x + 2x = 180°
As per condition,
ICSE Math Class VII
7
Question Bank
180°
= 36°
5
Hence, required angle = 36°
29. Find an angle which is 5% more than four times its supplement.
Ans. Let the required angle to be to x then its supplement be 180° – x.
As per condition,
x = 4 (180° – x) + 5°
⇒ x = 720° – 4x + 5
⇒ 5x = 180° ⇒ x =
⇒ x + 4x = 725 ⇒ 5x = 725 ⇒ x =
725
= 145°
5
Hence, the required angle = 145°
30. Two angles are supplementary and the larger is 20° less than three times the smaller.
Find the angles
Ans. Let the smaller angle be x°, then the larger angle be 3x° – 20°
As per condition
∴ x° + 3x° – 20° = 180°
⇒ 4x° – 20° = 180°
⇒ 4x° = 180° + 20° = 200°
Z
B
L
A
A
N
L IO
U AT
D
N
© E ER
T
IN
200
= 50°
4
Hence, smaller angle = x ° = 50° and larger angle = 3x° – 20° = 3 × 50° – 20°
= 150° – 20° = 130°
x
31. From the adjoining, find x and y if x = 2y.
y
Ans. a = x (corresponding angles)
x
Thus, a = 2y, Now, a + y = 180°
a y
2y + y = 180° and 3y = 180° ⇒ y = 60°
∴ y = 60°, x = 2y, 2× 60° = 120°
E
1 2
32. In the given figure AB || CD and EF is a transversal. If ∠ 8 = A
3 4
110°, find each one of the unknown angles marked in the fig5 6
ure. Given reasons.
∴ x° =
Ans. ∠ 8 = 110°
(Given)
(Linear pair)
⇒ ∠ 7 + 110° = 180°
⇒ ∠ 7 = 180° – 110° = 70°, ∠ 5 = 110°
∠6 = ∠7
⇒ ∠ 6 = 70°, ∠ 4 = ∠ 8 = 110°
ICSE Math Class VII
8
C
7 8
F
110°
B
D
(Vertically opposite angle)
(Corresponding angle)
Question Bank
∠ 3 = ∠ 7 = 70°
(Corresponding angle)
(Vertically opposite angle)
∠ 1 = ∠ 4 = 110°
∠ 2 = ∠ 3 = 70°
(Vertically opposite angle)
33. In each of the following figures AB || CD and E F is a transversal. Find one of the
unkwon angles x, y, z in each case.
(i)
(ii)
E
A
E
B
x°
D
B
55°
z°
130°
y°
y°
C
Z
B
D
z°
F
x°
A
L
A
A
N
L IO
U AT
D
N
© E ER
T
IN
C
F
E
(iii)
x°
A
B
y°
z°
D
C 40°
F
Ans. (i) x = 55°
(Vertically opposite angles)
y = 55°
(Interior alternate angles)
z + y = 180° ⇒ z + 55° = 180° z = 180° – 55° = 125°
(ii) z = 40°
(Vertically opposite angles)
y = z = 40°
(Interior alternate angles)
x + y = 180°
(Linear pair)
E
⇒ x + 40° = 180° x = 180° – 40° = 140°
y°
D
C
34. In the given figure, AB || DC and BC || AD.
z°
x°
Find the values of x, y and z.
Ans. AB || CD and BE is a transversal
110°
⇒
⇒
x + 110° = 180°
x + y = 180°
ICSE Math Class VII
⇒
⇒
x = 180° – 110° = 70°
70° + y = 180°
9
A
B
(Linear pair)
Question Bank
y = 180° – 70° = 110°
Then, z = y = 110°
35. Find x, y and p in the given figures
270°
40°
x
(i)
(alternate angles)
(ii)
x
y
p
z
p 110°
25°
y
Ans. In figure (i)
(corresponding angles)
∴ x=z
y = 40°
(corresponding angles)
But x + 40° + 270° = 360°
(angles at a point)
⇒ x + 310° = 360° ⇒ x = 360° – 310° = 50°
∴ z = x = 50°
But p + z = 180°
(linear pair)
⇒ p + 50° = 180° ⇒ p = 180° – 50° = 130°
Hence, x = 50°, y = 40°, z = 50° and p = 130°
(ii) In figure (ii) Lines are parallel
(corresponding angles)
∴ y = 110°
But 25° + p + 110° = 180°
(angles on a line)
⇒ p + 135° = 180°
⇒ p = 180° – 135° = 45°
(sum of angles of a triangle)
⇒ x + y + 25° = 180°
⇒ x + 110° + 25 = 180° ⇒ x + 135° = 180°
D
⇒ x = 180° – 135° = 45°
A
Hence, x = 45°, y = 110° and p = 45°
B
O
36. In the given figure, two lines AB and CD intersect at point O.
C
If ∠ AOC + ∠ BOD = 70°, find the measure of ∠ AOD.
Z
B
L
A
A
N
L IO
U AT
D
N
© E ER
T
IN
Ans. ∠ AOC + ∠ BOD = 70°
But ∠ AOC = ∠ BOD
(given)
(Vertically opposite angles)
∴ ∠ AOC + ∠ AOC = 70° ⇒ 2 ∠ AOC = 70°
 70° 
⇒ ∠ AOC = 
 = 35° ⇒ ∠ AOC = ∠ BOD = 35°
 2 
Now ∠ AOD + ∠ BOD = 180°
ICSE Math Class VII
10
[Linear pair angles]
Question Bank
⇒
∠ AOD + 35° = 180°
E
∠ AOD = 180° – 35° = 145°
∴
37. In the given figure, the lines AB, CD and EF intersect at
point O. If ∠ BOD = x° ∠ AOE = 2x° and ∠ COF = 90°
find ∠ AOE and ∠ AOC.
2x° O
A
D
x°
B
90°
C
F
(Vertically opposite angles)
Ans. ∠ DOE = ∠ COF
∴ ∠ DOE = 90°
∠ BOD + ∠ DOE + ∠ EOA = 180°
⇒ x + 90° + 2x = 180° ⇒ 3x = 180° – 90° = 90°
Z
B
 90 
⇒ x =   ° = 30°
 3 
∴ ∠ AOE = 2x = (2 × 30)° = 60° ∠ AOC = ∠ BOD (Vertically opposite angles)
= x° = 30°.
Hence, ∠AOC = 30°
38. Calculate the size of each literal angle in the following figures :
(i)
L
A
A
N
L IO
U AT
D
N
© E ER
T
IN
130°
(ii)
110° P
r
q
s
p
r
q
t
110°
Ans. (i) p = 135°
(Alternate interior angles)
Now,
p + r + 110° = 360°
(Sum of angles at a point)
135° + r + 110° = 360°
⇒ r = 360° – 245° = 115°
(∵ Co-interior angles are supplementary)
⇒ 115° + q = 180°
⇒ q = 180° – 115° = 65°
(ii) Mark an angle t as shown then, t = 110°
(Alternate interior angles)
Now, p + t = 180°
(Linear pair angles)
(Corresponding angles)
⇒ p = 180° – 110° = 70° ⇒ s = t = 110°
B
(Alternate A
⇒ r = p = 70°
interior angles)
140°
(Corre⇒ q = p = 70°
70°
x
sponding angles)
ICSE Math Class VII
11
C
D
Question Bank
39. From the adjoining diagram find the value of x and complete the following :
(i) Reflex ∠ A
(ii) Reflex ∠ C
Ans. Here, 70° + 140° + x = 360° (Sum of angles at point)
210° + x = 360°
x = 360° – 210° = 150°
140° + ∠ A = 180°
⇒ ∠ A = 180° – 140° = 40°
Reflex ∠ A = 360° – 40 = 320°
Now, x + ∠ c = 180°
(∵ Co-interior angles are supplementary)
63°
x y z
(∵ Co-interior angles are supplementary)
Z
B
55°
∴
∠ c = 180° – 150° = 30°.
40. From the adjoining diagram, find the values of x, y and z.
z°
40°
y°
Ans. In the given diagram,
63° + 55° + x = 180° (Sum of angles of a triangle)
105°
x°
118° + x = 180° ⇒ x = 180° – 118° = 62°
Now, ∠ y = 55°
(Alternate interior angles)
Thus, x + y + z = 180°
(Angles on a line)
62° + 55° + z = 180° ⇒ 117° + z = 180° z = 180° – 117° = 63°
41. From the adjoining figure, find the values of x, y and z.
L
A
A
N
L IO
U AT
D
N
© E ER
T
IN
Ans. In the given figure, mark an angle p° as shown
z°
40°
y°
Here,
x° = 40°
(Alternate interior
angle)
Now,
p° + 105° = 180°
(Linear pair angles)
x°
P° 105°
p° = 180° – 105° = 75°
∴
Now, x° + y° + p° = 180° sum of angles of a triangle is 180°
40° + y° + 75° = 180° ⇒ y° + 115° = 180°
⇒
y° = 180° – 115° = 65°
⇒
A
B
Now, 40° + y° + z° = 180° (Angles on a
x
line)
E 35°
y
40° + 65° + z = 180°
⇒
48°
C
D
z = 180° – 105° = 75°
⇒
42. From the adjoining figure, find the values of x, y and z. Also find the reflex ∠ E.
Ans. x + 35° = 180° (∵ Co-interior angles are supplementary)
∴ x = 180° – 35° = 145°
= 48°VII
(Alternate interior
ICSE MathyClass
12
Questionangle)
Bank
Now, 48° + z = 180°
∴ z = 180° – 48° = 132°
Reflex ∠ E = 360° – (35° + 48°) = 360° – 83° = 277°.
(Linear pair angles)
x
y
125°
44. From the adjoining figure, find the values of x, y and z.
Ans.
x
y
125°
Here, z + 125° = 180°
(∵Co-interior angles are supplementary)
z = 180° – 125° = 55°
∴
y
Now, x = z = 55°
(Vertically opposite angles)
3x – 10° 2x + 15°
z
90° + z + y = 180° (Angles on a line)
90° + 55° + y = 180°
⇒
145° + y = 180° ⇒ y = 180° – 145° = 35°.
⇒
44. The adjoining diagram shows two intersecting lines.
Find the values of x, y and z.
Z
B
L
A
A
N
L IO
U AT
D
N
© E ER
T
IN
Ans. 3x – 10° = 2x + 15°
(Vertically opposite angles)
⇒ 3x – 2x = 15° + 10° ⇒ x = 25°
Now 3x – 10° + y = 180°
⇒ 75 – 10° + y = 180° ⇒ 65° + y = 180°, y = 180° – 65° = 115°
45. In each of the following figures AB || CD. Find the unknown angles giving reasons
y = z = x = 115°.
(i)
y°
A
E
G B
x°
120°
z°
G
x°
t°
C z
F
H
B
(ii)
D
E
A
D
50°
H
y°
t°
120°
F
C
Ans. (i) x + 100° = 180° ⇒ x = 180° – 100° = 80°
(Linear pair)
y = 120°
(vertically opposite angles)
(vertically opposite and co-interior angles)
⇒ z + 120° = 180°
⇒ z = 180° – 120° = 60°
AB || CD and GH is a transversal t + 100° = 180° = 180° – 100° = 80°
(ii) AB || CD and GH is a transveral x = 50°
ICSE Math Class VII
13
Question Bank
(exterior alternate angle)
⇒ x + z = 180° ⇒ 50° + z = 180°
(Linear pair)
⇒ z = 180° – 50° = 130
t = 120 = t
(vertically opposite angles)
y = 120°
(interior alternate angles)
46. In the given figure l || m and p || q. Find the angles x , y and z.
m
l
Ans. p || q and m is a transversal, x = 70° (alternate angles)
x + z = 180°, 70° + z = 180°
⇒
z = 180° – 70° = 110°
⇒
l || m and q is a transversal, y = x = 70° (alternate angles)
47. In the given figure, AB || CD || EF. Find x, y and z.
Ans. In the given figure,
y°
z°
p
x°
70°
Z
B
q
140°
x°
120°
y° D
z°
F
L
A
A
N
L IO
U AT
D
N
© E ER
T
IN
AB || CD || EF, ∠ BAD = 140°, ∠ ADF = 120°
∵ AB || DC and AD intersects them
B
A
C
E
∠ ADC = ∠ DAB ⇒ x° = 140°
(Alternate angles)
∴
(Angles at a point)
But ∠ ADF + ∠ FDC + ∠ ADC = 360°
120° + y° + 140° = 360°
⇒
260° + y° = 360°, y° = 360° – 260° = 100°
⇒
∵ DC || EF and DF intersects them
(co-interior angles)
∠ FDC + ∠ DFE = 180°
∴
100° + z = 180° ⇒ z = 180° – 100 = 80°
⇒
Hence,
x = 140°, y = 100° and z = 80°
48. In the given figure, AB || CD || EF and AD || BE. Find x, y and z.
Ans. In the given figure,
AB || CD || EF and AD || BE and ∠ DAB = 115°,
∠ GBE = 30°
∵ ∠ EBA + ∠ DAB = 180° (co-interior angles)
⇒ 30° + x° + 115° = 180° ⇒ x° + 145° = 180°
x = 180° – 154° = 35°
⇒
∵ AB || CD and AD intersects them
∠ ADC = ∠ BAD ⇒ y° = 115°
∴
∴ AB || EF and BE intersects them So, ∠ ABE = ∠ BEF
x + 30° = z = 35° + 30° = 65°
⇒
ICSE Math Class VII
14
A
115°
C
y°
D
x° B
30°
G
E
z°
F
(alternate angles)
(alternate angles)
Question Bank
49.
Ans.
50.
Ans.
51.
Hence, x = 35°, y = 115° and z = 65°
In the given figure l || m || n and
l
100° r
p || q || r. Find the angles x, y, z and t.
x° y° q
l || n and r is a transversal, x + 100° = 180°
z°h
p
t°
x = 180° – 100° = 80°
⇒
m
(sum of interior angles)
r || q and n is a transversal, then x = y = 80°
r || p and n is a transversal, then z = x = 80°
(alternate angles)
l || m and r is a transversal, then t = 100°
(alternate angles)
In the given figure, l || m and p || q. Find the angles x ,y , z and t. 270°
l
40°
x°
In the figure,
q
x + 40° + 270° = 360° (Angles at a point)
m
t° y°
⇒ x° = 360° – 40° – 270° = 50°
z°
p
(corresponding angles)
∵ l || m and q || p
r
z° = x° = 50°
(corresponding angles)
∵ and
But z° + t° = 180°
⇒ 50° + t° = 180°
⇒ t° = 180° – 50° = 130°
Hence, x = 50°, y = 40, z = 50° and t = 130°
In the given figure,
AB || CD. Find the angles x, y and z.
Z
B
L
A
A
N
L IO
U AT
D
N
© E ER
T
IN
Ans. In ∆ CEF, ∠ C + ∠ E + ∠ F = 180° (Sum of angles of a triangle)
⇒ 90° + 30° + x° = 180° ⇒ 120° + x° = 180°
⇒ x° = 180° – 120° = 60°, ∠ x° + y° + 80° = 180°
(angles on one side
E of a straight line)
⇒ 140° + y° = 180°
⇒ 60° + y° + 80° = 180°
B
A
30°
⇒ y° = 180° – 140° = 40° ⇒ AB || FD
z°
⇒ ∠ BAF + ∠ AFD = 180° ⇒ z° + 80° = 180°
y°
90° x° 80°
(interior angles)
C
F
D
⇒ z° = 180° – 80° = 100°
Hence, x = 60°, y = 40° and z = 100°.
52. Draw a line segment AB = 6.2 cm. Draw a perpendicular to it at a point on AB. By
using.
Q
Ans.
Given : A line segment AB = 6.2 and a point P on it.
Required : To draw a perpendicular to AB at the point P.
A
ICSE Math Class VII
15
C
P
D
B
Question Bank
Steps of construction :
(i) With P as centre and any suitable radius, draw an arc to cut the line AB at
points C and D.
 1

(ii) With C and D as centres, draw two arcs of equal radius  > CD  cutting each
 2

other at Q.
(iii) Join P and Q. Then QP is the required perpendicular to the line AB at the point
P.
53. Draw any triangle ABC. Through A, draw a line parallel to
Q A
BC by using
P
(i) ruler and compass
(ii) ruler and set square.
Ans. (i) Using ruler and compass
N
Given : To draw a line parallel to BC passing through A.
Z
B
B
M
C
L
A
A
N
L IO
U AT
D
N
© E ER
T
IN
Steps of Construction:
(i) With B as centre and any suitable radius draw an arc to meet BC at M and AB at
N.
(ii) With A as centre and same radius as in step I, draw an arc to meet AB at P.
(iii) With P as centre and radius equal to MN draw an arc to cut the previous arc at
Q.
(iv) Join A, Q and product either side, then AQ is the required line || to BC and
passing throught A.
(ii) Using ruler and set square.
Given : A triangle ABC.
Required : To draw a line || to BC and passing through A.
Steps of construction :
(i) Place your set square so that one of its shorter edge XY just lies along the line
BC
A REQUIRED LINE
(ii) Place your ruler so that one edge of the ruler lies
along the other shorter edge XZ of the set square.
Hold the ruler frimly and slide the set square along
the ruler until the edge XY of the set square passes
through A.
C
(iii) Draw the line along the edge XY of the set square. B
This is the required line parallel to the line BC and passing through A.
54. Draw a line segment PQ = 6 cm. By using ruler and compass, construct ∠ RPQ =
30° and ∠ RQP = 45°. From R draw altitude (perpendicular) to PQ.
Ans. Given : A line segment PQ = 6 cm.
ICSE Math Class VII
16
Question Bank
Required : To construct ∠ RPQ = 30° and ∠ RQP = 45°.
To draw a ⊥ ar to PQ from a point R outside the line PQ.
Steps of construction :
T
S
R
(i) Construct ∠ SPQ = 60° and
Bisect ∠ SPQ, then ∠ RPQ = 30°.
P
(ii) Construct ∠ PQT = 90°.
56.
Ans.
57.
Ans.
C
Y
6 cm
D Q
(iii) Bisect ∠ PQT, then ∠ RQP = 45°.
X
(iv) Now, with R as centre and any suitable radius draw an arc to cut the line PQ at
C and D.
 1

(v) With C and D as centres, draw two arcs of equal radius  > CD  cutting each
 2

other at X.
(vi) Join R and X to meet the line PQ at Y, then RY is the required ⊥ ar to the line
PQ from a point R.
Draw a circle with centre O and radius 4.5 cm. Draw a chord AB of lengths 4.5 cm.
Indicate by marking points X and Y, the minor arc AXB and the major arc AYB of the
X
circle. Shade the major segments of the circle AB.
A
Steps of construction :
B
(i) Take a point O
O 4.5 cm
(ii) With O as circle and radius 4.5 cm , draw a circle.
(iii) Draw a chord AB = 4.5 cm.
Y
(iv) This chord divides circle into two segments.
322
Draw a circle of radius 4 cm with C as its centre. Draw two radii CP and CQ such that
∠ PCQ = 45° Shade the minor sector of the circle.
Steps of construction :
(i) Draw a circle with centre C and radius CP = 4 cm.
F
E L Q
(ii) With C as centre and any radius, draw an arc to meet CP a
B
D N
A.
45°
P
C MA
(iii) With A as centre and same radius (as in step 2) draw an arc
to meet the previous arc at B. With B as centre and same
radius, draw an arc to meet the first arc at D.
Z
B
L
A
A
N
L IO
U AT
D
N
© E ER
T
IN
 1

(iv) With B and D as centres draw two arcs of equal radius  > AB  cutting each
 2

ICSE Math Class VII
17
Question Bank
other at E.
(v) Join C, E and produce it to F, then ∠ PCF. = 90°.
(vi) Now, bisect ∠ PCF.
(vii) With C as centre and any radius draw an arc to meet CP at M and CF at N.
(viii) Now with M and N as centres, draw two arcs of equal radius (greater than
MN ) cutting each other at L.
(ix) Join CL and produce it to meet the circle at Q then ∠ PCQ = 45°.
(x) The minor sector of the circle is shown, shaded in the diagram.
1
2
Z
B
L
A
A
N
L IO
U AT
D
N
© E ER
T
IN
ICSE Math Class VII
18
Question Bank