11
LINES AND ANGLES
1. In the adjoining figure, write
(i) all pairs of parallel lines. Ans. l, m ; l, n ; m, n.
(ii) all pairs of intersecting lines.
Ans. l, p ; m, p ; n, p; l, q ; m, q ; n, q ; l, r ; m, r ; n, r ; p, r ; q, r.
(iii) concurent lines. Ans. n, r, q; l, r, p
r q
p
G
n
F m
E
A
B
H
I
C
D
(iv) collinear points. Ans. A, B, C ; A, H I, D ; D, E, F, G ;
C, I, E ; B, H, F.
A
B
E
2. With the help of a figure, find the maximum and minimum
number of points of intersection of four lines in a plane.
D
C
Ans. Six points; zero.
m
3. From the adjoining figure, write
n
C
(i) all pairs of intersecting lines.
B
A
Ans. (l, n); (l, p); (m, n); (m, p); (n, p)
E
(ii) one set of non-collinear points. Ans. {A, B, E}
D
P
(iii) all sets of collinear points. Ans. {A, B, C} and {A, E, D}
4. From the adjoining figure write
p
q
(i) collinear points. Ans. A, D, C ; B, D, E
m
D
A
(ii) concurrent lines and their points of concurrence.
n
E
B
l
C
Ans. l, n, p at point B and m, p, q at point A
5. Look at the following figures and count the number of triangles in each case.
(i)
(ii)
(iii)
(iv)
(v)
(vi)
Ans. (i) 6 (ii) 5 (iii) 8 (iv) 6 (v) 12 (vi) 8
1
Class - VI Mathematics
Question Bank
6. Look at the following figures and count the number of rectangles in each case :
(i)
(ii)
(iii)
Ans. (i) 3 (ii) 6 (iii) 18
7. (A) Add the following measures :
(i) 23° 30′ 45′′ and 32° 35′ 48′′
Ans. 45′′ + 48′′ = 93′′ = 60′′ + 33′′ = 1′ 33′′
Now, 30′ + 35′ + 1′ 33′′ = 66′ 33′′ = 1° 6′ 33′′
Now, 23° + 32° + 1° 6′ 33′′ = 56° 6′ 33′′
∴ 23° 30′ 45′′ + 32° 35′ 48′′ = 56° 6′ 33′′
(ii) 41° 24′ 56′′ and 39° 36′ 24′′
Ans. 56′′ + 24′′ = 80′′ = 60′′ + 20′′ = 1′ 20′′
Now, 24′ + 36′ + 1′ 20′′ = 61′ 20′′ = 1° 1′ 20′′
Now, 41° + 39° + 1° 1′ 20′′ = 81° 1′ 20′′
∴ 41° 24′ 56′′ + 39° 36′ 24′′ = 81° 1′ 20′′
(iii) 37° 52′ 55′′ and 16° 29′ 36′′
55′ + 36′′ = 91′′ = 60′′ + 31′′ = 1′ 31′′
Ans.
Now, 52′ + 29′ + 1′ 31′′ = 82′ 31′′ = 1° 22′ 31′′
Now, 37° + 16° + 1° 22′ 31′′ = 54° 22′ 31′′
∴ 37° 52′ 55′′ + 16° 29′ 36′′ = 54° 22′ 31′′
(B) Subtract the following measures :
(i) 25° 30′ 30′′ form 57° 15′ 28′′
57° 15′ 28′′ − 25° 30′ 30′′
Ans.
56° 74′ 88′′ − 25° 30′ 30′′ = 31° 44′ 58′′
(ii) 9° 36′ 52′′ from 16°
16° − 9° 36′ 52′′ = 15° 60′ − 9° 36′ 52′′
Ans.
= 15° 59′ 60′′ − 9° 36′ 52′′ = 6° 23′ 8′′
8. (A) Write the complement angles of :
1
of 80°
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Class - VI Mathematics
(i)
(ii)
2
of 70°
5
2
Question Bank
1

1

Ans. (i) Complement angle of  of 80°  = 90° −  of 80°  = 90° – 20° = 70°
4

4

2

2

(ii) Complement angle of  of 70°  = 90° −  of 70° 
5

5

= 90° – 28° = 62°
(B) Write the supplement angles of :
(i)
1
of 150°
3
(ii)
3
of 160°
4
1

(i) Supplement angle of  of 150° 
3

1

= 180° −  of 150°  = 180° – 50°
3

= 130°
3
3

of 160 = 180° −  of 160° 
4
4

= 180° – 120° = 60°
9. Two complementary angles are in the ratio 7 : 8. Find the angles.
Ans. Let two complementary angles are 7x and 8x
∴ 7x + 8x = 90° ⇒ 15x = 90°
(ii) Supplement of
90°
= 6°
15
∴ Two complementary angles are 7x = 7 × 6° = 42° and 8x = 8 × 6° = 48°
10. Two supplementary angles are in the ratio 7 : 11. Find the angles.
Ans. Let two supplementary angles are 7x and 11 x ∴ 7x + 11x = 180°
∴
x=
180
= 10°
18
∴ Two supplementary angles are 7x = 7 × 10° = 70° and 11x = 11 × 10° = 110°.
11. The measures of two supplementary angles are (3x + 15)° and (2x + 5)°. Find the
value of x.
Ans. We know that, sum of two supplementary angles = 180°
∴ (3x + 15)° + (2x + 5)° = 180°
⇒ 3x + 2x +15 + 5 = 180° ⇒ 5x° + 20° = 180°
⇒ 5x = 180° – 20° = 5x = 160°
⇒ 18x = 180° ⇒ x =
160
∴x=
= 32°
5
Class - VI Mathematics
3
Question Bank
12. If two angles of measures (x + 3)° and (2x – 9)° are complementary, find the value
of x.
Ans. (x + 3)° + (2x –9)° = 90° ⇒ x° + 3 + 2x – 9 = 90°
⇒ 3x = 90 + 6 = 96
⇒ x=
96
= 32°
3
Thus, the value of x = 32°
13. If two angles of measures (x + 27)° and (3x – 39)° are supplementary, find the
value of x.
Ans. (x + 27)° + (3x – 39)° = 180° ⇒ x + 27 + 3x – 39 = 180°
⇒ 4x – 12 = 180 ⇒ 4x = 192
192
= 48°
4
Thus, the value of x is 48°.
14. Find the measure of an angle which is 36° more than its complement.
Ans. Let angle be ‘x’. then x = 90 – x + 36
⇒
x + x = 90 + 36 ⇒ 2x = 126
∴
x=
126
a = 63°
2
The required angle is = 63°
∴
15. Find the measure of an angle which is 25° less than its supplement.
Ans. Let the angle be the ‘x’. x = 180 – x – 25
x + x = 180 – 25 ⇒ 2x = 155
∴
x=
155
⇒
2
1°
2
1°
∴
The required angle = 77
2
16. In the adjoining figure, AOB is a straight line. Find the value of x. Hence, find
∠ AOC and ∠ BOD.
C
⇒
x=
x = 77
D
Ans. ∠ BOD + ∠ COD + ∠ AOC = 180°
55°
x + 20 + 55 + 3x – 5 = 180 ⇒ 4x + 70 = 180
(3 x – 5)°
(x + 20 )°
O
⇒ 4x = 180 – 70
⇒ = 110
A
B
110
1
x=
= 27
⇒
4
2
165 − 10
1°
1
55
165
= 77
∠ AOC = 3 x − 5 = 3 × 27 − 5 = 3 × − 5 =
−5 =
2
2
2
2
2
Question Bank
4
Class - VI Mathematics
1
∠ BOD = x + 20 = 27 + 20
2
55
55 + 40 95
1°
+ 20 =
=
= 47
2
2
2
2
17. In the adjoining figure, AOB is a straight line. Find the value of x.Hence find
∠ AOC , ∠ COD and ∠ BOD.
C
=
Ans. ∠ BOD + ∠ COD + ∠AOC = 180° (straight line)
x + 2x – 19 + 3x + 7 = 180
⇒ 6x – 12 = 180 ⇒ 6x = 180 + 12
∴
⇒
∴
(2 x
(3 x + 7 )°
)°
– 19 D
x°
O
A
B
192
= 32°
6
∠ AOC = 3x + 7 = 3 × 32 + 7 = 96 + 7 = 103°
∠ COD = 2x – 19 = 2 × 32 – 19 = 64 – 19 = 45° and ∠ BOD = x = 32°
18. Find the value of x in each of the figures given below :
(Complete angle)
Ans. (i) ∠ AOB + ∠ BOC + ∠ COD + ∠ DOE + ∠ EOA = 360°
50° + x + 80° + 85° + 70° = 360° ⇒ x + 285° = 360°
x = 360° – 285° = 75°
⇒
(ii) ∠ POQ + ∠ QOR + ∠ ROS + ∠ SOP = 360°
(Complete angle)
R
B
C
x°
80°
85°
Q
50°
70°
D
2 x°
O
x°
3 x°
4 x°
A
P
S
E
(i )
( ii )
x + 2x + 3x + 4x = 360° ⇒ 10x = 360°
360
x=
= 36°
⇒
10
19. In the adjoining figure, AOB is a straight line. Calculate ∠ COE.
Ans. ∠ AOB = 180° = ∠ COB + ∠ COD + ∠ DOE + ∠ EOA
⇒ ∠ AOB = x + x + y + y
⇒ 180 = 2x + 2y
⇒ 180 = 2(x + y )
180
= (x + y)
⇒
⇒ ∠ COE = 90°
2
5
Class - VI Mathematics
D
E
A
C
y
y x
O
x
B
Question Bank
20. Find the value of x in each of the following diagrams :
x
41°
(i)
x
96°
105°
130°
(ii)
120°
2x
Ans.(i) Sum of angles at a pointt = 360° = 130° + 105° + 41° + x
360° = 276° + x ⇒
84° = x
⇒
(ii) Sum of ∠ s at a point = 360° = 96° + 120° + 2x + x + 90°
360° – 306° = 3x
54° = 3x
⇒
⇒
54
=x
18 = x
⇒
3
⇒
x = 18°
21. At 2 o′ clock the hands of a clock make an acute and a reflex angle. Calculate the
size, in degrees, of each of these angles.
[vertically opp. ∠s]
⇒
2
2
of a turn. = 360 × = 60°
Ans. The acute angle =
12
12
Thus, reflex angle = 360° – 60° = 300°
11 12
1
2
10
9
8
3
4
7
6
5
22. In the given figure, l || m and p and q are two transversals. Find the values of x, y,
z and w.
q
p
Ans. x = 65°
(vertically opp. angles)
y = 65°
(Interior alternate angles)
120 °
x°
l
w = 120°
(Exterior alternate angles)
65°
z + w = 180°
(Linear pair of angles)
y°
m
z + 120° = 180°
w°
⇒
z = 180° – 120° ⇒ z = 60°
23. In the given figure, l || m, p || q. Find the values of x, y and z.
p
Ans. y = 110°
(Interior alternate angles)
q
x + y = 180° (allied angles)
1 10 °
⇒
x + 110° = 180°
l
y°
x = 180° – 110° = 70°
∴
z°
m
Now, z = 180° – x(Interior alternate angles)
x°
⇒
= 180° – 70° = 110°
Question Bank
6
Class - VI Mathematics
24. In the given figure, DE || BC. Find the values of x, y and z.
Ans. In ∆ ABC , ∠ BAC + ∠ ABC + ∠ ACB = 180°
⇒
⇒
D
E
A
x z
y
z + 65° + 45° = 180° ⇒ z + 110° = 180°
z = 180° – 110° = 70°
∠ DAB = ∠ ABC ⇒ x = 65°
(Alternate angles)
B
Similarly, ∠ EAC = ∠ ACB ⇒ y = 45°
y=
45°
C
(Alternate angles)
25. In the given diagram, l || m. Find x and y if x = 2y.
Ans. y = 180° – x
x + y = 180°
⇒
⇒ 2y + y = 180° ⇒ 3y = 180°
⇒
65°
(Interior alternate angles)
(∵ x = 2y given)
180°
= 60°
3
Now, x + y = 180°
x + 60° = 180°
⇒
x = 180° – 60° = 120°
∴
26. Calculate the size of each lettered angle in the following sketches :
x
z
x y
(i)
y
z
60°
38°
(ii)
45°
Ans. (i) ∠x = 38° (alternate angles)
∠y = 60° − 38° = 22°
55°
(Property of Exterior angle)
∠z = 180° − (38° + 22°) = 120°
(ii) ∠x = 45°
(alternate angles)
∠y = 180° − (45° + 55°) = 180° − 100° = 80°
(alternate angles)
∠z = 55°
27. From the adjoining diagram, calculate the values of a, b, c and d.
(corresponding angles)
Ans. ∠a = 180° − 110° = 70°
110 °
a
b
∠b = 180° − 70° = 110° (linear pair)
(corresponding angles)
∠c = ∠a = 70°
∠d = ∠b = 110° (alternate angles)
d c
Class - VI Mathematics
7
Question Bank
28. In the adjoining figure, lines l, m, and n are parallel. Calculate the values of x, y and
z. Hence find the (i)obtuse angle ABC, (ii) reflex angle ABC.
A
Ans.
(alternate angles)
∠x = 52°
∠y = 180° – 105° = 75°
n
52°
(co-interior angles)
z
x
m
∠z = 180° – 52° = 128°
B y
(i) ∠ABC = ∠x + ∠y = 52 + 75 = 127°
105°
l
C
(ii) reflex ∠ABC = 360 – 127 = 233°
29. In the adjoining figure, FG || AB, FG || CD. If ∠ABE = 120° and ∠DCE = 100°,
find the value of x.
Ans. ∠ ABE + ∠ BEF = 180°
(pair of allied angles)
120° + ∠ BEF = 180°
⇒
A
100°
120°
⇒
∠ BEF = 180° − 120° ∠ BEF = 60°
⇒
∠ DCE + ∠CEG = 180° ⇒ 100° + ∠CEG = 180°
∠CEG = 180° –100° = 80°
∠ BEC + ∠BEF + ∠ CEG = 180°
D
C
B
x
E
F
G
⇒
x + 60° + 80° = 180° ⇒ x + 140° = 180°
⇒
x = 180° – 140° = 40°
30. Two parallel lines l and m are cut by a transversal p. If the interior angles on the
same side of p be (3x – 13)° and (5x – 15)°, find the measure of each of these
angles.
Ans. ∠ LRS + ∠ RSM = 180°
⇒
⇒
⇒
R
3x – 13 + 5x – 15 = 180
8x – 28 = 180
8x = 180 + 28
(3 x – 13 )°
L
(5 x – 15 )°
S
208
x=
= 26°
8
∴
p
(pair of allied angles)
m
R
M
∠ LRS = (3x – 13)° = 3 × 26 –13 = 78 – 13 = 65°
∠ RSM = (5x – 15)° = 5 × 26 – 15
= 130 – 15 = 115°
Class - VI Mathematics
8
l
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