| |
|
|
| |
|
Question (1):
Can two vertically opposite angles be supplement to each other? Draw a diagram to illustrate your answer.
|
Answer:
Let AB and CD bisect each other at right angles as shown in figure.
 
 
( Vertically opposite angles can be
supplements to each other)
|
|
Question (1):
|
Answer:
x and y
 146 + y = 180
 y = 180 - 146
= 340
 (vertically opposite angles)
2x
= 34
 x = 17o
|
|
Question (3):
|
Answer:
5x + 10 = 90o
5x = 90o - 10o
= 80o
|
|
Question (4):
In the following diagrams parallel lines are marked by arrows in the same direction. Transversal is also drawn. Find the missing angles and provide the reason for each.
Note: (No proof is required but the essential steps of working must be given)
 |
Answer:
(i) Find a, b, c, d, e, f, g.
(ii) In the adjoining diagram AB||CD; PQ and RS are transversals. Find angles 1, 2, .....16.
|
|
Question (5):
If two straight lines are each perpendicular to a third straight line then they are parallel to each other. Prove.
|
Answer:
Straight line AB and CD are perpendicular to line l.
AB||CD
|
|
Question (6):
|
Answer:
From (i), (ii) and (iii)
 are supplementary angles.
|
|
Question (7):
|
Answer:
AB||CD, BC||DE
AB||CD (given)
BC is a transversal
BC||DE (given)
DC is a transversal
|
|
Question (8):
In the figure, show that AB||EF.
|
Answer:
 [given]
 
i.e., 
|
|
Question (9):
If a line is perpendicular to one of two given parallel lines, show that it is also perpendicular to the other line.
|
Answer:
 [corresponding angles]
But,
 
|
|
Question (10):
are the bisectors of the corresponding angles BGE and DHE respectively.
Show that
|
Answer:
angles BGE and DHE respectively.
i)
|
|
Question (11):
Two lines 'l', 'm' are perpendicular to the lines 'n', 'p'. If 'n' and 'p' intersect, prove that 'l' and 'm' must also intersect.
|
Answer:
l ^ 'n' and 'm' ^ 'p' [given]
Let 'n' and 'p' intersect at 'A'
' l' and 'm' intersect.
Let us assume that 'l' and 'm' do not intersect. Then 'l' must be parallel to 'm'.
But l ^ n [given]
\m ^ n ..... (i)
now m ^ p ..... (ii)
from (i) and (ii) we conclude that n||p.
But it is given that 'n' and 'p' are intersecting lines. So our assumption that 'l' and 'm' do not intersect is wrong.
Hence 'l' and 'm' intersect.
|
|
Question (12):
In the given figure 'm' and 'n' are two plane mirrors parallel to 
|
Answer:
m||n [given]
|
|
Question (13):
|
Answer:
 (Angle sum property)
|
|
Question (14):
The sum of two angles of a triangle is 80o and their difference is 20o. Find all the angles of the triangle.
|
Answer:
Let the two angles of the triangle be xo and yo.
Then xo+ yo= 80o ...(i)
xo- yo= 20o ...(ii)
From equations (i) + (ii) we get 2xo= 100o
 
xo= 50o
Substituting xo= 50o in (i)
xo+ yo= 80o
50o+ yo= 80o
yo = 80o- 50o
yo = 30o
The two angles of the triangle are 50o and 30o.
The third angle = 180o - (50o +30o) (sum of angles in a triangle is 180o)
= 180o - 80o
= 100o
|
|
Question (15):
|
Answer:
 [angle sum property]
 ...(i)
 ...(ii)
|
|
Question (16):
|
Answer:
Thus the three angles are
|
|
Question (17):
An exterior angle of a triangle is 110o and one of the interior opposite angles is 30o. Find the other two angles of the triangle.
|
Answer:
 
= 80o
 
= 70o
|
|
Question (18):
|
Answer:
 
 [ Sum of angle of a triangle = 180o]
|
|
Question (19):
|
Answer:
Adding (i), (ii) and (iii)
=360o
 
|
|
Question (20):
|
Answer:
Multiplying both sides by 2.
Subtracting (i) from (ii)
|
|
Question (21):
Three angles of a quadrilateral are 100o, 48o and 92o. Find the 4th angle.
|
Answer:
quadrilateral is 360o]
=120o
The 4th angle measures 120o.
|
|
Question (22):
|
Answer:
Adding (i) and (ii)
 ...(iii)
 ...(iv) [  sum of the angles of a
quadrilateral equals 360o]
Equating (iii) and (iv)
 
|
|
Question (23):
ABCDE is a regular pentagon. The bisectors of  of the pentagon meets
[Hint: Each angle of a regular pentagon is 108o]
|
Answer:
ABCDE is a regular pentagon.
Number of sides of the regular pentagon, n=5.
 Each interior angle
=108o
=54o
In quadrilateral AMCD,
|
|
Question (24):
In the figure, lines 'l' and 'm' intersect at O, forming angles as shown in the figure. If x=45o, find the values of y, z and u.
|
Answer:
xo= 45o [given]
 zo= 45o [ vertically opposite angles]
xo = yo = 180o [Linear pair]
45o + yo = 180o
yo = 180o - 45o
yo = 135o
If yo = 135o, then xo = 135o [ vertically opposite angles]
Hence the values are: y=135o, z=45o and x=135o
|
|
Question (25):
In the adjoining figure, determine the value of y.
|
Answer:
ao = 5yo [ vertically opposite angles]
Now 5yo + ao + 2yo = 180o
i.e., 5yo + 5yo + 2yo = 180o [  Linear pair]
12yo = 180o
yo = 15o
|
|
Question (26):
In the adjoining figure, three coplanar lines intersect at a common point, forming angles as shown. Given a=50o and b=90o. Find the values of c, d, e and f.
|
Answer:
d = 50o [ ao=50o, vertically opposite angles]
e = 90o [ bo=90o, vertically opposite angles]
b + c + d = 180o [  They form a linear pair]
90 + c + 50 = 180o
140o+ c = 180o
c = 40o
Similarly a + e + f = 180o [ They form a linear pair]
50o+ 90o+ f = 180o
140o+ f = 180o
f = 40o [ c=40o, vertically opposite angles]
Hence c = 40o; d = 50o; e = 90o and f = 40o
|
|
