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Published on: 30/07/2018
Some of the important questions are prepared from this chapter Lines and Angles.
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1.
In figure lines AB and CD intersect at O.if \(\frac { \angle AOD }{ \angle DOB } =\frac { 4 }{ 5 } \angle COB\)
\(80^{ 0 }\)
\(100^{ 0 }\)
\(90^{ 0 }\)
\(70^{ 0 }\)
2.
In figure value of x is
\(30^{ 0 }\)
\(40^{ 0 }\)
\(30^{ 0 }\)
\(50^{ 0 }\)
3.
In the given figure the value of x which makes POQ a straight line is:
35
30
25
40
4.
The value of x in figure is:
\(80^{ 0 }\)
\(20^{ 0 }\)
\(25^{ 0 }\)
\(40^{ 0 }\)
5.
Which one of the following pairs is not a pair of adjacent angles?
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6.
The angle complementary to \(90^{ 0 }\)-\(9^{ 0 }\) is
\(90^{ 0 }\)+\(9^{ 0 }\)
\(9^{ 0 }\)
\(180^{ 0 }\)-\(9^{ 0 }\)
\(360^{ 0 }\)-\(9^{ 0 }\)
7.
An angle which measures between \(0^{ 0 }\) and \(90^{ 0 }\) is called
a straight angle
an acute angle
a right angle
an obtuse angle
8.
In how many points two intersecting lines intersect?
4
3
2
1
9.
In figure \(\angle \)PQR =\(\angle \)PRQ then prove that \(\angle \) PQS =\(\angle \)PRT
10.
In the figure below \(l_{ 1 }||l_{ 2 }\) and \(a_{ 1 }||a_{ 2 }\) find the value of x.
11.
In figure OP bisects \(\angle \) AOC,OQ bisects \(\angle \) BOC and OP \(\bot \) OQ Show that points A O and B are collinear.
12.
In the given figure , two straight lines PQ and RS intersect each other at O
If \(\angle \)POT =\(75^{ 0 }\) Find the values of a,b,c
13.
In figure, if y=\(20^{ 0 }\) , prove that the line AOB is a straight line.
14.
Ray OE bisects \(\angle \)AOB and OF the ray opposite to OE Show that \(\angle \)FOB=\(\angle \)FOA
15.
In figure given below l || m Before of RQB and DRQ intersect at P find the measure of RPQ
16.
(i) In figure ,AO \(\bot \)OB Find \(\angle \)AOC and \(\angle \) BOC
.png)
(ii) In figure,\(\angle \) AOB ;\(\angle \)BOC=2:3
If \(\angle \)AOC=\(75^{ 0 }\) then find the measure of ,\(\angle \) AOB ;\(\angle \)BOC
.png)
17.
In figure OA ,OB are opposite rays and \(\angle \)AOC +\(\angle \)BOD=\(90^{ 0 }\) Find \(\angle \) COD
18.
Find the value of x and y in the figure
19.
In the given figure a is greater than b, by \(\frac { 1 }{ 6 } \) th of a straight angle Find the angles of a and b.
20.
Find the supplement of \(\frac { 4 }{ 3 } \) of right angle
1.
\((\angle AOD+\angle DOB=180^{ 0 }\)
\(4+5=9\)
\(\therefore \angle AOD=\frac { 4 }{ 9 } X180^{ 0 }=80^{ 0 }\)
\(\angle COB= 80^0\)
2.
\(4x+2x=180^{ 0 }\)
3.
\(4x+(2x+30)=180^{ 0 }\)
4.
\(5x+4x=180^{ 0 }\)
5.
\(\angle COA\) and \(\angle BOA\) are not a pair of adjacent angles
6.
Required angle=\(180^{ 0 }\)-(\(90^{ 0 }\)+\(9^{ 0 }\)) = \(90^{ 0 }\)+\(9^{ 0 }\)
7.
Definition of an acute angle
8.
(d)
1
9.
\(\therefore \)Ray QP stands on line ST
\(\therefore \) \(\angle \)PQS +\(\angle \)PQR = \(180^{ 0 }\)
|Linear Pair Axiom
Ray RP stands on line ST
\(\therefore \) \(\angle \)PRQ +\(\angle \)PRT= \(180^{ 0 }\)
| Linear Pair Axiom
From (1) and (2) we obtain
\(\angle \)PQS+\(\angle \)PQR =\(\angle \)PRQ +\(\angle \)PRT
\(\Rightarrow \)\(\angle \) PQS =\(\angle \)PRT
\(\therefore \) \(\angle \)PQR=\(\angle \)PRQ(Given)
10.
\(\angle \)1=4x+15 | Corresponding angles
2x=180-\(\angle \)1 |Corresponding angles
\(\Rightarrow \) 2x=\(180^{ 0 }\) -(4x+15)
\(\Rightarrow \) 2x=165-4x
\(\Rightarrow \) 6x=165 \(\Rightarrow \)\(x=\frac { 165 }{ 6 } =27\frac { 1^{ 0 } }{ 2 } \)
11.
Given OP bisects \(\angle \) AOC,OQ bisects\(\angle \) BOC and OP \(\bot \) OQ
To prove the points A,O and B are collinear.
\(\therefore \) OP bisects \(\angle \) AOC
\(\therefore \) \(\angle \)AOP= \(\angle \)COP ...(1)
\(\therefore \angle \)OQ bisects \(\angle \) BOC
\(\therefore \angle \)BOQ=\(\angle \)COQ ...(2)
Now ,\(\angle \)AOB
=\(\angle \)AOP+\(\angle \)COP+\(\angle \)COQ+\(\angle \)BOQ
=\(\angle \)COP+\(\angle \)COP+\(\angle \)COQ+\(\angle \)COQ
|From (1) and (2)
=2(\(\angle \)COP+\(\angle \) COQ)
\(2\angle \)POQ
2(\(90^{ 0 }\)) |\(\because OP\bot OQ\)
=\(180^{ 0 }\)
\(\therefore \) The points A, O and B are collinear
| By converse of Linear Pair Axiom.
12.
\(\therefore \) ROS is a line
\(\therefore \) 4b+\(75^{ 0 }\)+b=\(180^{ 0 }\)
\(\Rightarrow 5b=180^{ 0 }-75^{ 0 }=105^{ 0 }\)
\(\Rightarrow \) b=\(\frac { 105^{ 0 } }{ 5 } =21^{ 0 }\)
2c=\(75^{ 0 }\)+b
|Vertically opposite angles
\(\Rightarrow 2c=75^{ 0 }+21^{ 0 }\)
\(\Rightarrow 2c=96^{ 0 }\)
\(\Rightarrow \)\(c=\frac { 96^{ 0 } }{ 2 } =48^{ 0 }\)
a=4b
|Vertically opposite angkes
\(\Rightarrow a=4X21^{ 0 }=84^{ 0 }\)
Thus \(a=84^{ 0 },b=21^{ 0 },c=48^{ 0 }\)
13.
\(\because \) Sum of all the angle round a point is equal to \(360^{ 0 }\)
\(\therefore \) y+(3x-15)+(y+15)+2y+(4y+10)+x=\(360^{ 0 }\)
\(\Rightarrow \) 4x+8y=\(360^{ 0 }\)
\(\Rightarrow \) x+2y=\(90^{ 0 }\)
\(\Rightarrow \) x+2(\(20^{ 0 }\))=\(90^{ 0 }\)
\(\Rightarrow \) x+\(40^{ 0 }\)=\(90^{ 0 }\)
\(\Rightarrow x=50^{ 0 }\)
Now, y+3x-15+y+5=3x+2y-10
= 3(\(50^{ 0 }\))+2(\(20^{ 0 }\))-10
\(=150^{ 0 }+40^{ 0 }-10^{ 0 }\)
= \(180^{ 0 }\)
\(\therefore \) AOB is straight line.
14.
\(\angle \)FOB+\(\angle \)BOE=\(180^{ 0 }\) ...(1)
| Linear pair Axiom
\(\angle \)FOA+\(\angle \)AOE=\(180^{ 0 }\) .. (2)
| Linear Pair Axiom
From (1) and (2)
\(\angle \)FOB+\(\angle \)BOE=\(\angle \) FOA+\(\angle \)AOE ....(3)
But \(\angle \)BOE=\(\angle \)AOE
\(\therefore \) From (3)
\(\Rightarrow \) \(\angle \)FOB=\(\angle \)FOA
15.
\(90^{ 0 }\)
16.
Let \(\angle AOB=2x\) and \(\angle BOC=3x\)
\(\angle AOB+\angle BOC=\angle AOC\)
2x+3x=75o
\(\Rightarrow x=\frac{75^o}{5}=15^o\)
\(\angle AOB=2x=30^o\) and \(\angle BOC=3x=45^o\)
17.
\(90^{ 0 }\)
18.
X=\(35^{ 0 }\)
y=\(105^{ 0 }\)
19.
Given,
\(a-b=\frac{1}{6}\times180^o\)
a-b=30o ...(1)
a+b=180o (Linear pair)...(2)
Adding (1) & (2), 2a=210o
a=105o
b=180o-a=180o-105o
=75o
20.
\(\frac{4}{3}\) of a right angle=\(\frac{4}{3}\times90^o=120^o\)
(Sum of supplementary angles is 180o)
Supplement of 120o=180o-120o=60o
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