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Published on: 09/12/2019
Lines and Angles
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Questions + Answers key
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1.
The length of the common perpendiculars at different points on parallel lines is the same and is called
the distance between the parallel lines
the altitude
the median
None of these
2.
Which of the following is not a pair of complementary angles?
\(60^{ 0 },30^{ 0 }\)
\(56^{ 0 },34^{ 0 }\)
\(0^{ 0 },90^{ 0 }\)
\(150^{ 0 },30^{ 0 }\)
3.
The angle of supplementary to \(90^{ 0 }\)+\(9^{ 0 }\) is
\(90^{ 0 }\)+\(9^{ 0 }\)
\(90^{ 0 }\)-\(9^{ 0 }\)
\(180^{ 0 }\)-+\(9^{ 0 }\)
\(180^{ 0 }\)-\(9^{ 0 }\)
4.
An right angle
measures between \(0^{ 0 }\) and \(90^{ 0 }\)
is exactly equal to \(90^{ 0 }\)
is greater than \(90^{ 0 }\) but less than \(180^{ 0 }\)
is equal to \(180^{ 0 }\)
5.
An obtuse angle
measures between \(0^{ 0 }\) and \(90^{ 0 }\)
is greater than \(90^{ 0 }\) but less than \(180^{ 0 }\)
is exactly equal to \(90^{ 0 }\)
is exactly equal to \(180^{ 0 }\)
6.
In figure if AB || CD then find the value of y.
7.
In the figure below \(l_{ 1 }||l_{ 2 }\) and \(a_{ 1 }||a_{ 2 }\) find the value of x.
8.
In figure OP bisects \(\angle \) AOC,OQ bisects \(\angle \) BOC and OP \(\bot \) OQ Show that points A O and B are collinear.
9.
In figure if AOB is a line OP bisects \(\angle \)BOC and OQ \(\angle \) AOC ,show that \(\angle \)POQ is a right angle
10.
If two lines intersect each other, then the vertically opposite angles are equal. prove it
11.
In the given figure if l || m then find the value of x.
12.
In the given figure find x,if AB || CD
13.
Lines PQ and Rs Intersect each other at O (see figure) If \(\angle \)POR;\(\angle \)ROQ=3:7 Find all the angles a,b,c and d.
14.
In figure \(\angle \) DOB =\(87^{ 0 }\) and \(\angle \) COA =\(82^{ 0 }\) If \(\angle \)BOA \(35^{ 0 }\) and Find \(\angle \) COB and \(\angle \)COD
15.
(a) In the figure , what value of x will make POQ a straight line:
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(b)In the given figure find the value of x,If AOB is a line
.png)
16.
In the given figure, l||m||n. From the figure, find the ratio of (x+y):(y-x).
17.
In the given figure, \(PO\bot AB\), If x:y:z=1:3:5, then find the degree measure of x,y and z.
1.
(a)
the distance between the parallel lines
2.
\(150^{ 0 },30^{ 0 }=180^{ 0 }\neq 90^{ 0 }\)
3.
Required angle=\(180^{ 0 }\)-(\(90^{ 0 }\)+\(9^{ 0 }\)) =\(90^{ 0 }\)+\(9^{ 0 }\)
4.
Definition of an right angle
5.
Definition of an obtuse angle
6.
Through O draw OE || AB || CD
Now y= \(\angle FOG\)
=\(\angle \)FOE + \(\angle GOE\)
=\(\angle \)CFO+\(\angle \) AGO
=\(\angle \)FOE ==\(\angle \) CFO (Alternate Interior angles)
=\(\angle \)GOE=\(\angle \)AGO (Alternate Interior Angles)
=\(45^{ 0 }\)+\(40^{ 0 }\)=\(85^{ 0 }\)
7.
\(\angle \)1=4x+15 | Corresponding angles
2x=180-\(\angle \)1 |Corresponding angles
\(\Rightarrow \) 2x=\(180^{ 0 }\) -(4x+15)
\(\Rightarrow \) 2x=165-4x
\(\Rightarrow \) 6x=165 \(\Rightarrow \)\(x=\frac { 165 }{ 6 } =27\frac { 1^{ 0 } }{ 2 } \)
8.
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.
9.
\(\angle \)POQ=\(\angle \)POC+\(\angle \)COQ
=\(\frac { 1 }{ 2 } \angle BOC+\frac { 1 }{ 2 } \angle AOC\)
\(\because \) OP bisects \(\angle \)BOC
\(\therefore \angle POC=\frac { 1 }{ 2 } \angle BOC\)
\(\therefore \) OQ bisects \(\angle \)AOC
\(\therefore \angle COQ=\frac { 1 }{ 2 } \angle AOC\)
\(\frac { 1 }{ 2 } (\angle BOC+\angle AOC)\)
\(\frac { 1 }{ 2 } (180^{ 0 })\)
|Linear Pair Axiom
=\(90^{ 0 }\)
10.
Let AB and CD two lines intersecting at O
This leads to two pairs of vertically opposite angles,namely
(i) \(\angle \)AOC and \(\angle \)BOD
(II) \(\angle \)AOD and \(\angle \)BOC
We are to prove that
(i) \(\angle \)AOC and \(\angle \)BOD
(II) \(\angle \)AOD and \(\angle \)BOC
\(\therefore \) Ray OA stands on line CD
Therefore (i) \(\angle \)AOC and \(\angle \)AOD=\(180^{ 0 }\)
|Linear Pair Axiom
From (1) and (2)
\(\angle \)AOC and \(\angle \)AOD= \(\angle \)AOD and \(\angle \)BOD \(\angle \)AOC and \(\angle \)BOD
\(\Rightarrow \) Similarly we can prove that
\(\angle \)AOD and \(\angle \)BOC
11.
\(20^{ 0 }\)
12.
\(255^{ 0 }\)
13.
a =\(126^{ 0 }\), b =\(54^{ 0 }\), c =\(126^{ 0 }\), d =\(54^{ 0 }\)
14.
\(\angle \)COB=\(47^{ 0 }\) ,\(\angle \)COD=\(40^{ 0 }\)
15.
(a) 28
(b) 26
16.
y=180o-(30o+20o)=130o
l||m \(\Rightarrow\)x+100o=180o\(\Rightarrow\)x=80o
x+y=210o,y-x=50o
(x+y):(y-x)=21:5
17.
\(OP\bot AB\)
\(\Rightarrow \angle POA=90^o\)
Let \(\angle POQ=a\)
\(\therefore \angle QOR=3a\)
\(\angle ROA=5a\)
\(\Rightarrow\) a+3a+5a=90o
\(\Rightarrow\) 9a=90o
\(\Rightarrow\)a=10o
\(\therefore\)x=10o
and y=3x10o=30o
z=5x10o=50o
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