Circles - Exercise 15.4

Circles - Exercise 15.4 | Extra Documents & Tests for Class 9 PDF Download

``` Page 1

Q u e s t i o n : 2 0
In the given figure, O is the centre of the circle. If ?APB
= 50°, find ?AOB and ?OAB.
S o l u t i o n :
This question seems to be incorrect.
Q u e s t i o n : 2 1
In the given figure, O is the centre of the circle. Find ?BAC.

S o l u t i o n :
It is given that
And  given
We have to find
In given triangle
Given
Therefore,  is an isosceles triangle.
So, ?OBA = ?OAB
..… 1
(Given )

From(1)
So
Again from figure,  is given triangle and
Now in ,
?OAC = ?OCA
(Given that )
Page 2

Q u e s t i o n : 2 0
In the given figure, O is the centre of the circle. If ?APB
= 50°, find ?AOB and ?OAB.
S o l u t i o n :
This question seems to be incorrect.
Q u e s t i o n : 2 1
In the given figure, O is the centre of the circle. Find ?BAC.

S o l u t i o n :
It is given that
And  given
We have to find
In given triangle
Given
Therefore,  is an isosceles triangle.
So, ?OBA = ?OAB
..… 1
(Given )

From(1)
So
Again from figure,  is given triangle and
Now in ,
?OAC = ?OCA
(Given that )
Then,
Since
Hence
Q u e s t i o n : 2 2
If O is the centre of the circle, find the value of x in each of the following figures.
i
ii
iii
iv
v
vi
Page 3

Q u e s t i o n : 2 0
In the given figure, O is the centre of the circle. If ?APB
= 50°, find ?AOB and ?OAB.
S o l u t i o n :
This question seems to be incorrect.
Q u e s t i o n : 2 1
In the given figure, O is the centre of the circle. Find ?BAC.

S o l u t i o n :
It is given that
And  given
We have to find
In given triangle
Given
Therefore,  is an isosceles triangle.
So, ?OBA = ?OAB
..… 1
(Given )

From(1)
So
Again from figure,  is given triangle and
Now in ,
?OAC = ?OCA
(Given that )
Then,
Since
Hence
Q u e s t i o n : 2 2
If O is the centre of the circle, find the value of x in each of the following figures.
i
ii
iii
iv
v
vi
vii
viii
ix
x
xi
xii
S o l u t i o n :
We have to find  in each figure.
i
It is given that
?AOC + ?COB = 180°   [Linear pair]
Page 4

Q u e s t i o n : 2 0
In the given figure, O is the centre of the circle. If ?APB
= 50°, find ?AOB and ?OAB.
S o l u t i o n :
This question seems to be incorrect.
Q u e s t i o n : 2 1
In the given figure, O is the centre of the circle. Find ?BAC.

S o l u t i o n :
It is given that
And  given
We have to find
In given triangle
Given
Therefore,  is an isosceles triangle.
So, ?OBA = ?OAB
..… 1
(Given )

From(1)
So
Again from figure,  is given triangle and
Now in ,
?OAC = ?OCA
(Given that )
Then,
Since
Hence
Q u e s t i o n : 2 2
If O is the centre of the circle, find the value of x in each of the following figures.
i
ii
iii
iv
v
vi
vii
viii
ix
x
xi
xii
S o l u t i o n :
We have to find  in each figure.
i
It is given that
?AOC + ?COB = 180°   [Linear pair]

As we know the angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point on the remaining part of the circle.
Now ,x =
1
2
?COB = 22
1
2
°
Hence
ii
As we know that  = x
Anglesinthesamesegment
line  is diameter passing through centre,
So,
?BCA = 90°       [Angle inscribed in a semicircle is a right angle ]

?CAB + ?ABC + ?BCA = 180°    [Angle sum property] ? x +40°+90° = 180° ? x = 50°
iii
It is given that
?ABC =
1
2
(Reflex ?AOC)

So
And
Then
Hence
iv
Linearpair

And
x =

Hence,
Page 5

Q u e s t i o n : 2 0
In the given figure, O is the centre of the circle. If ?APB
= 50°, find ?AOB and ?OAB.
S o l u t i o n :
This question seems to be incorrect.
Q u e s t i o n : 2 1
In the given figure, O is the centre of the circle. Find ?BAC.

S o l u t i o n :
It is given that
And  given
We have to find
In given triangle
Given
Therefore,  is an isosceles triangle.
So, ?OBA = ?OAB
..… 1
(Given )

From(1)
So
Again from figure,  is given triangle and
Now in ,
?OAC = ?OCA
(Given that )
Then,
Since
Hence
Q u e s t i o n : 2 2
If O is the centre of the circle, find the value of x in each of the following figures.
i
ii
iii
iv
v
vi
vii
viii
ix
x
xi
xii
S o l u t i o n :
We have to find  in each figure.
i
It is given that
?AOC + ?COB = 180°   [Linear pair]

As we know the angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point on the remaining part of the circle.
Now ,x =
1
2
?COB = 22
1
2
°
Hence
ii
As we know that  = x
Anglesinthesamesegment
line  is diameter passing through centre,
So,
?BCA = 90°       [Angle inscribed in a semicircle is a right angle ]

?CAB + ?ABC + ?BCA = 180°    [Angle sum property] ? x +40°+90° = 180° ? x = 50°
iii
It is given that
?ABC =
1
2
(Reflex ?AOC)

So
And
Then
Hence
iv
Linearpair

And
x =

Hence,
v
It is given that
is an isosceles triangle.

Therefore
And,
In ? AOB, ?AOB + ?OBA + ?BAO = 180° ? 70°+ ?BAO = 180° ? ?BAO = 110°
?AOB = 2(Reflex ?ACB)

Hence,
vi
It is given that

And
?COA + ?AOB = 180° ? ?COA = 180°-60° ? ?COA = 120°
? OCA is an isosceles triangle.
So
Hence,
vii
Angleinthesamesegment

In  we have
Hence
viii

Therefore,  is an isosceles triangle.
```

Extra Documents & Tests for Class 9

1 videos|228 docs|21 tests

Extra Documents & Tests for Class 9

1 videos|228 docs|21 tests

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