Class 9 Maths - Chapter 10 Question Answers Circles

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Q u e s t i o n : 6 0
In the given figure, two circles intersect at A and B. The centre of the smaller circle is O and it lies on the circumference of the larger circle. If ?APB = 70°, find ?ACB.
 
S o l u t i o n :
Consider the smaller circle whose centre is given as ‘O’.
The angle subtended by an arc at the centre of the circle is double the angle subtended by the arc in the remaining part of the circle.
So, here we have
Page 2


                                      
               
                
                              
                      
     
             
         
      
 
Q u e s t i o n : 6 0
In the given figure, two circles intersect at A and B. The centre of the smaller circle is O and it lies on the circumference of the larger circle. If ?APB = 70°, find ?ACB.
 
S o l u t i o n :
Consider the smaller circle whose centre is given as ‘O’.
The angle subtended by an arc at the centre of the circle is double the angle subtended by the arc in the remaining part of the circle.
So, here we have
Now consider the larger circle and the points ‘A’, ‘C’, ‘B’ and ‘O’ along its circumference. ‘ACBO’ form a cyclic quadrilateral.
In a cyclic quadrilateral it is known that the opposite angles are supplementary, meaning that the opposite angles add up to 180°.
Hence ,the measure of  is .
Q u e s t i o n : 6 1
In the given figure, two congruent circles with centres O and O' intersect at A and B. If ?AOB = 50°, then find ?APB.
 
S o l u t i o n :
Since both the circles are congruent, they will have equal radii. Let their radii be ‘r’.
So, from the given figure we have,
Now, since all the sides of the quadrilateral OBO’A are equal it has to be a rhombus.
One of the properties of a rhombus is that the opposite angles are equal to each other.
So, since it is given that , we can say that the angle opposite it, that is to say that  should also have the same value.
Hence we get 
Now, consider the first circle with the centre ‘O’ alone. ‘AB’ forms a chord and it subtends an angle of 50° with its centre, that is .
A property of a circle is that the angle subtended by an arc at the centre of the circle is double the angle subtended by the arc in the remaining part of the circle.
This means that,
Hence the measure of  is 
Q u e s t i o n : 6 2
In the given figure, ABCD is a cyclic quadrilateral in which ?BAD = 75°, ?ABD = 58° and ?ADC = 77°, AC and BD intersect at P. Then, find ?DPC.
S o l u t i o n :
In a cyclic quadrilateral it is known that the opposite angles are supplementary, meaning that the opposite angles add up to .
Here we have a cyclic quadrilateral ABCD. The centre of this circle is given as ‘O’.
Since in a cyclic quadrilateral the opposite angles are supplementary, here
Whenever a chord is drawn in a circle two segments are formed. One is called the minor segment while the other is called the major segment. The angle that the chord forms with any
point on the circumference of a particular segment is always the same.
Here, ‘CD’ is a chord and ‘A’ and ‘B’ are two points along the circumference on the major segment formed by the chord ‘CD’.
Page 3


                                      
               
                
                              
                      
     
             
         
      
 
Q u e s t i o n : 6 0
In the given figure, two circles intersect at A and B. The centre of the smaller circle is O and it lies on the circumference of the larger circle. If ?APB = 70°, find ?ACB.
 
S o l u t i o n :
Consider the smaller circle whose centre is given as ‘O’.
The angle subtended by an arc at the centre of the circle is double the angle subtended by the arc in the remaining part of the circle.
So, here we have
Now consider the larger circle and the points ‘A’, ‘C’, ‘B’ and ‘O’ along its circumference. ‘ACBO’ form a cyclic quadrilateral.
In a cyclic quadrilateral it is known that the opposite angles are supplementary, meaning that the opposite angles add up to 180°.
Hence ,the measure of  is .
Q u e s t i o n : 6 1
In the given figure, two congruent circles with centres O and O' intersect at A and B. If ?AOB = 50°, then find ?APB.
 
S o l u t i o n :
Since both the circles are congruent, they will have equal radii. Let their radii be ‘r’.
So, from the given figure we have,
Now, since all the sides of the quadrilateral OBO’A are equal it has to be a rhombus.
One of the properties of a rhombus is that the opposite angles are equal to each other.
So, since it is given that , we can say that the angle opposite it, that is to say that  should also have the same value.
Hence we get 
Now, consider the first circle with the centre ‘O’ alone. ‘AB’ forms a chord and it subtends an angle of 50° with its centre, that is .
A property of a circle is that the angle subtended by an arc at the centre of the circle is double the angle subtended by the arc in the remaining part of the circle.
This means that,
Hence the measure of  is 
Q u e s t i o n : 6 2
In the given figure, ABCD is a cyclic quadrilateral in which ?BAD = 75°, ?ABD = 58° and ?ADC = 77°, AC and BD intersect at P. Then, find ?DPC.
S o l u t i o n :
In a cyclic quadrilateral it is known that the opposite angles are supplementary, meaning that the opposite angles add up to .
Here we have a cyclic quadrilateral ABCD. The centre of this circle is given as ‘O’.
Since in a cyclic quadrilateral the opposite angles are supplementary, here
Whenever a chord is drawn in a circle two segments are formed. One is called the minor segment while the other is called the major segment. The angle that the chord forms with any
point on the circumference of a particular segment is always the same.
Here, ‘CD’ is a chord and ‘A’ and ‘B’ are two points along the circumference on the major segment formed by the chord ‘CD’.
So, 
Now,
In any triangle the sum of the interior angles need to be equal to .
Consider the triangle ?ABP,
?PAB + ?ABP + ?APB = 180° ? ?APB = 180°-30°-58° ? ?APB = 92°
From the figure, since ‘AC’ and ‘BD’ intersect at ‘P’ we have,
Hence the measure of  is .
Q u e s t i o n : 6 3
In the given figure, if ?AOB = 80° and ?ABC = 30°, then find ?CAO.
S o l u t i o n :
Consider the given circle with the centre ‘O’. Let the radius of this circle be ‘r’. ‘AB’ forms a chord and it subtends an angle of 80° with its centre, that is .
The angle subtended by an arc at the centre of the circle is double the angle subtended by the arc in the remaining part of the circle.
So, here we have
In any triangle the sum of the interior angles need to be equal to 180°.
Consider the triangle 
Since,  , we have . So the above equation now changes to
Considering the triangle ?ABC now,
Hence, the measure of  is .
Q u e s t i o n : 6 4
In the given figure, A is the centre of the circle. ABCD is a parallelogram and CDE is a straight line. Find ?BCD : ?ABE.
 
S o l u t i o n :
It is given that ‘ABCD’ is a parallelogram. But since ‘A’ is the centre of the circle, the lengths of ‘AB’ and ‘AD’ will both be equal to the radius of the circle.
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Q u e s t i o n : 6 0
In the given figure, two circles intersect at A and B. The centre of the smaller circle is O and it lies on the circumference of the larger circle. If ?APB = 70°, find ?ACB.
 
S o l u t i o n :
Consider the smaller circle whose centre is given as ‘O’.
The angle subtended by an arc at the centre of the circle is double the angle subtended by the arc in the remaining part of the circle.
So, here we have
Now consider the larger circle and the points ‘A’, ‘C’, ‘B’ and ‘O’ along its circumference. ‘ACBO’ form a cyclic quadrilateral.
In a cyclic quadrilateral it is known that the opposite angles are supplementary, meaning that the opposite angles add up to 180°.
Hence ,the measure of  is .
Q u e s t i o n : 6 1
In the given figure, two congruent circles with centres O and O' intersect at A and B. If ?AOB = 50°, then find ?APB.
 
S o l u t i o n :
Since both the circles are congruent, they will have equal radii. Let their radii be ‘r’.
So, from the given figure we have,
Now, since all the sides of the quadrilateral OBO’A are equal it has to be a rhombus.
One of the properties of a rhombus is that the opposite angles are equal to each other.
So, since it is given that , we can say that the angle opposite it, that is to say that  should also have the same value.
Hence we get 
Now, consider the first circle with the centre ‘O’ alone. ‘AB’ forms a chord and it subtends an angle of 50° with its centre, that is .
A property of a circle is that the angle subtended by an arc at the centre of the circle is double the angle subtended by the arc in the remaining part of the circle.
This means that,
Hence the measure of  is 
Q u e s t i o n : 6 2
In the given figure, ABCD is a cyclic quadrilateral in which ?BAD = 75°, ?ABD = 58° and ?ADC = 77°, AC and BD intersect at P. Then, find ?DPC.
S o l u t i o n :
In a cyclic quadrilateral it is known that the opposite angles are supplementary, meaning that the opposite angles add up to .
Here we have a cyclic quadrilateral ABCD. The centre of this circle is given as ‘O’.
Since in a cyclic quadrilateral the opposite angles are supplementary, here
Whenever a chord is drawn in a circle two segments are formed. One is called the minor segment while the other is called the major segment. The angle that the chord forms with any
point on the circumference of a particular segment is always the same.
Here, ‘CD’ is a chord and ‘A’ and ‘B’ are two points along the circumference on the major segment formed by the chord ‘CD’.
So, 
Now,
In any triangle the sum of the interior angles need to be equal to .
Consider the triangle ?ABP,
?PAB + ?ABP + ?APB = 180° ? ?APB = 180°-30°-58° ? ?APB = 92°
From the figure, since ‘AC’ and ‘BD’ intersect at ‘P’ we have,
Hence the measure of  is .
Q u e s t i o n : 6 3
In the given figure, if ?AOB = 80° and ?ABC = 30°, then find ?CAO.
S o l u t i o n :
Consider the given circle with the centre ‘O’. Let the radius of this circle be ‘r’. ‘AB’ forms a chord and it subtends an angle of 80° with its centre, that is .
The angle subtended by an arc at the centre of the circle is double the angle subtended by the arc in the remaining part of the circle.
So, here we have
In any triangle the sum of the interior angles need to be equal to 180°.
Consider the triangle 
Since,  , we have . So the above equation now changes to
Considering the triangle ?ABC now,
Hence, the measure of  is .
Q u e s t i o n : 6 4
In the given figure, A is the centre of the circle. ABCD is a parallelogram and CDE is a straight line. Find ?BCD : ?ABE.
 
S o l u t i o n :
It is given that ‘ABCD’ is a parallelogram. But since ‘A’ is the centre of the circle, the lengths of ‘AB’ and ‘AD’ will both be equal to the radius of the circle.
So, we have .
Whenever a parallelogram has two adjacent sides equal then it is a rhombus.
So ‘ABCD’ is a rhombus.
Let .
We know that in a circle the angle subtended by an arc at the centre of the circle is double the angle subtended by the arc in the remaining part of the circle.
By this property we have
In a rhombus the opposite angles are always equal to each other.
So, 
Since the sum of all the internal angles in any triangle sums up to in triangle , we have
In the rhombus ‘ABCD’ since one pair of opposite angles are ‘ ’ the other pair of opposite angles have to be 
From the figure we see that,
So now we can write the required ratio as,
Hence the ratio between the given two angles is .
Q u e s t i o n : 6 5
In the given figure, AB is a diameter of the circle such that ?A = 35° and ?Q = 25°, find ?PBR.
 
S o l u t i o n :
Let us first consider the triangle ?ABQ.
It is known that in a triangle the sum of all the interior angles add up to 180°.
So here in our triangle ?ABQ we have,
By a property of the circle we know that an angle formed in a semi-circle will be 90°..
In the given circle since ‘AB’ is the diameter of the circle the angle which is formed in a semi-circle will have to be 90°.
So, we have 
Now considering the triangle  we have,
From the given figure it can be seen that,
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Q u e s t i o n : 6 0
In the given figure, two circles intersect at A and B. The centre of the smaller circle is O and it lies on the circumference of the larger circle. If ?APB = 70°, find ?ACB.
 
S o l u t i o n :
Consider the smaller circle whose centre is given as ‘O’.
The angle subtended by an arc at the centre of the circle is double the angle subtended by the arc in the remaining part of the circle.
So, here we have
Now consider the larger circle and the points ‘A’, ‘C’, ‘B’ and ‘O’ along its circumference. ‘ACBO’ form a cyclic quadrilateral.
In a cyclic quadrilateral it is known that the opposite angles are supplementary, meaning that the opposite angles add up to 180°.
Hence ,the measure of  is .
Q u e s t i o n : 6 1
In the given figure, two congruent circles with centres O and O' intersect at A and B. If ?AOB = 50°, then find ?APB.
 
S o l u t i o n :
Since both the circles are congruent, they will have equal radii. Let their radii be ‘r’.
So, from the given figure we have,
Now, since all the sides of the quadrilateral OBO’A are equal it has to be a rhombus.
One of the properties of a rhombus is that the opposite angles are equal to each other.
So, since it is given that , we can say that the angle opposite it, that is to say that  should also have the same value.
Hence we get 
Now, consider the first circle with the centre ‘O’ alone. ‘AB’ forms a chord and it subtends an angle of 50° with its centre, that is .
A property of a circle is that the angle subtended by an arc at the centre of the circle is double the angle subtended by the arc in the remaining part of the circle.
This means that,
Hence the measure of  is 
Q u e s t i o n : 6 2
In the given figure, ABCD is a cyclic quadrilateral in which ?BAD = 75°, ?ABD = 58° and ?ADC = 77°, AC and BD intersect at P. Then, find ?DPC.
S o l u t i o n :
In a cyclic quadrilateral it is known that the opposite angles are supplementary, meaning that the opposite angles add up to .
Here we have a cyclic quadrilateral ABCD. The centre of this circle is given as ‘O’.
Since in a cyclic quadrilateral the opposite angles are supplementary, here
Whenever a chord is drawn in a circle two segments are formed. One is called the minor segment while the other is called the major segment. The angle that the chord forms with any
point on the circumference of a particular segment is always the same.
Here, ‘CD’ is a chord and ‘A’ and ‘B’ are two points along the circumference on the major segment formed by the chord ‘CD’.
So, 
Now,
In any triangle the sum of the interior angles need to be equal to .
Consider the triangle ?ABP,
?PAB + ?ABP + ?APB = 180° ? ?APB = 180°-30°-58° ? ?APB = 92°
From the figure, since ‘AC’ and ‘BD’ intersect at ‘P’ we have,
Hence the measure of  is .
Q u e s t i o n : 6 3
In the given figure, if ?AOB = 80° and ?ABC = 30°, then find ?CAO.
S o l u t i o n :
Consider the given circle with the centre ‘O’. Let the radius of this circle be ‘r’. ‘AB’ forms a chord and it subtends an angle of 80° with its centre, that is .
The angle subtended by an arc at the centre of the circle is double the angle subtended by the arc in the remaining part of the circle.
So, here we have
In any triangle the sum of the interior angles need to be equal to 180°.
Consider the triangle 
Since,  , we have . So the above equation now changes to
Considering the triangle ?ABC now,
Hence, the measure of  is .
Q u e s t i o n : 6 4
In the given figure, A is the centre of the circle. ABCD is a parallelogram and CDE is a straight line. Find ?BCD : ?ABE.
 
S o l u t i o n :
It is given that ‘ABCD’ is a parallelogram. But since ‘A’ is the centre of the circle, the lengths of ‘AB’ and ‘AD’ will both be equal to the radius of the circle.
So, we have .
Whenever a parallelogram has two adjacent sides equal then it is a rhombus.
So ‘ABCD’ is a rhombus.
Let .
We know that in a circle the angle subtended by an arc at the centre of the circle is double the angle subtended by the arc in the remaining part of the circle.
By this property we have
In a rhombus the opposite angles are always equal to each other.
So, 
Since the sum of all the internal angles in any triangle sums up to in triangle , we have
In the rhombus ‘ABCD’ since one pair of opposite angles are ‘ ’ the other pair of opposite angles have to be 
From the figure we see that,
So now we can write the required ratio as,
Hence the ratio between the given two angles is .
Q u e s t i o n : 6 5
In the given figure, AB is a diameter of the circle such that ?A = 35° and ?Q = 25°, find ?PBR.
 
S o l u t i o n :
Let us first consider the triangle ?ABQ.
It is known that in a triangle the sum of all the interior angles add up to 180°.
So here in our triangle ?ABQ we have,
By a property of the circle we know that an angle formed in a semi-circle will be 90°..
In the given circle since ‘AB’ is the diameter of the circle the angle which is formed in a semi-circle will have to be 90°.
So, we have 
Now considering the triangle  we have,
From the given figure it can be seen that,
Now, we can also say that,
Hence the measure of the angle  is 115°.
Q u e s t i o n : 6 6
In the given figure, P and Q are centres of two circles intersecting at B and C. ACD is a straight line. Then, ?BQD =
 
S o l u t i o n :
Consider the circle with the centre ‘P’.
The angle subtended by an arc at the centre of the circle is double the angle subtended by the arc in the remaining part of the circle.
So, here we have
Since ‘ACD’ is a straight line, we have
Now let us consider the circle with centre ‘Q’. Here let ‘E’ be any point on the circumference along the major arc ‘BD’. Now ‘CBED’ forms a cyclic quadrilateral.
In a cyclic quadrilateral it is known that the opposite angles are supplementary, meaning that the opposite angles add up to 180°.
So here,
The angle subtended by an arc at the centre of the circle is double the angle subtended by the arc in the remaining part of the circle.
So, now we have
Hence, the measure of is .
Q u e s t i o n : 6 7
In the given figure, if O is the circumcentre of ?ABC, then find the value of ?OBC + ?BAC.
S o l u t i o n :