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Page 1
Question:24
i
In Figure 1
, O is the centre of the circle. If ?OAB = 40° and ?OCB = 30°, find ?AOC.
ii In Figure
2, A, B and C are three points on the circle with centre O such that ?AOB = 90° and ?AOC = 110°. Find ?BAC.
Solution:
i
Join BO.
In ?BOC, we have:
OC = OB Radiiofacircle
? ?OBC = ?OCB
?OBC = 30° ...i
In ?BOA, we have:
OB = OA Radiiofacircle
? ?OBA = ?OAB [ ? ?OAB = 40°]
? ?OBA = 40° ...
ii
Now, we have:
Page 2
Question:24
i
In Figure 1
, O is the centre of the circle. If ?OAB = 40° and ?OCB = 30°, find ?AOC.
ii In Figure
2, A, B and C are three points on the circle with centre O such that ?AOB = 90° and ?AOC = 110°. Find ?BAC.
Solution:
i
Join BO.
In ?BOC, we have:
OC = OB Radiiofacircle
? ?OBC = ?OCB
?OBC = 30° ...i
In ?BOA, we have:
OB = OA Radiiofacircle
? ?OBA = ?OAB [ ? ?OAB = 40°]
? ?OBA = 40° ...
ii
Now, we have:
?ABC = ?OBC + ?OBA
= 30° + 40°
From(i)and(ii)
? ?ABC = 70°
The angle subtended by an arc of a circle at the centre is double the angle subtended by the arc at any point on the
circumference.
i.e., ?AOC = 2 ?ABC
= (2 × 70°) = 140°
ii
Here, ?BOC = {360° - (90° + 110°)}
= (360° - 200°) = 160°
We know that ?BOC = 2 ?BAC
? ?BAC =
?BOC
2
=
160°
2
= 80°
Hence, ?BAC = 80°
Question:25
In the given figure, O is the canter of the circle and ?AOB = 70°.
Calculate the values of
i ?OCA,
ii ?OAC.
Solution:
i
The angle subtended by an arc of a circle at the centre is double the angle subtended by the arc at any point on the
circumference.
Thus, ?AOB = 2 ?OCA
? ?OCA =
?AOB
2
=
70°
2
= 35°
ii
OA = OC Radiiofacircle
( )
( ) ( )
Page 3
Question:24
i
In Figure 1
, O is the centre of the circle. If ?OAB = 40° and ?OCB = 30°, find ?AOC.
ii In Figure
2, A, B and C are three points on the circle with centre O such that ?AOB = 90° and ?AOC = 110°. Find ?BAC.
Solution:
i
Join BO.
In ?BOC, we have:
OC = OB Radiiofacircle
? ?OBC = ?OCB
?OBC = 30° ...i
In ?BOA, we have:
OB = OA Radiiofacircle
? ?OBA = ?OAB [ ? ?OAB = 40°]
? ?OBA = 40° ...
ii
Now, we have:
?ABC = ?OBC + ?OBA
= 30° + 40°
From(i)and(ii)
? ?ABC = 70°
The angle subtended by an arc of a circle at the centre is double the angle subtended by the arc at any point on the
circumference.
i.e., ?AOC = 2 ?ABC
= (2 × 70°) = 140°
ii
Here, ?BOC = {360° - (90° + 110°)}
= (360° - 200°) = 160°
We know that ?BOC = 2 ?BAC
? ?BAC =
?BOC
2
=
160°
2
= 80°
Hence, ?BAC = 80°
Question:25
In the given figure, O is the canter of the circle and ?AOB = 70°.
Calculate the values of
i ?OCA,
ii ?OAC.
Solution:
i
The angle subtended by an arc of a circle at the centre is double the angle subtended by the arc at any point on the
circumference.
Thus, ?AOB = 2 ?OCA
? ?OCA =
?AOB
2
=
70°
2
= 35°
ii
OA = OC Radiiofacircle
( )
( ) ( )
?OAC = ?OCA
Baseanglesofanisoscelestriangleareequal
= 35°
Question:26
In the given figure, O is the centre of the circle. if ?PBC = 25° and ?APB = 110°, find the value of ?ADB.
Solution:
From the given diagram, we have:
?ACB = ?PCB
?BPC = (180° - 110°) = 70° Linearpair
Considering ?PCB, we have:
?PCB + ?BPC + ?PBC = 180° Anglesumproperty
? ?PCB + 70° + 25° = 180°
? ?PCB = (180° – 95°) = 85°
? ?ACB = ?PCB = 85°
We know that the angles in the same segment of a circle are equal.
? ?ADB = ?ACB = 85°
Question:27
In the given figure, O is the centre of the circle. If ?ABD = 35° and ?BAC = 70°, find ?ACB.
Solution:
Page 4
Question:24
i
In Figure 1
, O is the centre of the circle. If ?OAB = 40° and ?OCB = 30°, find ?AOC.
ii In Figure
2, A, B and C are three points on the circle with centre O such that ?AOB = 90° and ?AOC = 110°. Find ?BAC.
Solution:
i
Join BO.
In ?BOC, we have:
OC = OB Radiiofacircle
? ?OBC = ?OCB
?OBC = 30° ...i
In ?BOA, we have:
OB = OA Radiiofacircle
? ?OBA = ?OAB [ ? ?OAB = 40°]
? ?OBA = 40° ...
ii
Now, we have:
?ABC = ?OBC + ?OBA
= 30° + 40°
From(i)and(ii)
? ?ABC = 70°
The angle subtended by an arc of a circle at the centre is double the angle subtended by the arc at any point on the
circumference.
i.e., ?AOC = 2 ?ABC
= (2 × 70°) = 140°
ii
Here, ?BOC = {360° - (90° + 110°)}
= (360° - 200°) = 160°
We know that ?BOC = 2 ?BAC
? ?BAC =
?BOC
2
=
160°
2
= 80°
Hence, ?BAC = 80°
Question:25
In the given figure, O is the canter of the circle and ?AOB = 70°.
Calculate the values of
i ?OCA,
ii ?OAC.
Solution:
i
The angle subtended by an arc of a circle at the centre is double the angle subtended by the arc at any point on the
circumference.
Thus, ?AOB = 2 ?OCA
? ?OCA =
?AOB
2
=
70°
2
= 35°
ii
OA = OC Radiiofacircle
( )
( ) ( )
?OAC = ?OCA
Baseanglesofanisoscelestriangleareequal
= 35°
Question:26
In the given figure, O is the centre of the circle. if ?PBC = 25° and ?APB = 110°, find the value of ?ADB.
Solution:
From the given diagram, we have:
?ACB = ?PCB
?BPC = (180° - 110°) = 70° Linearpair
Considering ?PCB, we have:
?PCB + ?BPC + ?PBC = 180° Anglesumproperty
? ?PCB + 70° + 25° = 180°
? ?PCB = (180° – 95°) = 85°
? ?ACB = ?PCB = 85°
We know that the angles in the same segment of a circle are equal.
? ?ADB = ?ACB = 85°
Question:27
In the given figure, O is the centre of the circle. If ?ABD = 35° and ?BAC = 70°, find ?ACB.
Solution:
It is clear that BD is the diameter of the circle.
Also, we know that the angle in a semicircle is a right angle.
i.e., ?BAD = 90°
Now, considering the ?BAD, we have:
?ADB + ?BAD + ?ABD = 180° Anglesumpropertyofatriangle
? ?ADB + 90° + 35° = 180°
? ?ADB = (180° - 125°) = 55°
Angles in the same segment of a circle are equal.
Hence, ?ACB = ?ADB = 55°
Question:28
In the given figure, O is the centre of the circle. If ?ACB = 50°, find ?OAB.
Solution:
We know that the angle subtended by an arc of a circle at the centre is double the angle subtended by the arc at
any point on the circumference.
?AOB = 2 ?ACB
= 2 × 50°
Given
?AOB = 100° ...i
Let us consider the triangle ?OAB.
OA = OB Radiiofacircle
Thus, ?OAB = ?OBA
In ?OAB, we have:
?AOB + ?OAB + ?OBA = 180°
? 100° + ?OAB + ?OAB = 180°
? 100° + 2 ?OAB = 180°
? 2 ?OAB = 180° – 100° = 80°
? ?OAB = 40°
Hence, ?OAB = 40°
Question:29
In the given figure, ?ABD = 54° and ?BCD = 43°, calculate
i ?ACD
ii ?BAD
iii ?BDA.
Page 5
Question:24
i
In Figure 1
, O is the centre of the circle. If ?OAB = 40° and ?OCB = 30°, find ?AOC.
ii In Figure
2, A, B and C are three points on the circle with centre O such that ?AOB = 90° and ?AOC = 110°. Find ?BAC.
Solution:
i
Join BO.
In ?BOC, we have:
OC = OB Radiiofacircle
? ?OBC = ?OCB
?OBC = 30° ...i
In ?BOA, we have:
OB = OA Radiiofacircle
? ?OBA = ?OAB [ ? ?OAB = 40°]
? ?OBA = 40° ...
ii
Now, we have:
?ABC = ?OBC + ?OBA
= 30° + 40°
From(i)and(ii)
? ?ABC = 70°
The angle subtended by an arc of a circle at the centre is double the angle subtended by the arc at any point on the
circumference.
i.e., ?AOC = 2 ?ABC
= (2 × 70°) = 140°
ii
Here, ?BOC = {360° - (90° + 110°)}
= (360° - 200°) = 160°
We know that ?BOC = 2 ?BAC
? ?BAC =
?BOC
2
=
160°
2
= 80°
Hence, ?BAC = 80°
Question:25
In the given figure, O is the canter of the circle and ?AOB = 70°.
Calculate the values of
i ?OCA,
ii ?OAC.
Solution:
i
The angle subtended by an arc of a circle at the centre is double the angle subtended by the arc at any point on the
circumference.
Thus, ?AOB = 2 ?OCA
? ?OCA =
?AOB
2
=
70°
2
= 35°
ii
OA = OC Radiiofacircle
( )
( ) ( )
?OAC = ?OCA
Baseanglesofanisoscelestriangleareequal
= 35°
Question:26
In the given figure, O is the centre of the circle. if ?PBC = 25° and ?APB = 110°, find the value of ?ADB.
Solution:
From the given diagram, we have:
?ACB = ?PCB
?BPC = (180° - 110°) = 70° Linearpair
Considering ?PCB, we have:
?PCB + ?BPC + ?PBC = 180° Anglesumproperty
? ?PCB + 70° + 25° = 180°
? ?PCB = (180° – 95°) = 85°
? ?ACB = ?PCB = 85°
We know that the angles in the same segment of a circle are equal.
? ?ADB = ?ACB = 85°
Question:27
In the given figure, O is the centre of the circle. If ?ABD = 35° and ?BAC = 70°, find ?ACB.
Solution:
It is clear that BD is the diameter of the circle.
Also, we know that the angle in a semicircle is a right angle.
i.e., ?BAD = 90°
Now, considering the ?BAD, we have:
?ADB + ?BAD + ?ABD = 180° Anglesumpropertyofatriangle
? ?ADB + 90° + 35° = 180°
? ?ADB = (180° - 125°) = 55°
Angles in the same segment of a circle are equal.
Hence, ?ACB = ?ADB = 55°
Question:28
In the given figure, O is the centre of the circle. If ?ACB = 50°, find ?OAB.
Solution:
We know that the angle subtended by an arc of a circle at the centre is double the angle subtended by the arc at
any point on the circumference.
?AOB = 2 ?ACB
= 2 × 50°
Given
?AOB = 100° ...i
Let us consider the triangle ?OAB.
OA = OB Radiiofacircle
Thus, ?OAB = ?OBA
In ?OAB, we have:
?AOB + ?OAB + ?OBA = 180°
? 100° + ?OAB + ?OAB = 180°
? 100° + 2 ?OAB = 180°
? 2 ?OAB = 180° – 100° = 80°
? ?OAB = 40°
Hence, ?OAB = 40°
Question:29
In the given figure, ?ABD = 54° and ?BCD = 43°, calculate
i ?ACD
ii ?BAD
iii ?BDA.
Solution:
i
We know that the angles in the same segment of a circle are equal.
i.e., ?ABD = ?ACD = 54°
ii
We know that the angles in the same segment of a circle are equal.
i.e., ?BAD = ?BCD = 43°
iii
In ?ABD, we have:
?BAD + ?ADB + ?DBA = 180° Anglesumpropertyofatriangle
? 43° + ?ADB + 54° = 180°
? ?ADB = (180° – 97°) = 83°
? ?BDA = 83°
Question:30
In the adjoining figure, DE is a chord parallel to diameter AC of the circle with centre O. If ?CBD = 60°, calculate
?CDE.
Solution:
Angles in the same segment of a circle are equal.
i.e., ?CAD = ?CBD = 60°
We know that an angle in a semicircle is a right angle.
i.e., ?ADC = 90°
In ?ADC, we have:
?ACD + ?ADC + ?CAD = 180° Anglesumpropertyofatriangle
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FAQs on RS Aggarwal Solutions: Circles- 2 - Mathematics (Maths) Class 9
| 1. What are the properties of circles? |  |
Ans. Circles have several properties, such as:
- All points on the circumference of a circle are equidistant from its center.
- The diameter of a circle is twice the length of its radius.
- The radius of a circle is a line segment that connects the center of the circle to any point on its circumference.
- The circumference of a circle is the distance around its outer edge and can be calculated using the formula 2πr, where r is the radius.
- The area of a circle can be calculated using the formula πr², where r is the radius.
| 2. How do you find the equation of a circle? | |
Ans. The equation of a circle can be found using the coordinates of its center and the length of its radius. The standard equation of a circle is (x - h)² + (y - k)² = r², where (h, k) represents the coordinates of the center and r represents the radius.
| 3. Can a circle have a negative radius? | |
Ans. No, a circle cannot have a negative radius. The radius of a circle represents the distance from its center to any point on its circumference. Since distance cannot be negative, the radius of a circle is always a positive value or zero.
| 4. What is the relationship between the radius and diameter of a circle? | |
Ans. The radius and diameter of a circle are related as follows:
- The radius is half the length of the diameter.
- The diameter is twice the length of the radius.
- The radius and diameter are always proportional to each other.
| 5. How do you find the circumference of a circle? | |
Ans. The circumference of a circle can be calculated using the formula 2πr, where r is the radius. Simply multiply the radius by 2 and then multiply it by π (pi) to find the circumference.
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Dec 27, 2024 Last updated |
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