Maths is a subject that always concerns the students most. Concepts, theories, processes, exercises, and word problems are part of a Maths book in general view. 10 Class is considered a foundational class for many higher exams like NTSE, Olympiad, NEET 2022, JEE 2022, International Mathematics Olympiad, etc. How to prepare well in the case of a subject like Maths? A very simple answer with the help of a sample paper. CBSE Class 10th Maths is full of concepts revolving around shapes, and one such shape is a circle. So, in this article, we will find out about the CBSE class 10th Maths sample question on circles. One can take help from professionals and practice more questions from NCERT Solutions.
Introduction to Circles
The circles are a collection of all points in a plane surface at a continual distance and length from a fixed point. There is a famous quote from Ralph Waldo Emerson on circles: “Circles are like the soul that is never-ending and turns round and round without a stop”. In the CBSE class, 10th Circles is in chapter number 10, which usually falls under the 2nd term of the session of the Class 10th Maths. The Circles chapter is very significant in the Olympiads in India, KVPY syllabus, NTSE syllabus, and CUCETpreparation. The Greeks are known and considered as the inventor of geometry. Professor Kaoru Ishikawa was considered the father of the QC Circle in 1962. For more clearance on the topic of circles, one can take reference from the following solutions: RD Sharma Solutions, HC Verma Solutions, and RS Aggarwal Solutions.
Parts of a Circle
A circle consists of 6 main parts; they are-: A Centre, Radius, Chord, Diameter, and Tangent.
- Centre-: The fixed point in the exact middle of the circle is called a center.
- Radius-: The constant distance or length from the center is the radius. There can be many radii in number.
- Chord-: A line segment connecting any two points on the circle is a chord. Just like radius, chords can be many in numbers on a circle.
- Diameter -: A chord or line segment passing through the circle’s center is called the diameter. It is also known as the longest chord.
- Tangent-: When the line meets the circle at one or two coincidences. The line is called a point, a tangent. Practically Tangent to a circle is perpendicular to the radius through the point of the connection.
- Arc-: An arc is a curve formed on the circle’s boundary between two points.
What is the perimeter and area of a circle?
Perimeter is the distance defined around a given area. Perimeter is usually calculated as the total of all given lengths or sides. But since circles don’t have any straight lines that can be easily measured, they have a special formula to determine the perimeter.
First of all, circles have a special term used in place of the perimeter, ‘circumference’. The symbol used to denote the term is C. The formula to calculate the circumference consists of a Pi×Diameter or Pi×2R, equal to the circle’s C. Pi is the value that always remains the same no matter the length of the radius and diameter. Pi always remains either 3.14 or 22/ 7. The symbolic representation of Pi is π. Therefore,
- Circumference(C) of the circle = “π× d” or “π× 2r”
- For example, Let’s assume ‘d’ is 8 then what will be the circumference?
C of the circle = π×d
= 3.14 × 8 = 25.12
- Let’s assume ‘r’ is 4 then calculate the circumference.
C of the circle = π×2r
= 3.14× 2× 4 = 25.12 (unit)
- Why 2r not 2d? Well, the Radius is half of the Diameter’s length therefore while calculating with a radius the formula will be ‘π×2r’ and with a diameter or will change to ‘π×d’ where the value of π will remain the same.
The area of the circle is again different from any other figure consisting of straight lengths. Rhind Papyrus was the first to determine the area of the circle that corresponds to the value π, which was 256/81, nearly around 3.16. The area of a circle is pi times the radius squared, which means A= π×r². Here ‘A’ is the symbolic representation of the area. Therefore,
- The area of the circle = “π×r²”
- For example, let’s assume ‘r’ is equal to 4 then determine the area
Area (A) of the circle = π×r²
= 3.14×4² = 3.14×16 = 50.24 (unit²)
In CBSE Class 10th Maths, In the chapter on circles, there are several most important theorems which consist of tangents.
- First Theorem -: Tangent perpendicular to the radius at the point of contact.
- Theorem- The theorem remarks that the constructed tangent to the given circle at any marked point is perpendicular to the presented or marked radius of the circle that passes through the marked point of connection.
- Given- YX is tangent at the point named P to the given circle with center O.
- To prove- OP is perpendicular to YX
- Construction- Take a point somewhere on YX, name it Q, and join it with the center O to form OQ.
- Proof- if the point named Q is marked inside the given circle with a center named O, then YX will serve as a secant but not a tangent in respect to the circle with a center named O.
- Therefore, OQ> OP.
- TThe same process will occur with every other point on the marked line YX except the point named P. OP is marked as the shortest length between the given points YX and the center point O.
- Therefore, OP is considered perpendicular to line YX because OP is the shortest side, referred to as the perpendicular.
YX is the line in the above picture, with O as the center of the center, OP as the radius, and OP as perpendicular to YX.
- Second Theorem-: The Line was drawn to the endpoints of the given radius and marked perpendicular to it, which is the tangent to the circle.
- Theorem- The Theorem states that a line is marked through the endpoint of the presented radius and the presented perpendicular to it, which is the tangent to the given circle.
- Given – A circle named C with the center named O and the lone AB with a middle point P is perpendicular to the marked radius of the circle OP.
- To prove- AB is tangent at P.
- Construction- Mark a point named Q on the marked line AB, other than the point P, and further join OQ.
- Proof- since OP is perpendicular to line AB.
- Therefore, OP<OQ.
- The point named Q falls outside the circle with the given center named O. This, every other point on the marked line AB except the point P that falls exterior to circle C. This proves that the line named AB meets outside the circle with the center I at point P. Hence, AP is the recognized tangent to a circle named C at point P.
AB is the line in the above picture, with O as the center of the center, OP as the radius, and OP as perpendicular to AB.
- The third theorem-: The lengths of the tangents formed from an exterior point to a presented circle are equal.
- Theorem- The theorem suggests that the lengths/ sides of the tangents marked from an exterior point to a circle are equal.
- Given – PT and PS
- Construction- Draw a line and join O to P, T, and S points.
- Proof- In the constructed triangle, OTP and OSP.
OT= OS ( radius constructed in the same circle)
OP= OP (common)
Angle OTP = Angle OSP (each at 90°)
Therefore, ∆ OTP= ∆ OSP
Hence, PT= PS.
Note related to tangents-: If the constructed two tangents are marked to a given circle from an exterior point, then,
- They subtend a proportional angle at the given center of the circle.
- The angles are inclined to the segment equally that are joining the center to the point.
In the above-given figure, a circle is given with a center named C and two points on the circle named T and S. Also, P at the exterior of the circle. Joining the points forms two triangles named OTP and the other one as OSP.
Facts and Properties
- The tangents constructed at the very ends of the diameter of any circle are parallel.
- The angle formed between the two tangents drawn from an exterior point of any circle is supplementary to the formed angle by the presented line segment joining the points of the connection at the marked center of the circle.
- In presented two concentric circles, the marked chord of the bigger circle touches and the smaller one intersect at the point of their connection.
- The perpendicular is formed at the point of connection with the passing tangents to the presented circle through its center.
- The parallelogram circumscribed in a circle is known as a rhombus.
- The adverse sides of any quadrilateral circumscribed around the given circle forms supplementary angles at the center of the presented circle.
Question 1) In the figure below, C is the center of a circle, AB is a chord, and AD is the tangent at point A. If ∠AOB = 100°, then calculate ∠BAD.
In the mentioned figure let’s construct an angle named E and another angle named F at points A and B respectively.
∠E = ∠F
∠E + ∠F + 100° = 180°
∠E + ∠E = 80°
⇒ 2∠E = 80°
⇒ ∠E = 40°
∠E + ∠BAD = 90°
∠BAD = 90° – 40° = 50°
Question 2) Prove that the line segment connecting the points of contact between two parallel tangents of a circle with center C passes through its center.
Given: PQ and RS are 2 parallel tangents at points A and T of a circle with center C.
To prove: AB passes through the circle with center C or ACT is the diameter of the circle.
Construction: Join CA and CB. Draw CU || PQ, Mark ∠a, ∠b and ∠c
Proof: ∠a= 90° … (i)
[∵ Tangent is parallel to the radius through the point of contact
CU || PQ
∴ ∠a + ∠b = 180° …(due to Co-interior angles
90° + ∠b = 180° …[From (i)
∠b = 180° – 90 = 90°
Similarly, ∠c = 90°
∠c + ∠b = 90° + 90° = 180°
∴ ACT is a straight line.
Hence ACT can be determined as the diameter of the circle with center C.
∴ AT passes through the circle with center C.
Question 3) From an exterior point D, two tangents DA and DB are drawn to a circle with center C. If ∠DAB = 50°, then find ∠ACB.
DA = DB …[∵ Tangents drawn from an exterior point are equal to each other
∠DBA = ∠DAB = 50° …[ property of Angles equal to opposite sides
∠DBA + ∠DAB + ∠ADB = 180° …[ due to Angle-sum-property of a triangle
50° + 50° + ∠ADB = 180°
∠ADB is equal to 180° – 50° – 50° = 80°
In cyclic quadrilateral CADB
∠ACB + ∠ADB = 180° ……[the property of the Sum of opposite angles of a cyclic quadrilateral is 180°
∠ACB + 80o = 180°
∠ACB = 180° – 80° = 100°
Question 4) In the given figure, AD and BD are the 2 tangents to a circle with center C, such that AD is 5 cm and ∠ADB is 60°. Find the length of the marked chord with the name AB.
DA = DB …[Tangents drawn from an external point D are equal
∠ADB = 60°
∠PDB = ∠DBA … (i) …( Property of Angles opposite to equal sides
In ∆DAB, ∠DAB + ∠DBA + ∠ADB = 180° …[ Due to the Angle-sum-property of a triangle
⇒ ∠DAB + ∠DAB + 60° = 180°
⇒ 2∠DAB = 180° – 60° = 120°
⇒ ∠DAB = 60°
⇒ ∠DAB = ∠DBA = ∠ADB = 60°
Therefore, ADB is an equilateral triangle
Hence, AB = AD = 5 cm …[∵ All sides of an equilateral triangle are equal to each other
A famous quote on Maths is, “The only way to learn Mathematics is to do Mathematics. Sample papers are like a field that offers practice until one succeeds in a particular concept. Such a topic is circles that need a lot of practice and guidance. CBSE Class 10th Maths will play a significant role in JEE Advanced 2022 and NEET-UG 2022 as a foundational course. Maths is a very important key in JEE Advanced 2022 Syllabus. So, students who are in 10th should consider an important CBSE class 10th Maths sample question on circles and other topics to form a solid base for their future goals.