{"id":136559,"date":"2022-04-08T15:30:00","date_gmt":"2022-04-08T10:00:00","guid":{"rendered":"https:\/\/www.aakash.ac.in\/blog\/?p=136559"},"modified":"2023-04-03T13:22:56","modified_gmt":"2023-04-03T07:52:56","slug":"cbse-class-10-maths-sample-questions-on-circles","status":"publish","type":"post","link":"https:\/\/www.aakash.ac.in\/blog\/cbse-class-10-maths-sample-questions-on-circles\/","title":{"rendered":"CBSE Class 10 Maths sample questions on Circles"},"content":{"rendered":"<p><span style=\"font-weight: 400;\">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 <a href=\"https:\/\/www.aakash.ac.in\/national-talent-search-examination-ntse\" target=\"_blank\" rel=\"noopener\">NTSE<\/a>, Olympiad,<a href=\"https:\/\/www.aakash.ac.in\/neet-exam\" target=\"_blank\" rel=\"noopener\"> NEET 2022,<\/a> <a href=\"https:\/\/www.aakash.ac.in\/jee-main-exam\" target=\"_blank\" rel=\"noopener\">JEE 2022<\/a>, 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 <\/span><a href=\"https:\/\/www.aakash.ac.in\/ncert-solutions\" target=\"_blank\" rel=\"noopener\"><span style=\"font-weight: 400;\">NCERT Solutions<\/span><\/a><span style=\"font-weight: 400;\">.\u00a0<\/span><\/p>\n<p><strong>Introduction to Circles<\/strong><\/p>\n<p><span style=\"font-weight: 400;\">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: &#8220;Circles are like the soul that is never-ending and turns round and round without a stop&#8221;. 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, <a href=\"https:\/\/www.aakash.ac.in\/kvpy-exam\" target=\"_blank\" rel=\"noopener\">KVPY syllabus<\/a>, 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: <\/span><a href=\"https:\/\/www.aakash.ac.in\/book-solutions\/rd-sharma-solutions\"><span style=\"font-weight: 400;\">RD Sharma Solutions<\/span><\/a><span style=\"font-weight: 400;\">, <\/span><a href=\"https:\/\/www.aakash.ac.in\/book-solutions\/hc-verma-solutions\" target=\"_blank\" rel=\"noopener\"><span style=\"font-weight: 400;\">HC Verma Solutions<\/span><\/a><span style=\"font-weight: 400;\">, and <\/span><a href=\"https:\/\/www.aakash.ac.in\/book-solutions\/rs-aggarwal-solutions\" target=\"_blank\" rel=\"noopener\"><span style=\"font-weight: 400;\">RS Aggarwal Solutions<\/span><\/a><span style=\"font-weight: 400;\">.\u00a0<\/span><\/p>\n<h3>Parts of a Circle<\/h3>\n<p><span style=\"font-weight: 400;\">A circle consists of 6 main parts; they are-: A Centre, Radius, Chord, Diameter, and Tangent.<\/span><\/p>\n<ul>\n<li style=\"font-weight: 400;\" aria-level=\"1\"><span style=\"font-weight: 400;\">Centre-: The fixed point in the exact middle of the circle is called a center.<\/span><\/li>\n<li style=\"font-weight: 400;\" aria-level=\"1\"><span style=\"font-weight: 400;\">Radius-: The constant distance or length from the center is the radius. There can be many radii in number.<\/span><\/li>\n<li style=\"font-weight: 400;\" aria-level=\"1\"><span style=\"font-weight: 400;\">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.<\/span><\/li>\n<li style=\"font-weight: 400;\" aria-level=\"1\"><span style=\"font-weight: 400;\">Diameter -: A chord or line segment passing through the circle&#8217;s center is called the diameter. It is also known as the longest chord.<\/span><\/li>\n<li style=\"font-weight: 400;\" aria-level=\"1\"><span style=\"font-weight: 400;\">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.<\/span><\/li>\n<li style=\"font-weight: 400;\" aria-level=\"1\"><span style=\"font-weight: 400;\">Arc-: An arc is a curve formed on the circle&#8217;s boundary between two points.<\/span><\/li>\n<\/ul>\n<p><span style=\"font-weight: 400;\">To learn more about such concepts you can check out <\/span><a href=\"https:\/\/www.aakash.ac.in\/important-concepts\/maths\" target=\"_blank\" rel=\"noopener\"><span style=\"font-weight: 400;\">Maths Concepts<\/span><\/a><span style=\"font-weight: 400;\"> and\u00a0 <\/span><a href=\"https:\/\/www.aakash.ac.in\/important-concepts\" target=\"_blank\" rel=\"noopener\"><span style=\"font-weight: 400;\">Important Concepts<\/span><\/a><span style=\"font-weight: 400;\">.\u00a0<\/span><\/p>\n<h3>What is the perimeter and area of a circle?<\/h3>\n<p><span style=\"font-weight: 400;\">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&#8217;t have any straight lines that can be easily measured, they have a special formula to determine the perimeter.<\/span><\/p>\n<p><span style=\"font-weight: 400;\">First of all, circles have a special term used in place of the perimeter, &#8216;circumference&#8217;. The symbol used to denote the term is C. The formula to calculate the circumference consists of a Pi\u00d7Diameter or Pi\u00d72R, equal to the circle&#8217;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 \u03c0. Therefore,\u00a0<\/span><\/p>\n<ol>\n<li style=\"font-weight: 400;\" aria-level=\"1\"><span style=\"font-weight: 400;\">Circumference(C) of the circle = \u201c\u03c0\u00d7 d\u201d or \u201c\u03c0\u00d7 2r\u201d<\/span><\/li>\n<\/ol>\n<ul>\n<li style=\"font-weight: 400;\" aria-level=\"1\"><span style=\"font-weight: 400;\">For example, Let&#8217;s assume \u2018d&#8217; is 8 then what will be the circumference?<\/span><\/li>\n<\/ul>\n<p><span style=\"font-weight: 400;\">C of the circle = \u03c0\u00d7d<\/span><\/p>\n<p><span style=\"font-weight: 400;\">\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0= 3.14 \u00d7 8 = 25.12<\/span><\/p>\n<ul>\n<li style=\"font-weight: 400;\" aria-level=\"1\"><span style=\"font-weight: 400;\">Let&#8217;s assume \u2018r\u2019 is 4 then calculate the circumference.<\/span><\/li>\n<\/ul>\n<p><span style=\"font-weight: 400;\">C of the circle = \u03c0\u00d72r<\/span><\/p>\n<p><span style=\"font-weight: 400;\">\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0= 3.14\u00d7 2\u00d7 4 = 25.12 (unit)<\/span><\/p>\n<ul>\n<li style=\"font-weight: 400;\" aria-level=\"1\"><span style=\"font-weight: 400;\">Why 2r not 2d? Well, the Radius is half of the Diameter&#8217;s length therefore while calculating with a radius the formula will be \u2018\u03c0\u00d72r&#8217; and with a diameter or will change to \u2018\u03c0\u00d7d\u2019 where the value of \u03c0 will remain the same.<\/span><\/li>\n<\/ul>\n<p><span style=\"font-weight: 400;\">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 \u03c0, which was 256\/81, nearly around 3.16. The area of a circle is pi times the radius squared, which means A= \u03c0\u00d7r\u00b2. Here \u2018A\u2019 is the symbolic representation of the area. Therefore,<\/span><\/p>\n<ol>\n<li style=\"font-weight: 400;\" aria-level=\"1\"><span style=\"font-weight: 400;\">The area of the circle = \u201c\u03c0\u00d7r\u00b2\u201d<\/span><\/li>\n<\/ol>\n<ul>\n<li style=\"font-weight: 400;\" aria-level=\"1\"><span style=\"font-weight: 400;\">For example, let&#8217;s assume \u2018r\u2019 is equal to 4 then determine the area<\/span><\/li>\n<\/ul>\n<p><span style=\"font-weight: 400;\">Area (A) of the circle = \u03c0\u00d7r\u00b2<\/span><\/p>\n<p><span style=\"font-weight: 400;\">\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0= 3.14\u00d74\u00b2 = 3.14\u00d716 = 50.24 (unit\u00b2)<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Theorems<\/span><\/p>\n<p><span style=\"font-weight: 400;\">In CBSE Class 10th Maths, In the chapter on circles, there are several most important theorems which consist of tangents.<\/span><\/p>\n<ol>\n<li style=\"font-weight: 400;\" aria-level=\"1\"><span style=\"font-weight: 400;\">First Theorem -: Tangent perpendicular to the radius at the point of contact.<\/span><\/li>\n<\/ol>\n<ul>\n<li style=\"font-weight: 400;\" aria-level=\"1\"><span style=\"font-weight: 400;\">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.<\/span><\/li>\n<li style=\"font-weight: 400;\" aria-level=\"1\"><span style=\"font-weight: 400;\">Given- YX is tangent at the point named P to the given circle with center O.<\/span><\/li>\n<li style=\"font-weight: 400;\" aria-level=\"1\"><span style=\"font-weight: 400;\">To prove- OP is perpendicular to YX<\/span><\/li>\n<li style=\"font-weight: 400;\" aria-level=\"1\"><span style=\"font-weight: 400;\">Construction- Take a point somewhere on YX, name it Q, and join it with the center O to form OQ.<\/span><\/li>\n<li style=\"font-weight: 400;\" aria-level=\"1\"><span style=\"font-weight: 400;\">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.<\/span><\/li>\n<\/ul>\n<ul>\n<li style=\"font-weight: 400;\" aria-level=\"1\"><span style=\"font-weight: 400;\">Therefore, OQ&gt; OP.<\/span><\/li>\n<li style=\"font-weight: 400;\" aria-level=\"1\"><span style=\"font-weight: 400;\">T<\/span><span style=\"font-weight: 400;\">The 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.<\/span><\/li>\n<li style=\"font-weight: 400;\" aria-level=\"1\"><span style=\"font-weight: 400;\">Therefore, OP is considered perpendicular to line YX because OP is the shortest side, referred to as the perpendicular.<\/span><\/li>\n<\/ul>\n<p><span style=\"font-weight: 400;\">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.<\/span><\/p>\n<ol>\n<li style=\"font-weight: 400;\" aria-level=\"1\"><span style=\"font-weight: 400;\">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.<\/span><\/li>\n<\/ol>\n<ul>\n<li style=\"font-weight: 400;\" aria-level=\"1\"><span style=\"font-weight: 400;\">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.<\/span><\/li>\n<li style=\"font-weight: 400;\" aria-level=\"1\"><span style=\"font-weight: 400;\">Given &#8211; 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.<\/span><\/li>\n<li style=\"font-weight: 400;\" aria-level=\"1\"><span style=\"font-weight: 400;\">To prove- AB is tangent at P.<\/span><\/li>\n<li style=\"font-weight: 400;\" aria-level=\"1\"><span style=\"font-weight: 400;\">Construction- Mark a point named Q on the marked line AB, other than the point P, and further join OQ.<\/span><\/li>\n<li style=\"font-weight: 400;\" aria-level=\"1\"><span style=\"font-weight: 400;\">Proof- since OP is perpendicular to line AB.<\/span><\/li>\n<li style=\"font-weight: 400;\" aria-level=\"1\"><span style=\"font-weight: 400;\">Therefore, OP&lt;OQ.<\/span><\/li>\n<li style=\"font-weight: 400;\" aria-level=\"1\"><span style=\"font-weight: 400;\">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.<\/span><\/li>\n<\/ul>\n<p><span style=\"font-weight: 400;\">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.<\/span><\/p>\n<ol>\n<li style=\"font-weight: 400;\" aria-level=\"1\"><span style=\"font-weight: 400;\">The third theorem-: The lengths of the tangents formed from an exterior point to a presented circle are equal.<\/span><\/li>\n<\/ol>\n<ul>\n<li style=\"font-weight: 400;\" aria-level=\"1\"><span style=\"font-weight: 400;\">Theorem- The theorem suggests that the lengths\/ sides of the tangents marked from an exterior point to a circle are equal.<\/span><\/li>\n<li style=\"font-weight: 400;\" aria-level=\"1\"><span style=\"font-weight: 400;\">Given &#8211; PT and PS\u00a0<\/span><\/li>\n<li style=\"font-weight: 400;\" aria-level=\"1\"><span style=\"font-weight: 400;\">Construction- Draw a line and join O to P, T, and S points.<\/span><\/li>\n<li style=\"font-weight: 400;\" aria-level=\"1\"><span style=\"font-weight: 400;\">Proof- In the constructed triangle, OTP and OSP.<\/span><\/li>\n<\/ul>\n<p><span style=\"font-weight: 400;\">OT= OS ( radius constructed in the same circle)<\/span><\/p>\n<p><span style=\"font-weight: 400;\">OP= OP (common)<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Angle OTP = Angle OSP (each at 90\u00b0)<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Therefore, \u2206 OTP= \u2206 OSP\u00a0<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Hence, PT= PS.<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Note related to tangents-: If the constructed two tangents are marked to a given circle from an exterior point, then,<\/span><\/p>\n<ul>\n<li style=\"font-weight: 400;\" aria-level=\"1\"><span style=\"font-weight: 400;\">They subtend a proportional angle at the given center of the circle.<\/span><\/li>\n<li style=\"font-weight: 400;\" aria-level=\"1\"><span style=\"font-weight: 400;\">The angles are inclined to the segment equally that are joining the center to the point.<\/span><\/li>\n<\/ul>\n<p><span style=\"font-weight: 400;\">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\u00a0 OSP.\u00a0<\/span><\/p>\n<h3>Facts and Properties<\/h3>\n<ul>\n<li style=\"font-weight: 400;\" aria-level=\"1\"><span style=\"font-weight: 400;\">The tangents constructed at the very ends of the diameter of any circle are parallel.<\/span><\/li>\n<li style=\"font-weight: 400;\" aria-level=\"1\"><span style=\"font-weight: 400;\">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.<\/span><\/li>\n<li style=\"font-weight: 400;\" aria-level=\"1\"><span style=\"font-weight: 400;\">In presented two concentric circles, the marked chord of the bigger circle touches and the smaller one intersect at the point of their connection.<\/span><\/li>\n<li style=\"font-weight: 400;\" aria-level=\"1\"><span style=\"font-weight: 400;\">The perpendicular is formed at the point of connection with the passing tangents to the presented circle through its center.<\/span><\/li>\n<li style=\"font-weight: 400;\" aria-level=\"1\"><span style=\"font-weight: 400;\">The parallelogram circumscribed in a circle is known as a rhombus.<\/span><\/li>\n<li style=\"font-weight: 400;\" aria-level=\"1\"><span style=\"font-weight: 400;\">The adverse sides of any quadrilateral circumscribed around the given circle forms supplementary angles at the center of the presented circle.<\/span><\/li>\n<\/ul>\n<h3>Sample Questions<\/h3>\n<p><span style=\"font-weight: 400;\">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 \u2220AOB = 100\u00b0, then calculate \u2220BAD.\u00a0<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Solution:<\/span><\/p>\n<p><span style=\"font-weight: 400;\">In the mentioned figure let&#8217;s construct an angle named E and another angle named F at points A and B respectively.<\/span><\/p>\n<p><span style=\"font-weight: 400;\">\u2220E = \u2220F<\/span><\/p>\n<p><span style=\"font-weight: 400;\">\u2220E + \u2220F + 100\u00b0 = 180\u00b0<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Since \u2220E=\u2220F<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Therefore,<\/span><\/p>\n<p><span style=\"font-weight: 400;\">\u2220E + \u2220E = 80\u00b0<\/span><\/p>\n<p><span style=\"font-weight: 400;\">\u21d2 2\u2220E = 80\u00b0<\/span><\/p>\n<p><span style=\"font-weight: 400;\">\u21d2 \u2220E = 40\u00b0<\/span><\/p>\n<p><span style=\"font-weight: 400;\">\u2220E + \u2220BAD = 90\u00b0<\/span><\/p>\n<p><span style=\"font-weight: 400;\">\u2220BAD = 90\u00b0 \u2013 40\u00b0 = 50\u00b0<\/span><\/p>\n<p><span style=\"font-weight: 400;\">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.<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Solution:\u00a0<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Given: PQ and RS are 2\u00a0 parallel tangents at points A and T of a circle with center C.<\/span><\/p>\n<p><span style=\"font-weight: 400;\">To prove: AB passes through the circle with center C or ACT is the diameter of the circle.<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Construction: Join CA and CB. Draw CU || PQ, Mark \u2220a, \u2220b and \u2220c<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Proof: \u2220a= 90\u00b0 \u2026 (i)<\/span><\/p>\n<p><span style=\"font-weight: 400;\">[\u2235 Tangent is parallel to the radius through the point of contact<\/span><\/p>\n<p><span style=\"font-weight: 400;\">CU || PQ<\/span><\/p>\n<p><span style=\"font-weight: 400;\">\u2234 \u2220a + \u2220b = 180\u00b0 \u2026(due to Co-interior angles<\/span><\/p>\n<p><span style=\"font-weight: 400;\">90\u00b0 + \u2220b = 180\u00b0 \u2026[From (i)<\/span><\/p>\n<p><span style=\"font-weight: 400;\">\u2220b = 180\u00b0 \u2013 90 = 90\u00b0<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Similarly, \u2220c = 90\u00b0<\/span><\/p>\n<p><span style=\"font-weight: 400;\">\u2220c + \u2220b = 90\u00b0 + 90\u00b0 = 180\u00b0<\/span><\/p>\n<p><span style=\"font-weight: 400;\">\u2234 ACT is a straight line.<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Hence ACT can be determined as the diameter of the circle with center C.<\/span><\/p>\n<p><span style=\"font-weight: 400;\">\u2234 AT\u00a0 passes through the circle with center C.<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Question 3) From an exterior point D, two tangents DA and DB are drawn to a circle with center C. If \u2220DAB = 50\u00b0, then find \u2220ACB.\u00a0<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Solution:<\/span><\/p>\n<p><span style=\"font-weight: 400;\">DA = DB \u2026[\u2235 Tangents drawn from an exterior point are equal to each other<\/span><\/p>\n<p><span style=\"font-weight: 400;\">\u2220DBA = \u2220DAB = 50\u00b0 \u2026[ property of Angles equal to opposite sides<\/span><\/p>\n<p><span style=\"font-weight: 400;\">In \u2206ABD,\u00a0<\/span><\/p>\n<p><span style=\"font-weight: 400;\">\u2220DBA + \u2220DAB + \u2220ADB = 180\u00b0 \u2026[ due to Angle-sum-property of a triangle<\/span><\/p>\n<p><span style=\"font-weight: 400;\">50\u00b0 + 50\u00b0 + \u2220ADB = 180\u00b0<\/span><\/p>\n<p><span style=\"font-weight: 400;\">\u2220ADB is equal to 180\u00b0 \u2013 50\u00b0 \u2013 50\u00b0 = 80\u00b0<\/span><\/p>\n<p><span style=\"font-weight: 400;\">In cyclic quadrilateral CADB<\/span><\/p>\n<p><span style=\"font-weight: 400;\">\u2220ACB + \u2220ADB = 180\u00b0 \u2026\u2026[the property of the Sum of opposite angles of a cyclic quadrilateral is 180\u00b0<\/span><\/p>\n<p><span style=\"font-weight: 400;\">\u2220ACB + 80o = 180\u00b0<\/span><\/p>\n<p><span style=\"font-weight: 400;\">\u2220ACB = 180\u00b0 \u2013 80\u00b0 = 100\u00b0<\/span><\/p>\n<p><span style=\"font-weight: 400;\">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 \u2220ADB is 60\u00b0. Find the length of the marked chord with the name AB.\u00a0<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Solution:<\/span><\/p>\n<p><span style=\"font-weight: 400;\">DA = DB \u2026[Tangents drawn from an external point D are equal<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Given:<\/span><\/p>\n<p><span style=\"font-weight: 400;\">\u2220ADB = 60\u00b0<\/span><\/p>\n<p><span style=\"font-weight: 400;\">\u2220PDB = \u2220DBA \u2026 (i) \u2026( Property of Angles opposite to equal sides<\/span><\/p>\n<p><span style=\"font-weight: 400;\">In \u2206DAB, \u2220DAB + \u2220DBA + \u2220ADB = 180\u00b0 \u2026[ Due to the Angle-sum-property of a triangle<\/span><\/p>\n<p><span style=\"font-weight: 400;\">\u21d2 \u2220DAB + \u2220DAB + 60\u00b0 = 180\u00b0<\/span><\/p>\n<p><span style=\"font-weight: 400;\">\u21d2 2\u2220DAB = 180\u00b0 \u2013 60\u00b0 = 120\u00b0<\/span><\/p>\n<p><span style=\"font-weight: 400;\">\u21d2 \u2220DAB = 60\u00b0<\/span><\/p>\n<p><span style=\"font-weight: 400;\">\u21d2 \u2220DAB = \u2220DBA = \u2220ADB = 60\u00b0<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Therefore, ADB is an equilateral triangle<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Hence, AB = AD = 5 cm \u2026[\u2235 All sides of an equilateral triangle are equal to each other<\/span><\/p>\n<h3>Conclusion<\/h3>\n<p><span style=\"font-weight: 400;\">A famous quote on Maths is, &#8220;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.<\/span><\/p>\n<p>&nbsp;<\/p>\n","protected":false},"excerpt":{"rendered":"<p>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 [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":136634,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[3581],"tags":[1568,2749,2812,2126],"class_list":["post-136559","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-cbse","tag-cbse-class-10","tag-cbse-class-10-maths","tag-cbse-maths","tag-cbse-term-2"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v26.0 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>CBSE Class 10 Maths sample questions on Circles<\/title>\n<meta name=\"description\" content=\"Through this article you will 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