OR Prove that a parallelogram circumscribing a circle is a rhombus. A rhombus is a parallelogram with all sides equal, 5. These relationships are: 1. https://www.zigya.com/share/TUFFTjEwMDUwMDQw. So, This circle is called the incircle of the quadrilateral or its inscribed circle, its center is the incenter and its radius is called the inradius. 232, Block C-3, Janakpuri, New Delhi, Ex 10.2,8 A quadrilateral ABCD is drawn to circumscribe a circle (see figure). 3. So, ABCD is a parallelogram with all sides equal An example of a quadrilateral that cannot be cyclic is a non-square rhombus. AB + CD = AD + BC Let ABCD be a parallelogram and a circle with centre O. Jan 17, 2021 - NCERT Solutions - Chapter 10: Circles, Class 10, Maths Class 10 Notes | EduRev is made by best teachers of Class 10. A triangle has exactly one circumscribed circle. Given: ABCD be a parallelogram circumscribing a circle with centre O. AB + AB = AD + AD If all the side of a parallelogram touch a circle, show that the parallelogram is a rhombus. It can be observed that. BP = BQ (vi) The sum of the squares of the diagonals is equal to the sum of the squares of the four sides in the figure. The two heights in a rhombus are equal, that is, the rhombus arises out of the intersection of two congruent strips. Zigya App. A problem about orthocenter and circumscribed circle. Teachoo is free. If a quadrilateral is inscribed inside of a circle, then the opposite angles are supplementary. Let sides AB, BC, CD and AD of the parallelogram touch the circle at E, F, G and H respectively. Inscribed quadrilaterals are also called cyclic quadrilaterals. Ex 10.2,11Prove that the parallelogram circumscribing a circle is a rhombus.Given: A circle with centre O.A parallelogram ABCD touching the circle at points P,Q,R and STo prove: ABCD is a rhombus Proof:A rhombus is a parallelogram with all sides equal, So, we have to prove a Since ABCD is a parallelogram, AB = CD .....1. Adding the above equations, AP + BP + CR + DR = AS + BQ + CQ + DS. Adding (2) + (3) + (4) + (5) Hence, opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle. A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively (see figure). 120.4k VIEWS. 2AB = 2AD A cyclic polygon has each of its vertices on a particular circle, called the circumcircle or circumscribed circle. Don't spam and answer it as fast as possible 1 See answer ganeshkumar14 is waiting for your help. AB = AD Download the PDF Question Papers Free for off line practice and view the Solutions online. This circle is called the circumcircle or circumscribed circle, and the vertices are said to be concyclic. (iv) A parallelogram circumscribed about a circle is a rhombus. Let sides AB, BC, CD and AD of the parallelogram touch the circle at E, F, G and H respectively. DR = DS (Tangents on the circle from point D) CR = CQ (Tangents on the circle from point C) BP = BQ (Tangents on the circle from point B) AP = AS (Tangents on the circle from point A) Adding all these equations, we obtain. Also, the interior opposite angles of a parallelogram are equal in measure. Since, the length of two tangents drawn from an external point to a circle are equal.So,    AE = AH    ...(i)BE = BF    ....(ii)CG = CF    ...(iii)and    DG = DH    ....(iv)Adding (i), (ii), (iii) and (iv), we getAE + BE + GC + DG = AH + BF + CF + DH⇒ (AE + BE) + (GC + DG)= (AH + DH) + (BF + CF)⇒ AB + CD = AD + BC⇒    2 AB = 2BC[∵ ABCD is a || gm So, AB = CDand BC = AD]⇒    AB = BCSimilarly, BC = CD and    CD = ADThus,    AB = BC = CD = DAHence, ABCD is a rhombus. Prove that the parallelogram circumscribing a circle is a rhombus. Hence, We know that the tangents drawn to a circle from an exterior point are equal in length. The center of the circumcircle, called the circumcenter, can be considered a center of the polygon. Proof: Also, radius = … Prove that the parallelogram circumscribing a circle is a rhombus. A circumscribed circle and a trapezoid. How many circumscribed circles can a triangle have? Let ABCD be a parallelogram and a circle with centre O. The circle is called the circumscribing circle and the radius of the circle is the circumradius of the polygon. 2021 Zigya Technology Labs Pvt. The rhombus can be circumscribed by a circle. Since, the tangent at any point of a circle is perpendicular to the radius through the point of contact.∴ ∠OAP = 90°Now, in ΔOAP,∠OAP + ∠OPA + ∠POA = 180°⇒ 90° + 40° + ∠POA = 180°⇒    130 + ∠POA = 180°⇒    ∠POA = 50°So, right option is (A). Teachoo provides the best content available! This document is highly rated by … Therefore, AP = AS, BP = BQ, CR = CQ and DR = DS. Ltd. Download books and chapters from book store. A parallelogram is a two-dimensional geometrical shape, whose sides are parallel to each other. if a parallelogram is inscribed in a circle, it must be a square. Answer: Consider a parallelogram ABCD which is circumscribing a circle with a center O. Prove that the parallelogram circumscribing a circle is a rhombus. If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 80°, then ∠ POA is equal to, In ΔPOA and ΔPOBPA = PB(Tangents from external point P)OA = OB (Radii of a circle)and    OP = OP (common)∴ ΔPOA ≅ ΔPOB(by SSS congruency)⇒ ∠OPA = ∠OPB⇒ ∠OPA = ∠OPB = 40°. ... No, a circle can't be a parallelogram. On signing up you are confirming that you have read and agree to All triangles, all regular simple polygons, all rectangles, all isosceles trapezoids, and all right kites are cyclic. SOLUTION: Soln. AB = CD & AD = BC ... Parallelogram and circumscribed circle. 2. Hence proved. : 13) NCERT Solution for Class 10 math - circles 214 , Question 13 Now, since ABCD is a parallelogram, AB = CD and BC = AD. A circle many have ______ parallel tangents. (ii) Rectangles: A rectangle is a parallelogram with all angles 90º. 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