Theorems Related to Circle | Class 10 math wbbse | বৃত্ত সম্পর্কিত উপপাদ্য | প্রয়োগ : 9,10,13,14

Sohoj Ganit Pathshala shares content - WB Board class 10 math solution ? Theorems of Circle | Wbbse class 10 math book solution | Theorems about circle | Ganit Prakash book chapter 3 | How can we prove extra theorems ? What are the applications of theorems ? Here is the answers of extra theorem ( applications ) 9,10,13,14 of page 62, 63, 64 of WB madhyamik ganit Prakash book. Aplications of the theorem 31 , 32 , 33 are below : Application: 9. Prove that diameter is the largest chord of a circle. Given: MN is any chord not the diameter of the circle centred with O and AC is a diameter. We have to Prove: AC is greater than MN i.e. diameter is the largest chord of the circle. Drawing: we draw OD perpendicular to chord MN from O. Add O, M. Proof : OM is greater than MD [since ∆OMD is a right triangle and OM is hypotenuse] or, OA is greater than MD [since OA = OM is the radius of the same circle] or, ½AC is greater than ½ MN or, AC is greater than MN . So the diameter is the largest chord of the circle. (Proved) Application: 10. Prove by logic that two equal chords of a circle are equidistant from the centre of the circle. Given: AB and CD are two chords of equal length of a circle centred at O.Distances of AB and CD from the center are OE and OF respectively i.e. OE ⊥ AB and OF ⊥ CD To prove: OE = OF Drawing: O, A and O, C are added. Proof: OE ⊥ AB and OF ⊥ CD [Given] therefore, AE=½AB and CF=½CD [Since, a perpendicular bisects a chord on a chord not a diameter from the center of the circle.] Again, AB = CD [Given] hence, AE = CF..... (i) Now, in right angle triangles ∆ΑΕΟ and ∆OFC. ∠OEA = ∠OFC (each are right angle) Hypotenuse OA = Hypotenuse OC [radius of same circle] AE = CF [ from (i)] Thus, ∆AEO = ∆CFO [by R-H-S equality condition] Therefore, OE = OF [proved] Application: 13. If two circles intersect in two points, prove that their centers lie on the perpendicular bisector of their common chord. Given: Two circles with centers X and Y intersect at points C and D. So CD is its common chord. We have to Prove: Points X and Y lie on the perpendicular bisector of two common chords CD. Drawing: From point X, we draw XO perpendicular to CD. I added two points O and Y. Proof: CD is the chord of the circle centred with X and XO ⊥ CD Therefore, O is the mid-point of CD. Again, CD is the chord of the circle centred with Y and O, is the midpoint of CD. Thus, ΟΥ⊥ CD Since only one perpendicular to a point on a straight line is possible, therefore, XO and OY lie on the same straight line. Hence, XY is the perpendicular bisector of the common chord CD. .. The centers X and Y of two circles lie on the perpendicular bisector of their common chord CD. [proved] Application: 14. prove that the straight line joining the midpoints of two parallel chords of a circle is concentric. Ans: Given: Let two chords AB and CD of a circle with cente O be parallel to each other and the midpoints of AB and CD are respectively P , Q. To prove: PQ passes through O. Drawing: Join O, P and O, Q and draw straight line MN parallel to AB and CD through point O. Proof: P is the midpoint of chord AB. therefore, OP ⊥ AB . Again AB || MN, therefore OP ⊥ MN. Similarly, OQ ⊥CD [since Q, midpoint of CD] Thus,OQ⊥ MN [since MN || CD]. Both OP and OQ are perpendicular to MN at point O.Since only one perpendicular can be drawn at a point on a straight line, therefore, P, O and Q are collinear. .. PQ, passes through the center of the circle . [proved] So I hope that you learnt the applications of theorem 31, theorem 32, theorem 33 very wel. At the last I request you to subscribe my channel and shares the videos to your relatives.