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Math Assignment Ch-6 Class X | Triangle
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Internal bisector of angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle. In Triangle ABC, if AD is the bisector of ∠A then \[\frac{AB}{AC}=\frac{BD}{DC}\]
Question 1: Q. 1. PA, QB and RC are each perpendicular to AC then |
Question 2: If the bisector of an angle of triangle bisects the opposite side, then prove that the triangle is isosceles. |
Question 3: O is any point inside a triangle ABC. The bisectors of ∠AOB, ∠BOC and ∠COA meet the side AB, BC and CA in points D, E, F respectively. Show that AD X BE X CE = BD X EC X FA |
Question 4: In triangle ABC, AD is the internal bisector of ∠A which meets BC at point D. Prove that: |
Question 5: In the given figure express x in terms of a, b, c |
Question 6: Q6 The areas of two similar triangles are 81 cm2 and 49 cm2 respectively. Find the ratio of their corresponding heights? What is the ratio of their corresponding median. |
Question 7: In quadrilateral ABCD, ∠B = 90o If AD2 = AB2 + BC2 = CD2 Prove that ∠ACD = 90o |
Question 8: Prove that 8PT2 = 3PR2 + 5PS2
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Question 9: Triangle ABC is right angled at B , D is the mid- point of BC. Prove that AC2 = 4AD2 - 3AB2 |
Question 10: In triangle ABC, AD ⟂ BC and point D lies on BC such that 2BD = 3CD, Prove that 5AB2 = 5AC2 +BC2 |
Question 11: △ABC is right angled at C. Let BC = a, CA = b, BC = c, and let p be the length of perpendicular from C on AB Prove That :
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Question 12: In △ABC, P and Q are the points on sides CA and CB respectively, which divides these sides in 2 : 1. Prove that (i) 9AQ2 = 9AC2 + 4BC2 (ii) 9BP2 = 9BC2 + 4AC2 (iii) 13AB2 = 9(AQ2 + BP2)
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Question 13: In △ABC, AC > AB, D is the mid - point of BC and AE⟂BC . |
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