WebbBD = —BC I 9AD2 = 7AB2. ABC BC D BD = 1 —BC 9AD2 = 7AB2. 30. From a point on a bridge across a river, the angles of depression of the banks on opposite sides of the river are 300 and 450 respectively. If the bridge is at a height of 3 m from the bank, find the width of the river. 300 4501 3 I 300 4501 3 I 7tà ùà 300 3TTà 450, 3 WebbClick here👆to get an answer to your question ️ In an equilateral triangle ABC, D is a point on side BC such that BD = 13BC . Prove that 9 AD^2 = 7 AB^2 .

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WebbProve that 9AD 2 = 7AB 2 Solution: We know that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. In ΔABC as shown … WebbIn an Equilateral Triangle Abc, the Side Bc is Trisected at D. Prove that 9 Ad2 = 7 Ab2. - Mathematics Shaalaa.com. CISCE ICSE Class 9. Question Papers 10. Textbook … female tiefling bouncer

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WebbProve that 9AD2 = 7AB2. Given : An equilateral triangle ABC such that To Prove : 9AD2 = 7AB Given : An equilateral triangle ABC such that BD equals 1 third BC.To Prove : 9AD2 = 7AB2 Const: Draw AE ⊥ BC. Proof: In right triangles ABE and ACE, we have AE = AE [common] ∠AEB = ∠AEC [90°] and AB = AC [∆ABC is an equilateral] Therefore, by ... Webb31 jan. 2024 · To prove :- → 9AD² = 7AB² . Construction :- → Draw AL ⊥ BC . Proof :- In right triangles ALB and ALC, we have AB = AC ( given ) and AL = AL ( common ) ∴ ∆ALB ≅ ∆ ALC [ By RHS axiom ] . So, BL = CL . Thus, BD = ⅓BC and BL = ½BC . In ∆ALB, ∠ALB = 90° . ∴ AB² = AL² + BL² ....... (1) [ by Pythagoras' theorem ] . In ∆ALD , ∠ALD = 90° . WebbGiven : An equilateral triangle ABC such that To Prove : 9AD2 = 7AB2Const: Draw AE ⊥ BC.Proof: In right triangles ABE and ACE, we haveAE = AE [common]∠AEB = ∠AEC … definitive technology bp 8080