WebbIn an equilateral triangle ABC,D is a point on side BC such that BD= 31BC. Prove that 9AD 2=7AB 2. Hard. 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 …
In an equilateral triangle abc d is a point on side bc such that bd
Webb1. In an equilateral Δ ABC, the side BC is trisected at D. Prove that 9AD2 = 7AB2 2. P and Q are points on sides AB and AC respectively, of ΔABC. If AP = 3 cm,PB = 6 cm, AQ = 5 cm and QC = 10 cm, show that BC = 3 PQ. 3. The image of a tree on the film of a camera is of length 35 mm, the distance from the lens to the film is Webb28 mars 2024 · Prove that 9AD2 = 7 AB2 Given: Equilateral triangle ABC D is a point an BC Such that BD = 1/3 BC To prove: 9 AD2 = 7 AB2 Construction: Lets draw AE BC Proof: All … f3a akb
Ex 6.5, 15 - In equilateral triangle ABC, D is a point on BC - teachoo
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 … Webb30 mars 2024 · Transcript. Question 17 In given figure ∠1 = ∠2 and ∆NSQ ≅ ∆MTR , then prove that ∆PTS ~ ∆ PRQ . Given: ∠ 1 = ∠ 2 & ΔNSQ ≅ ΔMTR To Prove: ΔPTS ∼ PRQ Proof: Given ΔNSQ ≅ ΔMTR ∴ ∠ NQS = ∠ MRT i.e. ∠ PQR = ∠ PRQ Now, In Δ PST By angle sum property ∠ P + ∠ 1 + ∠ 2 = 180° Since ∠ 1 = ∠ 2 given ∠ P + ∠ 1 + ∠ 1 = 180° ∠ P + 2 ∠ 1 = … hindi gunjan class 8 pdf