# Example 5

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Example 5 Are the points A (3, 6, 9), B (10, 20, 30) and C (25, – 41, 5), the vertices of a right angled triangle? Lets first calculate distances AB, BC and AC and then apply Pythagoras theorem to determine whether it is right angle triangle Calculating AB A (3, 6, 9) B (10, 20, 30) AB = x2−x12+y2−y12+z2 −z12 Here, x1 = 3, y1 = 6, z1 = 9 x2 = 10, y2 = 20, z2 = 30 AB = 10−3)2+(20−62+30−92 = 7)2+(142+212 = 49+196+441 = 686 Calculating BC B (10, 20, 30) C (25, – 41, 5) BC = x2−x12+y2−y12+z2 −z12 Here x1 = 10, y1 = 20, z1 = 30 x2 = 25, y2 = – 41, z2 = 5 BC = 25−10)2+(−41−202+5−302 = 15)2+(−612+−252 = 225+3721+625 = 4571 Calculating CA C (25, –41, 5) A (3, 6, 9) CA = x2−x12+y2−y12+z2 −z12 x1 = 25, y1 = – 14, z1 = 5 x2 = 3, y2 = 6, z2 = 9 AB = 3−25)2+(6−(−41)2+9−52 = −22)2+(6+412+42 = 484+472+16 = 484+2209+16 = 2709 Now AB = 686 , BC = 4571 , CA = 2709 In Right angle tringle, (Hypotenuse)2 = (Height)2 + (Base)2 Since 4571 is the biggest of the three sides , we take Hypotenuse = 4571 Hence we have to prove (4571)2 = (686)2 + (2709)2 Thus, L.H.S ≠ R.H.S Hence, It is not a right angle triangle

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CA Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .