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18. In the given figure, Δ ABC is right angled at CIfBC-a, CA-b, AB-C & P is the length ofperpendicular drawn from C to AB then prove that-A Ki, cp = ab |
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Answer» (i) Area of ΔABC = 1/2 × Base × Height = 1/2 × BC × AC = 1/2 ab Area of ΔABC = 1/2 × Base × Height = 1/2 × AB × CD = 1/2 cp ⇒ 1/2 ab = 1/2 cp ⇒ ab = cp Hence proved.(ii) In right angled triangle ABC,AB2= BC2+ AC2 c2= a2+ b2(ab/p)2= a2+ b2a2b2/p2= a2+ b2-------- From proof (1)1/p2= (a2+ b2) / a2b21/p2= (a2/ a2b2+ b2/ a2b2) 1/p2= (1/b2+ 1/a2)1/p2= (1/a2+ 1/b2)Hence proved. |
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