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Question

In A B C comma straight angle C equals 90 to the power of ring operator. C D perpendicular A B. CD equals straight h. Bc equals straight a comma CA equals straight b text  and  end text AB equals straight ctext  Prove that  end text 1 over h squared equals 1 over a squared plus 1 over b squared

The correct answer is: Hence proved


    Solution :-
    Aim  :- Prove that 1 over h squared equals 1 over a squared plus 1 over b squared
    Hint :- Using pythagoras theorem in triangle ABC we get relation between a, b and c
    Now , using the relation between a, b, c and h found from the area of the triangle, substitute the value of c in terms of a, b and h to get the required relation.
    Explanation(proof ) :-
    Area of triangle ABC = 1 half × base × height
    Area of triangle taking AC as base = 1 half ×  ab
    Area of triangle taking AB as base = 1 half× c h
    As area of triangle ABC is constant so, we get  ab = ch not stretchy rightwards double arrow c equals fraction numerator a b over denominator h end fraction
    Applying pythagoras theorem in ABC ,we get
      c squared equals a squared plus b squared                                                                                                                  
    Substitute c equals fraction numerator a b over denominator h end fraction in the above equation
    open parentheses fraction numerator a b over denominator h end fraction close parentheses squared equals a squared plus b squared not stretchy rightwards double arrow 1 over h squared equals fraction numerator a squared plus b squared over denominator a squared b squared end fraction not stretchy rightwards double arrow 1 over h squared equals 1 over a squared plus 1 over b squared
    Hence proved

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