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    Subject

    Bank Exam

    Topic

    Trust Bank Ltd. - Management Trainee Officer - 2016

    In the figure below, the angle POQ is 90 degrees. What is the perimeter of the triangle OPQ ?

    ক)
    4\(\sqrt 2 \)
    খ)
    8\(\sqrt 2 \)
    গ)
    6 + 2\(\sqrt 5 \)
    ঘ)
    6
    ঙ)
    6\(\sqrt 2 \) + 2\(\sqrt {10} \)

    Explanation

    Solution: অনুবাদঃ POQ একটি সমকোণী ত্রিভুজ । ত্রিভুজটির পরিসীমা বের করতে হবে । (x 1 , y 1 ) এবং (x 2 , y 2 ) দূরত্ব = (x1−x2)2+(y1−y2)2\sqrt {{{({x_1} - {x_2})}^2} + {{({y_1} - {y_2})}^2}} PO = (−4−0)2+(4−0)2 =16+16 =32\sqrt {{{( - 4 - 0)}^2} + {{(4 - 0)}^2}}  = \sqrt {16 + 16}  = \sqrt {32} = 2×16 =42\sqrt {2 \times 16}  = 4\sqrt 2 QO = (2−0)2+(2−0)2 =22  +22  \sqrt {{{(2 - 0)}^2} + {{(2 - 0)}^2}}  = \sqrt {{2^{2\;}} + {2^{2\;}}} = 2+22   =22\sqrt {2 + {2^{2\;}}}  = 2\sqrt 2 PO = (−4−2)2+(4−2)2 =(−6)2+(2)2\sqrt {{{( - 4 - 2)}^2} + {{(4 - 2)}^2}}  = \sqrt {{{( - 6)}^2} + {{(2)}^2}} = 40 = =210\sqrt {40}  =  = 2\sqrt {10} Perimeter of OPQ will be = (42 +22 +210)(4\sqrt 2  + 2\sqrt 2  + 2\sqrt {10} ) = 62\sqrt 2 + 210\sqrt {10}

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