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17 September, 01:02

The ΔPQR is right-angled at P, and PN is an altitude. If QN = 12 in and NR = 6 in, find PN, PQ, PR.

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  1. 17 September, 03:50
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    We know that PN^{2} = QN. NR

    So, PN = / sqrt{QN. NR}

    = / sqrt{12. 6} = / sqrt{72}

    PN = 8.5

    Now, PNQ is a right angled triangle with PQ as the hypotenuse.

    So, PQ^{2} = PN^{2} + NQ^{2}

    That means PQ = / sqrt{ PN^{2} + NQ^{2} }

    = / sqrt{ 8.5^{2} + 12^{2} }

    PQ = 14.7

    Same way, PNR is also a right angled triangle with PR as the hypotenuse

    So, PR^{2} = PN^{2} + NR^{2}

    That means PR = / sqrt{ PN^{2} + NR^{2} }

    = / sqrt{ 8.5^{2} + 6^{2} }

    PR = 10.4
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