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In △JKL, KM is an altitude. What is the length, in units, of JK?
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Have u got any related diagram of the question?
just posted it
By using gepmetric mean property we have \[KM^2 =JM \times ML =5 \times (35-5)=5 \times 30 = 150\] i.e. \[KM = \sqrt{150}= \sqrt{25 \times 6} =5 \sqrt{6} units\] = 5*2.449=12.245 units
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Now since KM=5√6 units so condsider triangle JMK use here Pythagoras theorem we have \[JK^2=JM^2+MK^2 =5^2+ (5\sqrt{6})^2=25+150 =175 \] \[JK =\sqrt{175} \rightarrow JK = \sqrt{25 \times 7} \rightarrow JK =5 \sqrt{7} units\]
So 5sprt of 7 is the answer
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