Why would I divide by 4 instead of 2 in this equation?












2












$begingroup$


So I have this problem:




A vertical flag pole of height $h;text{meters}$ is erected exactly in the middle of the flat roof of a building. The roof is rectangular of width $w;text{meters}$ and depth $d;text{meters}$. The flag pole is stabilized by cables that join the corners of the roof top to the flag pole at a point $k;text{meters}$ below the top of the flagpole.



Let $ell;text{m}$ be the total length of cable required to stabilize the flag pole. Find a correct expression for $ell$ in terms of $w$, $d$, $h$ and $k$.




Illustration of the flag pole




After working it out several times I always came to the conclusion that the answer should be $$4sqrt{frac{w^2+d^2}{2}+(h-k)^2}$$



However the correct answer is actually
$$4sqrt{frac{w^2+d^2}{4}+(h-k)^2}$$ and I just don't understand why you would divide by 4 instead of 2?



The diagonal length of the roof is calculated using Pythagorean Theorem $sqrt{w^2+d^2}$. However, what's used for determining the hypotenuse is just half of that diagonal length right? If someone could explain this it would be of great assistance.






algebra-precalculus







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$endgroup$








  • 1




    $begingroup$
    $frac{sqrt{w^2+d^2}}{2} = sqrt{frac{w^2+d^2}{4}}$
    $endgroup$
    – Daniel Gendin
    Dec 10 '18 at 22:03
















2












$begingroup$


So I have this problem:




A vertical flag pole of height $h;text{meters}$ is erected exactly in the middle of the flat roof of a building. The roof is rectangular of width $w;text{meters}$ and depth $d;text{meters}$. The flag pole is stabilized by cables that join the corners of the roof top to the flag pole at a point $k;text{meters}$ below the top of the flagpole.



Let $ell;text{m}$ be the total length of cable required to stabilize the flag pole. Find a correct expression for $ell$ in terms of $w$, $d$, $h$ and $k$.




Illustration of the flag pole




After working it out several times I always came to the conclusion that the answer should be $$4sqrt{frac{w^2+d^2}{2}+(h-k)^2}$$



However the correct answer is actually
$$4sqrt{frac{w^2+d^2}{4}+(h-k)^2}$$ and I just don't understand why you would divide by 4 instead of 2?



The diagonal length of the roof is calculated using Pythagorean Theorem $sqrt{w^2+d^2}$. However, what's used for determining the hypotenuse is just half of that diagonal length right? If someone could explain this it would be of great assistance.






algebra-precalculus







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    $frac{sqrt{w^2+d^2}}{2} = sqrt{frac{w^2+d^2}{4}}$
    $endgroup$
    – Daniel Gendin
    Dec 10 '18 at 22:03














2












2








2





$begingroup$


So I have this problem:




A vertical flag pole of height $h;text{meters}$ is erected exactly in the middle of the flat roof of a building. The roof is rectangular of width $w;text{meters}$ and depth $d;text{meters}$. The flag pole is stabilized by cables that join the corners of the roof top to the flag pole at a point $k;text{meters}$ below the top of the flagpole.



Let $ell;text{m}$ be the total length of cable required to stabilize the flag pole. Find a correct expression for $ell$ in terms of $w$, $d$, $h$ and $k$.




Illustration of the flag pole




After working it out several times I always came to the conclusion that the answer should be $$4sqrt{frac{w^2+d^2}{2}+(h-k)^2}$$



However the correct answer is actually
$$4sqrt{frac{w^2+d^2}{4}+(h-k)^2}$$ and I just don't understand why you would divide by 4 instead of 2?



The diagonal length of the roof is calculated using Pythagorean Theorem $sqrt{w^2+d^2}$. However, what's used for determining the hypotenuse is just half of that diagonal length right? If someone could explain this it would be of great assistance.






algebra-precalculus







share|cite|improve this question











$endgroup$




So I have this problem:




A vertical flag pole of height $h;text{meters}$ is erected exactly in the middle of the flat roof of a building. The roof is rectangular of width $w;text{meters}$ and depth $d;text{meters}$. The flag pole is stabilized by cables that join the corners of the roof top to the flag pole at a point $k;text{meters}$ below the top of the flagpole.



Let $ell;text{m}$ be the total length of cable required to stabilize the flag pole. Find a correct expression for $ell$ in terms of $w$, $d$, $h$ and $k$.




Illustration of the flag pole




After working it out several times I always came to the conclusion that the answer should be $$4sqrt{frac{w^2+d^2}{2}+(h-k)^2}$$



However the correct answer is actually
$$4sqrt{frac{w^2+d^2}{4}+(h-k)^2}$$ and I just don't understand why you would divide by 4 instead of 2?



The diagonal length of the roof is calculated using Pythagorean Theorem $sqrt{w^2+d^2}$. However, what's used for determining the hypotenuse is just half of that diagonal length right? If someone could explain this it would be of great assistance.






algebra-precalculus




algebra-precalculus






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share|cite|improve this question













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edited Dec 10 '18 at 22:00









gt6989b

34k22455




34k22455










asked Dec 10 '18 at 22:00









Ballusha ThreeBallusha Three

132




132








  • 1




    $begingroup$
    $frac{sqrt{w^2+d^2}}{2} = sqrt{frac{w^2+d^2}{4}}$
    $endgroup$
    – Daniel Gendin
    Dec 10 '18 at 22:03














  • 1




    $begingroup$
    $frac{sqrt{w^2+d^2}}{2} = sqrt{frac{w^2+d^2}{4}}$
    $endgroup$
    – Daniel Gendin
    Dec 10 '18 at 22:03








1




1




$begingroup$
$frac{sqrt{w^2+d^2}}{2} = sqrt{frac{w^2+d^2}{4}}$
$endgroup$
– Daniel Gendin
Dec 10 '18 at 22:03




$begingroup$
$frac{sqrt{w^2+d^2}}{2} = sqrt{frac{w^2+d^2}{4}}$
$endgroup$
– Daniel Gendin
Dec 10 '18 at 22:03










1 Answer
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$begingroup$

The full diagonal of the rectangle is $sqrt{w^2+d^2}$, so half of this is



$$frac{sqrt{w^2+d^2}}{2}.$$



This length is one leg of a right triangle; the other leg is $(h-k)$.



Hence the total length of four identical lengths of wire that make up the four hypoteneuses is



$$4 sqrt{left(frac{sqrt{w^2+d^2}}{2}right)^2+(h-k)^2} = 4 sqrt{left(frac{w^2+d^2}{4}right)+(h-k)^2}.$$






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    $begingroup$

    The full diagonal of the rectangle is $sqrt{w^2+d^2}$, so half of this is



    $$frac{sqrt{w^2+d^2}}{2}.$$



    This length is one leg of a right triangle; the other leg is $(h-k)$.



    Hence the total length of four identical lengths of wire that make up the four hypoteneuses is



    $$4 sqrt{left(frac{sqrt{w^2+d^2}}{2}right)^2+(h-k)^2} = 4 sqrt{left(frac{w^2+d^2}{4}right)+(h-k)^2}.$$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      The full diagonal of the rectangle is $sqrt{w^2+d^2}$, so half of this is



      $$frac{sqrt{w^2+d^2}}{2}.$$



      This length is one leg of a right triangle; the other leg is $(h-k)$.



      Hence the total length of four identical lengths of wire that make up the four hypoteneuses is



      $$4 sqrt{left(frac{sqrt{w^2+d^2}}{2}right)^2+(h-k)^2} = 4 sqrt{left(frac{w^2+d^2}{4}right)+(h-k)^2}.$$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        The full diagonal of the rectangle is $sqrt{w^2+d^2}$, so half of this is



        $$frac{sqrt{w^2+d^2}}{2}.$$



        This length is one leg of a right triangle; the other leg is $(h-k)$.



        Hence the total length of four identical lengths of wire that make up the four hypoteneuses is



        $$4 sqrt{left(frac{sqrt{w^2+d^2}}{2}right)^2+(h-k)^2} = 4 sqrt{left(frac{w^2+d^2}{4}right)+(h-k)^2}.$$






        share|cite|improve this answer









        $endgroup$



        The full diagonal of the rectangle is $sqrt{w^2+d^2}$, so half of this is



        $$frac{sqrt{w^2+d^2}}{2}.$$



        This length is one leg of a right triangle; the other leg is $(h-k)$.



        Hence the total length of four identical lengths of wire that make up the four hypoteneuses is



        $$4 sqrt{left(frac{sqrt{w^2+d^2}}{2}right)^2+(h-k)^2} = 4 sqrt{left(frac{w^2+d^2}{4}right)+(h-k)^2}.$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 10 '18 at 22:11









        JohnJohn

        22.7k32450




        22.7k32450






























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