Why would I divide by 4 instead of 2 in this equation?
$begingroup$
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 $k$.
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
edited Dec 10 '18 at 22:00
gt6989b
33.9k22455
asked Dec 10 '18 at 22:00
Ballusha ThreeBallusha Three
132
$endgroup$
add a comment |
$begingroup$
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 $k$.
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
edited Dec 10 '18 at 22:00
gt6989b
33.9k22455
asked Dec 10 '18 at 22:00
Ballusha ThreeBallusha Three
132
$endgroup$
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
add a comment |
$begingroup$
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 $k$.
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
edited Dec 10 '18 at 22:00
gt6989b
33.9k22455
asked Dec 10 '18 at 22:00
Ballusha ThreeBallusha Three
132
$endgroup$
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 $k$.
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
edited Dec 10 '18 at 22:00
gt6989b
33.9k22455
asked Dec 10 '18 at 22:00
Ballusha ThreeBallusha Three
132
edited Dec 10 '18 at 22:00
gt6989b
33.9k22455
asked Dec 10 '18 at 22:00
Ballusha ThreeBallusha Three
132
edited Dec 10 '18 at 22:00
gt6989b
33.9k22455
edited Dec 10 '18 at 22:00
gt6989b
33.9k22455
edited Dec 10 '18 at 22:00
gt6989b
33.9k22455
33.9k22455
asked Dec 10 '18 at 22:00
Ballusha ThreeBallusha Three
132
asked Dec 10 '18 at 22:00
Ballusha ThreeBallusha Three
132
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
add a comment |
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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
$$4 sqrt{left(frac{sqrt{w^2+d^2}}{2}right)^2+(h-k)^2} = 4 sqrt{left(frac{w^2+d^2}{4}right)+(h-k)^2}.$$
answered Dec 10 '18 at 22:11
JohnJohn
22.7k32450
$endgroup$
add a comment |
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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
$$4 sqrt{left(frac{sqrt{w^2+d^2}}{2}right)^2+(h-k)^2} = 4 sqrt{left(frac{w^2+d^2}{4}right)+(h-k)^2}.$$
answered Dec 10 '18 at 22:11
JohnJohn
22.7k32450
$endgroup$
add a comment |
$begingroup$
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
$$4 sqrt{left(frac{sqrt{w^2+d^2}}{2}right)^2+(h-k)^2} = 4 sqrt{left(frac{w^2+d^2}{4}right)+(h-k)^2}.$$
answered Dec 10 '18 at 22:11
JohnJohn
22.7k32450
$endgroup$
add a comment |
$begingroup$
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
$$4 sqrt{left(frac{sqrt{w^2+d^2}}{2}right)^2+(h-k)^2} = 4 sqrt{left(frac{w^2+d^2}{4}right)+(h-k)^2}.$$
answered Dec 10 '18 at 22:11
JohnJohn
22.7k32450
$endgroup$
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
$$4 sqrt{left(frac{sqrt{w^2+d^2}}{2}right)^2+(h-k)^2} = 4 sqrt{left(frac{w^2+d^2}{4}right)+(h-k)^2}.$$
answered Dec 10 '18 at 22:11
JohnJohn
22.7k32450
answered Dec 10 '18 at 22:11
JohnJohn
22.7k32450
answered Dec 10 '18 at 22:11
JohnJohn
22.7k32450
answered Dec 10 '18 at 22:11
JohnJohn
22.7k32450
22.7k32450
add a comment |
add a comment |
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$begingroup$
$frac{sqrt{w^2+d^2}}{2} = sqrt{frac{w^2+d^2}{4}}$
$endgroup$
– Daniel Gendin
Dec 10 '18 at 22:03