Mixing
Why the number of pints is proportional to the square of the height of a statue?
$begingroup$
The main question:
If 1 pint of paint is needed to paint a statue 6 ft. high, then the
number of pints it will take to paint (to the same thickness) 540
statues similar to the original but only 1 ft. high, is
(A)
90
(B) 72
(C) 45
(D) 30
(E) 15
The matter I can not understand:
In the answer it is said that the number of pints is proportional to the square of height. Can anyone explain me why it is proportional to the square of height?
algebra-precalculus geometry word-problem
edited Feb 12 at 0:00
J. W. Tanner
4,7721420
asked Feb 2 at 3:58
ShromiShromi
3289
$endgroup$
add a comment |
$begingroup$
The main question:
If 1 pint of paint is needed to paint a statue 6 ft. high, then the
number of pints it will take to paint (to the same thickness) 540
statues similar to the original but only 1 ft. high, is
(A)
90
(B) 72
(C) 45
(D) 30
(E) 15
The matter I can not understand:
In the answer it is said that the number of pints is proportional to the square of height. Can anyone explain me why it is proportional to the square of height?
algebra-precalculus geometry word-problem
edited Feb 12 at 0:00
J. W. Tanner
4,7721420
asked Feb 2 at 3:58
ShromiShromi
3289
$endgroup$
add a comment |
$begingroup$
The main question:
If 1 pint of paint is needed to paint a statue 6 ft. high, then the
number of pints it will take to paint (to the same thickness) 540
statues similar to the original but only 1 ft. high, is
(A)
90
(B) 72
(C) 45
(D) 30
(E) 15
The matter I can not understand:
In the answer it is said that the number of pints is proportional to the square of height. Can anyone explain me why it is proportional to the square of height?
algebra-precalculus geometry word-problem
edited Feb 12 at 0:00
J. W. Tanner
4,7721420
asked Feb 2 at 3:58
ShromiShromi
3289
$endgroup$
The main question:
If 1 pint of paint is needed to paint a statue 6 ft. high, then the
number of pints it will take to paint (to the same thickness) 540
statues similar to the original but only 1 ft. high, is
(A)
90
(B) 72
(C) 45
(D) 30
(E) 15
The matter I can not understand:
In the answer it is said that the number of pints is proportional to the square of height. Can anyone explain me why it is proportional to the square of height?
algebra-precalculus geometry word-problem
algebra-precalculus geometry word-problem
edited Feb 12 at 0:00
J. W. Tanner
4,7721420
asked Feb 2 at 3:58
ShromiShromi
3289
edited Feb 12 at 0:00
J. W. Tanner
4,7721420
asked Feb 2 at 3:58
ShromiShromi
3289
edited Feb 12 at 0:00
J. W. Tanner
4,7721420
edited Feb 12 at 0:00
J. W. Tanner
4,7721420
edited Feb 12 at 0:00
J. W. Tanner
4,7721420
4,7721420
asked Feb 2 at 3:58
ShromiShromi
3289
asked Feb 2 at 3:58
ShromiShromi
3289
asked Feb 2 at 3:58
ShromiShromi
3289
3289
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Consider a small square area, say of side $x$, of any portion of a given figure. With a figure that is similar, this area would be stretched or compressed by a certain factor, say $s$, in both the width & height. Since the area of the original square is the product of its side lengths, $x^2$, the area of the similar square area would be $sx times sx = s^2 x^2$. Adding up all these small areas gives that the new surface area would be proportional to original surface area by the square of the scaling factor. In your particular case it would be $s = frac{1}{6}$, i.e., the new height of $1$ divided by the original height of $6$, so (as Rhys Hughes has stated in his answer) the proportional area value would be the square of it, i.e., $frac{1}{36}$.
For example, consider an original height of $h_0$ and an area of $A_0$. Then for a new height $h$, the area would be $A = A_0left(frac{h}{h_0}right)^2 = left(frac{A_0}{h_0^2}right)h^2$. As $A_0$ and $h_0$ are known, constant values, we can let $k = frac{A_0}{h_0^2}$ be a constant, giving that
$$A = kh^2 tag{1}label{eq1}$$
showing that the area is proportional (with a constant of proportionality of $k$ as given) to the square of the height. Similarly, as the amount of paint is proportional to the area, we get
$$P = kh^2 tag{21}label{eq2}$$
where $k = frac{P_0}{h_0^2}$ and $P_0$ is the amount of paint required for the original object.
Note the number of pints is also proportional to the square of the width, but as only the heights are provided, this alternate fact is not needed or used.
answered Feb 2 at 4:08
John OmielanJohn Omielan
4,9692218
$endgroup$
add a comment |
$begingroup$
It means that if the number of pints needed is $P$ and the height of the statue is $H$, then:
$$P=kH^2$$ for some constant $k$. You should be able to see that $k=frac{1}{36}$ in this case.
answered Feb 2 at 4:08
Rhys HughesRhys Hughes
7,0501630
$endgroup$
1
$begingroup$
could you please explain why it is $H^2$
$endgroup$
– Shromi
Feb 2 at 5:46
$begingroup$
Because it's proportional to the square of the height, i.e. $H^2$
$endgroup$
– Rhys Hughes
Feb 2 at 11:54
$begingroup$
My question is why it is proportional to the $H^2$? Why not only $H$?
$endgroup$
– Shromi
Feb 2 at 13:26
1
$begingroup$
Surface area (amount of paint needed) is proportional to the square of the height. Volume (if you had to fill it with water) would be proportional to the cube of the height.
$endgroup$
– J. W. Tanner
Feb 11 at 23:48
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Consider a small square area, say of side $x$, of any portion of a given figure. With a figure that is similar, this area would be stretched or compressed by a certain factor, say $s$, in both the width & height. Since the area of the original square is the product of its side lengths, $x^2$, the area of the similar square area would be $sx times sx = s^2 x^2$. Adding up all these small areas gives that the new surface area would be proportional to original surface area by the square of the scaling factor. In your particular case it would be $s = frac{1}{6}$, i.e., the new height of $1$ divided by the original height of $6$, so (as Rhys Hughes has stated in his answer) the proportional area value would be the square of it, i.e., $frac{1}{36}$.
For example, consider an original height of $h_0$ and an area of $A_0$. Then for a new height $h$, the area would be $A = A_0left(frac{h}{h_0}right)^2 = left(frac{A_0}{h_0^2}right)h^2$. As $A_0$ and $h_0$ are known, constant values, we can let $k = frac{A_0}{h_0^2}$ be a constant, giving that
$$A = kh^2 tag{1}label{eq1}$$
showing that the area is proportional (with a constant of proportionality of $k$ as given) to the square of the height. Similarly, as the amount of paint is proportional to the area, we get
$$P = kh^2 tag{21}label{eq2}$$
where $k = frac{P_0}{h_0^2}$ and $P_0$ is the amount of paint required for the original object.
Note the number of pints is also proportional to the square of the width, but as only the heights are provided, this alternate fact is not needed or used.
answered Feb 2 at 4:08
John OmielanJohn Omielan
4,9692218
$endgroup$
add a comment |
$begingroup$
Consider a small square area, say of side $x$, of any portion of a given figure. With a figure that is similar, this area would be stretched or compressed by a certain factor, say $s$, in both the width & height. Since the area of the original square is the product of its side lengths, $x^2$, the area of the similar square area would be $sx times sx = s^2 x^2$. Adding up all these small areas gives that the new surface area would be proportional to original surface area by the square of the scaling factor. In your particular case it would be $s = frac{1}{6}$, i.e., the new height of $1$ divided by the original height of $6$, so (as Rhys Hughes has stated in his answer) the proportional area value would be the square of it, i.e., $frac{1}{36}$.
For example, consider an original height of $h_0$ and an area of $A_0$. Then for a new height $h$, the area would be $A = A_0left(frac{h}{h_0}right)^2 = left(frac{A_0}{h_0^2}right)h^2$. As $A_0$ and $h_0$ are known, constant values, we can let $k = frac{A_0}{h_0^2}$ be a constant, giving that
$$A = kh^2 tag{1}label{eq1}$$
showing that the area is proportional (with a constant of proportionality of $k$ as given) to the square of the height. Similarly, as the amount of paint is proportional to the area, we get
$$P = kh^2 tag{21}label{eq2}$$
where $k = frac{P_0}{h_0^2}$ and $P_0$ is the amount of paint required for the original object.
Note the number of pints is also proportional to the square of the width, but as only the heights are provided, this alternate fact is not needed or used.
answered Feb 2 at 4:08
John OmielanJohn Omielan
4,9692218
$endgroup$
add a comment |
$begingroup$
Consider a small square area, say of side $x$, of any portion of a given figure. With a figure that is similar, this area would be stretched or compressed by a certain factor, say $s$, in both the width & height. Since the area of the original square is the product of its side lengths, $x^2$, the area of the similar square area would be $sx times sx = s^2 x^2$. Adding up all these small areas gives that the new surface area would be proportional to original surface area by the square of the scaling factor. In your particular case it would be $s = frac{1}{6}$, i.e., the new height of $1$ divided by the original height of $6$, so (as Rhys Hughes has stated in his answer) the proportional area value would be the square of it, i.e., $frac{1}{36}$.
For example, consider an original height of $h_0$ and an area of $A_0$. Then for a new height $h$, the area would be $A = A_0left(frac{h}{h_0}right)^2 = left(frac{A_0}{h_0^2}right)h^2$. As $A_0$ and $h_0$ are known, constant values, we can let $k = frac{A_0}{h_0^2}$ be a constant, giving that
$$A = kh^2 tag{1}label{eq1}$$
showing that the area is proportional (with a constant of proportionality of $k$ as given) to the square of the height. Similarly, as the amount of paint is proportional to the area, we get
$$P = kh^2 tag{21}label{eq2}$$
where $k = frac{P_0}{h_0^2}$ and $P_0$ is the amount of paint required for the original object.
Note the number of pints is also proportional to the square of the width, but as only the heights are provided, this alternate fact is not needed or used.
answered Feb 2 at 4:08
John OmielanJohn Omielan
4,9692218
$endgroup$
Consider a small square area, say of side $x$, of any portion of a given figure. With a figure that is similar, this area would be stretched or compressed by a certain factor, say $s$, in both the width & height. Since the area of the original square is the product of its side lengths, $x^2$, the area of the similar square area would be $sx times sx = s^2 x^2$. Adding up all these small areas gives that the new surface area would be proportional to original surface area by the square of the scaling factor. In your particular case it would be $s = frac{1}{6}$, i.e., the new height of $1$ divided by the original height of $6$, so (as Rhys Hughes has stated in his answer) the proportional area value would be the square of it, i.e., $frac{1}{36}$.
For example, consider an original height of $h_0$ and an area of $A_0$. Then for a new height $h$, the area would be $A = A_0left(frac{h}{h_0}right)^2 = left(frac{A_0}{h_0^2}right)h^2$. As $A_0$ and $h_0$ are known, constant values, we can let $k = frac{A_0}{h_0^2}$ be a constant, giving that
$$A = kh^2 tag{1}label{eq1}$$
showing that the area is proportional (with a constant of proportionality of $k$ as given) to the square of the height. Similarly, as the amount of paint is proportional to the area, we get
$$P = kh^2 tag{21}label{eq2}$$
where $k = frac{P_0}{h_0^2}$ and $P_0$ is the amount of paint required for the original object.
Note the number of pints is also proportional to the square of the width, but as only the heights are provided, this alternate fact is not needed or used.
answered Feb 2 at 4:08
John OmielanJohn Omielan
4,9692218
edited Feb 2 at 6:14
answered Feb 2 at 4:08
John OmielanJohn Omielan
4,9692218
answered Feb 2 at 4:08
John OmielanJohn Omielan
4,9692218
answered Feb 2 at 4:08
John OmielanJohn Omielan
4,9692218
4,9692218
add a comment |
add a comment |
$begingroup$
It means that if the number of pints needed is $P$ and the height of the statue is $H$, then:
$$P=kH^2$$ for some constant $k$. You should be able to see that $k=frac{1}{36}$ in this case.
answered Feb 2 at 4:08
Rhys HughesRhys Hughes
7,0501630
$endgroup$
1
$begingroup$
could you please explain why it is $H^2$
$endgroup$
– Shromi
Feb 2 at 5:46
$begingroup$
Because it's proportional to the square of the height, i.e. $H^2$
$endgroup$
– Rhys Hughes
Feb 2 at 11:54
$begingroup$
My question is why it is proportional to the $H^2$? Why not only $H$?
$endgroup$
– Shromi
Feb 2 at 13:26
1
$begingroup$
Surface area (amount of paint needed) is proportional to the square of the height. Volume (if you had to fill it with water) would be proportional to the cube of the height.
$endgroup$
– J. W. Tanner
Feb 11 at 23:48
add a comment |
$begingroup$
It means that if the number of pints needed is $P$ and the height of the statue is $H$, then:
$$P=kH^2$$ for some constant $k$. You should be able to see that $k=frac{1}{36}$ in this case.
answered Feb 2 at 4:08
Rhys HughesRhys Hughes
7,0501630
$endgroup$
1
$begingroup$
could you please explain why it is $H^2$
$endgroup$
– Shromi
Feb 2 at 5:46
$begingroup$
Because it's proportional to the square of the height, i.e. $H^2$
$endgroup$
– Rhys Hughes
Feb 2 at 11:54
$begingroup$
My question is why it is proportional to the $H^2$? Why not only $H$?
$endgroup$
– Shromi
Feb 2 at 13:26
1
$begingroup$
Surface area (amount of paint needed) is proportional to the square of the height. Volume (if you had to fill it with water) would be proportional to the cube of the height.
$endgroup$
– J. W. Tanner
Feb 11 at 23:48
add a comment |
$begingroup$
It means that if the number of pints needed is $P$ and the height of the statue is $H$, then:
$$P=kH^2$$ for some constant $k$. You should be able to see that $k=frac{1}{36}$ in this case.
answered Feb 2 at 4:08
Rhys HughesRhys Hughes
7,0501630
$endgroup$
It means that if the number of pints needed is $P$ and the height of the statue is $H$, then:
$$P=kH^2$$ for some constant $k$. You should be able to see that $k=frac{1}{36}$ in this case.
answered Feb 2 at 4:08
Rhys HughesRhys Hughes
7,0501630
answered Feb 2 at 4:08
Rhys HughesRhys Hughes
7,0501630
answered Feb 2 at 4:08
Rhys HughesRhys Hughes
7,0501630
answered Feb 2 at 4:08
Rhys HughesRhys Hughes
7,0501630
7,0501630
1
$begingroup$
could you please explain why it is $H^2$
$endgroup$
– Shromi
Feb 2 at 5:46
$begingroup$
Because it's proportional to the square of the height, i.e. $H^2$
$endgroup$
– Rhys Hughes
Feb 2 at 11:54
$begingroup$
My question is why it is proportional to the $H^2$? Why not only $H$?
$endgroup$
– Shromi
Feb 2 at 13:26
1
$begingroup$
Surface area (amount of paint needed) is proportional to the square of the height. Volume (if you had to fill it with water) would be proportional to the cube of the height.
$endgroup$
– J. W. Tanner
Feb 11 at 23:48
add a comment |
1
$begingroup$
could you please explain why it is $H^2$
$endgroup$
– Shromi
Feb 2 at 5:46
$begingroup$
Because it's proportional to the square of the height, i.e. $H^2$
$endgroup$
– Rhys Hughes
Feb 2 at 11:54
$begingroup$
My question is why it is proportional to the $H^2$? Why not only $H$?
$endgroup$
– Shromi
Feb 2 at 13:26
1
$begingroup$
Surface area (amount of paint needed) is proportional to the square of the height. Volume (if you had to fill it with water) would be proportional to the cube of the height.
$endgroup$
– J. W. Tanner
Feb 11 at 23:48
1
1
$begingroup$
could you please explain why it is $H^2$
$endgroup$
– Shromi
Feb 2 at 5:46
$begingroup$
could you please explain why it is $H^2$
$endgroup$
– Shromi
Feb 2 at 5:46
$begingroup$
Because it's proportional to the square of the height, i.e. $H^2$
$endgroup$
– Rhys Hughes
Feb 2 at 11:54
$begingroup$
Because it's proportional to the square of the height, i.e. $H^2$
$endgroup$
– Rhys Hughes
Feb 2 at 11:54
$begingroup$
My question is why it is proportional to the $H^2$? Why not only $H$?
$endgroup$
– Shromi
Feb 2 at 13:26
$begingroup$
My question is why it is proportional to the $H^2$? Why not only $H$?
$endgroup$
– Shromi
Feb 2 at 13:26
1
1
$begingroup$
Surface area (amount of paint needed) is proportional to the square of the height. Volume (if you had to fill it with water) would be proportional to the cube of the height.
$endgroup$
– J. W. Tanner
Feb 11 at 23:48
$begingroup$
Surface area (amount of paint needed) is proportional to the square of the height. Volume (if you had to fill it with water) would be proportional to the cube of the height.
$endgroup$
– J. W. Tanner
Feb 11 at 23:48
add a comment |
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