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# A rectangular box has dimensions of 8 feet, 8 feet, and z feet. In ter

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A rectangular box has dimensions of 8 feet, 8 feet, and z feet. In ter  [#permalink]

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02 Mar 2012, 15:53
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5% (low)

Question Stats:

85% (01:14) correct 15% (01:55) wrong based on 227 sessions

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A rectangular box has dimensions of 8 feet, 8 feet, and z feet. In terms of z, what is the greatest possible (straight-line) distance, in feet, between any two points on the box?

(A) 8 + z

(B) $$8\sqrt{2} + z$$

(C) $$8z\sqrt{2}$$

(D) $$\sqrt{p64 + z^2}$$

(E) $$\sqrt{128 + z^2}$$

Guy - I know that this will be solved using the right angle triangle, but struggling. For me the longest distance will be the diagonal of the box, but I don't get an answer "E". Can you please help?

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Re: A rectangular box has dimensions of 8 feet, 8 feet, and z feet. In ter  [#permalink]

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02 Mar 2012, 20:20
2
enigma123 wrote:
A rectangular box has dimensions of 8 feet, 8 feet, and z feet. In terms of z, what is the greatest possible (straight-line) distance, in feet, between any two points on the box?
(A) 8 + z
(B)$$8\sqrt{2} + z$$
(C) $$8z\sqrt{2}$$
(D)$$\sqrt{p64 + z^2}$$
(E) $$\sqrt{128 + z^2}$$

Guy - I know that this will be solved using the right angle triangle, but struggling. For me the longest distance will be the diagonal of the box, but I don't get an answer "E". Can you please help?

The longest distance is the diagonal of the box. Formula for the diagonal of a rectangular is $$\sqrt{a^2 + b^2 + c^2}$$.
Just use the information from the text and insert it in the formula:
$$\sqrt{8^2 + 8^2 + z^2} = \sqrt{64 + 64 + z^2} = \sqrt{128 + z^2}$$

E, correct.
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Re: A rectangular box has dimensions of 8 feet, 8 feet, and z feet. In ter  [#permalink]

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02 Mar 2012, 21:50
4
enigma123 wrote:
A rectangular box has dimensions of 8 feet, 8 feet, and z feet. In terms of z, what is the greatest possible (straight-line) distance, in feet, between any two points on the box?
(A) 8 + z
(B) $$8\sqrt{2} + z$$
(C) $$8z\sqrt{2}$$
(D) $$\sqrt{p64 + z^2}$$
(E) $$\sqrt{128 + z^2}$$

Guy - I know that this will be solved using the right angle triangle, but struggling. For me the longest distance will be the diagonal of the box, but I don't get an answer "E". Can you please help?

You are right the longest distance will be the diagonal of a rectangular box. Look at the diagram below:

Square of the diagonal of the face (base) is $$d^2=a^2+b^2$$ and the square of the diagonal of a rectangular box is $$D^2=d^2+c^2=(a^2+b^2)+c^2$$ --> $$D=\sqrt{a^2+b^2+c^2}$$.

Applying this to our question, we get: $$D=\sqrt{8^2+8^2+z^2}=\sqrt{128+z^2}$$.

Hope it's clear.
Attachment:

Diagonal.jpg [ 17.98 KiB | Viewed 19190 times ]

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Re: A rectangular box has dimensions of 8 feet, 8 feet, and z feet. In ter  [#permalink]

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18 Jul 2017, 06:38
Poorly worded question as it does not mention if z is the height, width or length of box. In order for (E) to be the answer, we have to assume (without warrant) that the height of the box is z.
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Re: A rectangular box has dimensions of 8 feet, 8 feet, and z feet. In ter  [#permalink]

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18 Jul 2017, 08:03
gatreya14 wrote:
Poorly worded question as it does not mention if z is the height, width or length of box. In order for (E) to be the answer, we have to assume (without warrant) that the height of the box is z.

How does it matter whether z is height, width or length? Does it affect the greatest possible distance?
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Re: A rectangular box has dimensions of 8 feet, 8 feet, and z feet. In ter  [#permalink]

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18 Jul 2017, 11:07
Bunuel wrote:
gatreya14 wrote:
Poorly worded question as it does not mention if z is the height, width or length of box. In order for (E) to be the answer, we have to assume (without warrant) that the height of the box is z.

How does it matter whether z is height, width or length? Does it affect the greatest possible distance?

Actually, you're right, it doesn't matter.
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Re: A rectangular box has dimensions of 8 feet, 8 feet, and z feet. In ter  [#permalink]

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20 Jul 2017, 16:41
enigma123 wrote:
A rectangular box has dimensions of 8 feet, 8 feet, and z feet. In terms of z, what is the greatest possible (straight-line) distance, in feet, between any two points on the box?
(A) 8 + z
(B) $$8\sqrt{2} + z$$
(C) $$8z\sqrt{2}$$
(D) $$\sqrt{p64 + z^2}$$
(E) $$\sqrt{128 + z^2}$$

We can use the following formula to determine the rectangular box’s diagonal, which is the longest line connecting two points on the box.

diagonal^2 = 8^2 + 8^2 + z^2

d^2 = 128 + z^2

d = √(128 + z^2)

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Re: A rectangular box has dimensions of 8 feet, 8 feet, and z feet. In ter  [#permalink]

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13 May 2019, 22:57
enigma123 wrote:
A rectangular box has dimensions of 8 feet, 8 feet, and z feet. In terms of z, what is the greatest possible (straight-line) distance, in feet, between any two points on the box?

(A) 8 + z

(B) $$8\sqrt{2} + z$$

(C) $$8z\sqrt{2}$$

(D) $$\sqrt{p64 + z^2}$$

(E) $$\sqrt{128 + z^2}$$

Longest straight line is the diagonal of the cuboid. sq. root(a^2 + b^2 + c^2), where a, b and c are sides.
OA is E.

Guy - I know that this will be solved using the right angle triangle, but struggling. For me the longest distance will be the diagonal of the box, but I don't get an answer "E". Can you please help?

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Diagonal.PNG [ 21.36 KiB | Viewed 95 times ]

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Re: A rectangular box has dimensions of 8 feet, 8 feet, and z feet. In ter   [#permalink] 13 May 2019, 22:57
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