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A rectangular box is 12 inches wide, 16 inches long, and 20 inches hig

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A rectangular box is 12 inches wide, 16 inches long, and 20 inches hig  [#permalink]

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09 Jul 2015, 04:21
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A rectangular box is 12 inches wide, 16 inches long, and 20 inches high. Which of the following is closest to the longest straight-line distance between any two corners of the box?

A. 22 inches
B. 25 inches
C. 28 inches
D. 30 inches
E. 34 inches

Kudos for a correct solution.

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Re: A rectangular box is 12 inches wide, 16 inches long, and 20 inches hig  [#permalink]

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09 Jul 2015, 04:28
1
Bunuel wrote:
A rectangular box is 12 inches wide, 16 inches long, and 20 inches high. Which of the following is closest to the longest straight-line distance between any two corners of the box?

A. 22 inches
B. 25 inches
C. 28 inches
D. 30 inches
E. 34 inches

Kudos for a correct solution.

Answer -

The longest straight-line distance between any two corners of the box = √(W^2+L^2+H^2) = √(144+256+400) = √800

ANS C
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A rectangular box is 12 inches wide, 16 inches long, and 20 inches hig  [#permalink]

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09 Jul 2015, 04:38
Bunuel wrote:
A rectangular box is 12 inches wide, 16 inches long, and 20 inches high. Which of the following is closest to the longest straight-line distance between any two corners of the box?

A. 22 inches
B. 25 inches
C. 28 inches
D. 30 inches
E. 34 inches

Kudos for a correct solution.

If L, B and H denote the Length, Breadth and Height of a cube or a cuboid, the longest straight line distance is equal to = $$\sqrt{(L^2+B^2+H^2)}$$

Thus, per the questions, as L=16, B=12 and H=20, substituting these values , we get Longets distance = 28 inches $$(=20 \sqrt{2})$$
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A rectangular box is 12 inches wide, 16 inches long, and 20 inches hig  [#permalink]

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09 Jul 2015, 06:48
Bunuel wrote:
A rectangular box is 12 inches wide, 16 inches long, and 20 inches high. Which of the following is closest to the longest straight-line distance between any two corners of the box?

A. 22 inches
B. 25 inches
C. 28 inches
D. 30 inches
E. 34 inches

Kudos for a correct solution.

The longest straight-line distance between any two corners of the box = $$\sqrt{L^2+B^2+H^2}$$ = $$\sqrt{12^2+16^2+20^2}$$ = $$\sqrt{144+256+400}$$ = $$\sqrt{800}$$ = 28.3 approx.

Answer: Option C

Please see the Derivation of Longest Diagonal in a Rectangular Solid. Longest diagonal in the figure is S
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Re: A rectangular box is 12 inches wide, 16 inches long, and 20 inches hig  [#permalink]

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09 Jul 2015, 10:58
1
Bunuel wrote:
A rectangular box is 12 inches wide, 16 inches long, and 20 inches high. Which of the following is closest to the longest straight-line distance between any two corners of the box?

A. 22 inches
B. 25 inches
C. 28 inches
D. 30 inches
E. 34 inches

Kudos for a correct solution.

We need the diagonal.

Diagonal = $$\sqrt{12^2+16^2+20^2}$$
= $$\sqrt{144+256+400}$$
=$$\sqrt{800}$$
=20$$\sqrt{2}$$
=20*1.41
=28.2

Answer: C
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Joined: 30 Jul 2008
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A rectangular box is 12 inches wide, 16 inches long, and 20 inches hig  [#permalink]

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10 Jul 2015, 09:03
1
Bunuel wrote:
A rectangular box is 12 inches wide, 16 inches long, and 20 inches high. Which of the following is closest to the longest straight-line distance between any two corners of the box?

A. 22 inches
B. 25 inches
C. 28 inches
D. 30 inches
E. 34 inches

Kudos for a correct solution.

longest diagonal =$$\sqrt{l^2+b^2+h^2} = \sqrt{16^2+12^2+20^2}$$
= $$\sqrt{800}$$
= $$20 X \sqrt{2}$$ = 28 (approx)
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Re: A rectangular box is 12 inches wide, 16 inches long, and 20 inches hig  [#permalink]

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10 Jul 2015, 09:14
1
Bunuel wrote:
A rectangular box is 12 inches wide, 16 inches long, and 20 inches high. Which of the following is closest to the longest straight-line distance between any two corners of the box?

A. 22 inches
B. 25 inches
C. 28 inches
D. 30 inches
E. 34 inches

Kudos for a correct solution.

although the answer is straight from formulae $$\sqrt{l^2+b^2+h^2}$$= $$\sqrt{12^2+16^2+20^2}$$
the other way to look at is 12, 16 ,hyp is a right angle triange of ratio 3:4:5..
so here hyp =20..
now the sides are 20,20,hyp in a right angle triangle in ratio 1:1:$$\sqrt{2}$$.. hyp=20$$\sqrt{2}$$=28
ans C
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Re: A rectangular box is 12 inches wide, 16 inches long, and 20 inches hig  [#permalink]

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11 Jul 2015, 10:35
1
Bunuel wrote:
A rectangular box is 12 inches wide, 16 inches long, and 20 inches high. Which of the following is closest to the longest straight-line distance between any two corners of the box?

A. 22 inches
B. 25 inches
C. 28 inches
D. 30 inches
E. 34 inches

Kudos for a correct solution.

Longest straight line distance between any two corners of the box=\sqrt{L^2+B^2+H^2}
=\sqrt{144+256+400}
=\sqrt{800}
Answer C
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Posts: 58381
Re: A rectangular box is 12 inches wide, 16 inches long, and 20 inches hig  [#permalink]

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13 Jul 2015, 02:49
Bunuel wrote:
A rectangular box is 12 inches wide, 16 inches long, and 20 inches high. Which of the following is closest to the longest straight-line distance between any two corners of the box?

A. 22 inches
B. 25 inches
C. 28 inches
D. 30 inches
E. 34 inches

Kudos for a correct solution.

800score Official Solution:

We must first determine which two corners are furthest apart in a rectangular box. To determine the corners that are furthest apart in a rectangular box, pick any corner, and think about drawing a line through the center of the box to the furthest other corner. Let's call corners that have this relationship "opposite corners".

The distance D between any two opposite corners in a rectangular solid (whether a perfect cube or not) can be determined very quickly using the Pythagorean Theorem in a 3-dimensional manner:
D² = a² + b² + c², where a, b, and c represent the dimensions of the solid.

Now let's use the formula to determine the distance:
D² = 12² + 16² + 20²
D² = 144 + 256 + 400 = 800.

So D = √800.

Rather than trying to approximate √800, let's square the answer choices and determine which is closest to 800.

We know that 30² = 900, so the correct answer must be either choice (C) or (D). Now let's square the number in choice (C).
28² = 784.

Since 784 is much closer to 800 than 900 is, the correct answer is choice (C).

Another way to solve this question is to use 3:4:5 triangles. Since 12:16:20 is a multiple of 3:4:5, the diagonal is 20. To find the answer, find the hypotenuse of a triangle with that diagonal and the third dimension of the box – 20 and 20, which has a hypotenuse of 20√2. √2 equals approximately 1.4, so the answer is about 28.
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Re: A rectangular box is 12 inches wide, 16 inches long, and 20 inches hig  [#permalink]

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13 Jul 2015, 02:50
Bunuel wrote:
Bunuel wrote:
A rectangular box is 12 inches wide, 16 inches long, and 20 inches high. Which of the following is closest to the longest straight-line distance between any two corners of the box?

A. 22 inches
B. 25 inches
C. 28 inches
D. 30 inches
E. 34 inches

Kudos for a correct solution.

800score Official Solution:

We must first determine which two corners are furthest apart in a rectangular box. To determine the corners that are furthest apart in a rectangular box, pick any corner, and think about drawing a line through the center of the box to the furthest other corner. Let's call corners that have this relationship "opposite corners".

The distance D between any two opposite corners in a rectangular solid (whether a perfect cube or not) can be determined very quickly using the Pythagorean Theorem in a 3-dimensional manner:
D² = a² + b² + c², where a, b, and c represent the dimensions of the solid.

Now let's use the formula to determine the distance:
D² = 12² + 16² + 20²
D² = 144 + 256 + 400 = 800.

So D = √800.

Rather than trying to approximate √800, let's square the answer choices and determine which is closest to 800.

We know that 30² = 900, so the correct answer must be either choice (C) or (D). Now let's square the number in choice (C).
28² = 784.

Since 784 is much closer to 800 than 900 is, the correct answer is choice (C).

Another way to solve this question is to use 3:4:5 triangles. Since 12:16:20 is a multiple of 3:4:5, the diagonal is 20. To find the answer, find the hypotenuse of a triangle with that diagonal and the third dimension of the box – 20 and 20, which has a hypotenuse of 20√2. √2 equals approximately 1.4, so the answer is about 28.

Similar question to practice: a-carpenter-wants-to-ship-a-copper-rod-to-his-new-construction-locatio-187870.html
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Re: A rectangular box is 12 inches wide, 16 inches long, and 20 inches hig  [#permalink]

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18 Sep 2019, 10:02
Bunuel wrote:
A rectangular box is 12 inches wide, 16 inches long, and 20 inches high. Which of the following is closest to the longest straight-line distance between any two corners of the box?

A. 22 inches
B. 25 inches
C. 28 inches
D. 30 inches
E. 34 inches

Kudos for a correct solution.

We can use the following formula to calculate the diagonal (or d):

d^2 = 12^2 + 20^2 + 16^2

d^2 = 144 + 400 + 256

d^2 = 800

d = √800 = 20√2, so d is about 20 * 1.4 = 28.

Answer: C
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Re: A rectangular box is 12 inches wide, 16 inches long, and 20 inches hig  [#permalink]

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18 Sep 2019, 10:54
Bunuel wrote:
A rectangular box is 12 inches wide, 16 inches long, and 20 inches high. Which of the following is closest to the longest straight-line distance between any two corners of the box?

A. 22 inches
B. 25 inches
C. 28 inches
D. 30 inches
E. 34 inches

Kudos for a correct solution.

$$\sqrt{12^2 + 16^2 + 20^2}$$

= $$\sqrt{144 + 256 + 400}$$

= $$\sqrt{800}$$

We know $$30^2 = 900$$, so our answer must be a bit less than 30 , among the given options only (C) 28 matches, so Answer must be (C)
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Re: A rectangular box is 12 inches wide, 16 inches long, and 20 inches hig   [#permalink] 18 Sep 2019, 10:54
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