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# A rectangular box is 10 inches wide, 10 inches long, and 5

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Re: A rectangular box is 10 inches wide, 10 inches long, and 5 [#permalink]
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One can also apply the 3-D or Deluxe Pythagorean theorem directly, which is D^2 = L^2 + W^2 + H^2, to get the value directly.
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A rectangular box is 10 inches wide, 10 inches long, and 5 [#permalink]
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A rectangular box is 10 inches wide, 10 inches long, and 5 inches high. What is the greatest possible (straight-line) distance, in inches, between any two points on the box?

(A) 15
(B) 20
(C) 25
(D) $$10\sqrt{2}$$
(E) $$10\sqrt{3}$$

The greatest distance between any two points in a CUBE/CUBOID is given by the formula
$$d=\sqrt{l^2+b^2+h^2}$$

$$d=\sqrt{100+100+25}$$; $${given===> l=10; b=10; h=5}$$

$$d=\sqrt{225}= 15$$

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Re: A rectangular box is 10 inches wide, 10 inches long, and 5 [#permalink]
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A rectangular box is 10 inches wide, 10 inches long, and 5 inches high. What is the greatest possible (straight-line) distance, in inches, between any two points on the box?

(A) 15
(B) 20
(C) 25
(D) $$10\sqrt{2}$$
(E) $$10\sqrt{3}$$

The greatest distance will be when the two points are in opposite corners.

In these instances, we have a nice rule that says:
If x, y, and z are the three measurements of a box, then the distance between two points in OPPOSITE CORNERS equals √(x² + y² + z²)

So, for your question, the distance = √(10² + 10² + 5²) = √225 = 15

Cheers,
Brent
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A rectangular box is 10 inches wide, 10 inches long, and 5 [#permalink]
This is such an easy question... but the question stem states - straight line between any two points "on" the box. And that means face diagonal, an i missing something? Wouldn't we use the space diagonal formula is the question stated "in" the box perhaps?
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A rectangular box is 10 inches wide, 10 inches long, and 5 [#permalink]
A rectangular box is 10 inches wide, 10 inches long, and 5 inches high. What is the greatest possible (straight-line) distance, in inches, between any two points on the box?

(A) 15
(B) 20
(C) 25
(D) $$10\sqrt{2}$$
(E) $$10\sqrt{3}$$

$$= \sqrt{10^2+10^2+5^2}$$

$$= \sqrt{225}$$

$$= 15$$, Answer must be (A)

dhruvmm wrote:
This is such an easy question... but the question stem states - straight line between any two points "on" the box. And that means face diagonal, an i missing something? Wouldn't we use the space diagonal formula is the question stated "in" the box perhaps?

Check the highlighted part the straight line is the diagonal inside the box...
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Re: A rectangular box is 10 inches wide, 10 inches long, and 5 [#permalink]
Hi, that's my exact point of confusion. The highlighted part says "between any two points ON the box?"
And you mentioned - "straight line is the diagonal INSIDE the box."

ON is not the same as IN - is it? minor error in the verbose but makes a world of a difference imo
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Re: A rectangular box is 10 inches wide, 10 inches long, and 5 [#permalink]
Using the Pythagorean theorem, we can find the length of the diagonal. Let's denote the length, width, and height of the box as L, W, and H, respectively.

The length of the diagonal (D) can be calculated as:

D = √(L^2 + W^2 + H^2) = √(10^2 + 10^2 + 5^2) = √(100 + 100 + 25) = √225 = 15

Therefore, the greatest possible straight-line distance between any two points on the box is 15 inches, which corresponds to option (A).
Re: A rectangular box is 10 inches wide, 10 inches long, and 5 [#permalink]
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