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ankoles [38]
2 years ago
7

What is the longest line segment that can be drawn in a right rectangular prism that is 15 cm​ long, 11 cm​ wide, and 10 cm​ tal

l?
Mathematics
1 answer:
sammy [17]2 years ago
8 0

Answer:

  √446 ≈ 21.12 cm

Step-by-step explanation:

The longest dimension of a rectangular prism is the length of the space diagonal from one corner to the opposite corner through the center of the prism. The Pythagorean theorm tells you the square of its length is the sum of the squares of the dimensions of the prism:

  d² = (15 cm)² +(11 cm)² +(10 cm)² = (225 +121 +100) cm² = 446 cm²

  d = √446 cm ≈ 21.12 cm

The longest line segment that can be drawn in a right rectangular prism is about 21.12 cm.

_____

<em>Additional comment</em>

The square of the face diagonal is the sum of the squares of the dimensions of that face. The square of the space diagonal will be the sum of that square and the square of the remaining prism dimenaion, hence the sum of squares of all three prism dimensions.

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One of our brainliest, Konrad509, made this:
Reil [10]

\dfrac{B_x \sqrt{74_x}}{1D_x}+J_x51_x=4G3_x

A=10, B=11, C=12, etc.

\dfrac{11\cdot x^0\cdot \sqrt{7\cdot x^1+4\cdot x^0}}{1\cdot x^1+13\cdot x^0}+19\cdot x^0\cdot (5\cdot x^1+1\cdot x^0)=4\cdot x^2+16\cdot x^1+3\cdot x^0\\\\\dfrac{11\sqrt{7x+4}}{x+13}+19(5x+1)=4x^2+16x+3\\\\\dfrac{11\sqrt{7x+4}}{x+13}+95x+19=4x^2+16x+3\\\\11\sqrt{7x+4}+95x(x+13)+19(x+13)=(4x^2+16x+3)(x+13)\\\\11\sqrt{7x+4}+95x^2+1235x+19x+247=4x^3+52x^2+16x^2+208x+3x+39\\\\11\sqrt{7x+4}=4x^3-27x^2-1043x-208\\\\121(7x+4)=(4x^3-27x^2-1043x-208)^2

121(7x+4)=(4x^3-27x^2-1043x-208)^2\\\\847x+484=16 x^6 - 216 x^5 - 7615 x^4 + 54658 x^3 + 1099081 x^2 + 433888 x + 43264\\\\16 x^6 - 216 x^5 - 7615 x^4 + 54658 x^3 + 1099081 x^2 + 433041 x +42780=0

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I think the best way to approach it is using the rational root theorem since we know that x\in\mathbb{N}. Moreover we can deduce that x\geq19 since there is J and J=19.

After you succesfully solve it, you should get the answer x=20.

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