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enyata [817]
3 years ago
10

if a model airplane is built at a scale of 1 inch to 6 feet. If the plane is 8 inches long how many feet is the airplane?

Mathematics
1 answer:
Lyrx [107]3 years ago
5 0

Answer:

The real plane is 48 ft long.

Step-by-step explanation:

Since the model airplane was built at a scale of 1 inch to 6 feet, this means that for every 1 inch on the model there is 6 feet on the real plane. We know that the model is 8 inches longs, therefore we need to multiply it's length by 6 and we will have the correct answer. We have:

real length(ft) = [model lengh (in)]*6

real length(ft) = 8*6 = 48 ft

The real plane is 48 ft long.

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Step-by-step explanation:

Given:

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\begin{aligned}\textsf{Area of a rectangle}&=\sf width \times length\\& = \sf 75 \times 105\\& = \sf 7875\:\: yd^2\end{aligned}

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In geometry, length pertains to the <u>longest side</u> of the rectangle while width is the <u>shorter side</u>.  Therefore, we should choose:

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<u>Define the variables</u>:

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\implies \sf width \times length < 7875\:\:yd^2

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Solve the inequality:

\begin{aligned}(75-x)(105+x) & < 7875\\7875-30x-x^2 & < 7875\\-x^2-30x & < 0\\-x(x+30) & < 0\\x(x+30) & > 0\\\implies x & > 0 \:\: \textsf{ or }\:\:x < - 30\end{aligned}

Therefore, as distance is positive only and the maximum width is 75 yd (since we are subtracting from the original width):

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⇒ Perimeter = 2(72 + 108) = 360 yd

⇒ Area = 72 × 108 = 7776 yd²

<u>Example 2</u>:

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⇒ Width = 75 - 74 = 1 yd

⇒ Length = 105 + 74 = 179 yd

⇒ Perimeter = 2(1 + 179) = 360 yd

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