AlieeR
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A blue print of a house has a scale of 1.5 in: 5 feet. Would a ping pong table that is a feet by 5 feet fit in a space that has drawing dimensions of 3 inches and 4.8 inches? If so, how much space (in ft2) will remain after the table is put in the room?​

Sagot :

Answer:

Yes, the ping pong table fits in the room.

There will be 115 ft² remaining.

Step-by-step explanation:

Given scale:  

1.5 in : 5 ft

Given dimensions:

  • Ping pong table = 9 ft × 5 ft
  • Room = 3 in × 4.8 in

To determine if the ping pong table can fit in the given space, convert the dimensions of the room to feet using the given scale.

First find how many feet is equal to one inch by dividing both sides of the ratio by 1.5:

[tex]\implies \sf \dfrac{1.5}{1.5} = \dfrac{5}{1.5}[/tex]

[tex]\implies \sf 1\;in=\dfrac{10}{3}\;ft[/tex]

To convert the dimensions of the room to feet, multiply each dimension by ¹⁰/₃:

[tex]\implies \sf 3\; in \times \dfrac{10}{3}=10\;ft[/tex]

[tex]\implies \sf 4.8\; in \times \dfrac{10}{3}=16\;ft[/tex]

Therefore, the dimensions of the room are 10 ft × 16 ft.

The ping pong table will easily fit in the room as its largest dimension of 9 ft is smaller than the smallest dimension of the room.

To find how much space in square foot will remain after the table is put in the room, calculate the area of the table and the room, then subtract the former from the latter.

Model the table and the room as rectangles.

[tex]\boxed{\textsf{Area of a rectangle}=\sf width \times length}[/tex]

[tex]\implies \textsf{Area of ping pong table}=\sf 5\times9=45\;ft^2[/tex]

[tex]\implies \textsf{Area of room}=\sf 10 \times 16=160\;ft^2[/tex]

Therefore:

[tex]\begin{aligned}\implies \textsf{Space remaining}&=\textsf{Area of room}-\textsf{Area of table}\\&=\sf 160-45\\&=\sf 115\;ft^2\end{aligned}[/tex]