Question

An architect makes a model of a new house. The model shows a tile patio in the backyard. In the? model, each tile has length one fourth in and width one sixth in. The actual tiles have length one sixth ft and width one ninth ft. What is the ratio of the length of a tile in the model to the length of an actual? tile? What is the ratio of the area of a tile in the model to the area of an actual? tile? Use pencil and paper. Describe two ways to find each ratio.

Answer 1

Answer: 1:8 and 1:64

Step-by-step explanation:

First way :

Since, Here Each tile has length one fourth in and width one sixth in.

That is $L_1 = 1/4$ inches and $W_1 = 1/6$ inches

And, the actual tiles have length one sixth ft and width one ninth ft.

That is $L_2 = 2$ inches and $W_2 = 4/3$ inches ( because 1 foot = 12 inches)

Thus the ratio of length of the actual length of tile =  $L_1/L_2$

= $\frac{1/4}{2}= \frac{1}{8}$

And, The ratio of tile in the model of the actual area = Area of the tile in model/ actual area

= $\frac{1/4\times 1/6}{2\times 4/3}$

= 1/64

Second way:

Since the ratio of the width of the tiles in the model and the actual width = 1/8

Therefore, there is dilation occurs with a scale factor 1/8 therefore,

By the property of dilation, The ratio of there area = $(1/8)^2$ = 1/64

Answer 2

Answer:

The required ratio of length of tile is $\frac{1}{8}$ and ratio of area $\frac{1}{64}$.

Step-by-step explanation:

Given: An architect makes a model of a new house. The model shows a tile patio in the backyard. In the model, each tile has length one fourth in and width one sixth in. The actual tiles have length one sixth ft and width one ninth ft.

According to question,

Length and width of each tiles are one fourth in and width one sixth in that is .

 And, the actual tiles have length one sixth ft and width one ninth ft.

$L_{2} =\frac{1}{6}\;\rm{ft}=2\;\rm{inches} \;\& \;W_{2}=\frac{1}{9}\;\rm{ft}=\frac{4}{3}\;\rm{inches}\;\;\;\; \{1\;\rm{ft}=12\;\rm{inches}\}$.  

Thus, the ratio of length of the actual length of tile $=\frac{L_1}{L_2}=\frac{1/4}{2}=\frac{1}{8}$

Now, Area of tile$=\;\rm{length\times\;breadth}$

Required ratio$=\frac{1/4\times1/6}{2\times4/3}=\frac{1}{64}$

Since the ratio of the width of the tiles in the model and the actual width

By the property of dilation, The ratio of there area

Hence, the required ratio of length of tile is $\frac{1}{8}$ and ratio of area $\frac{1}{64}$.

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