Solution
Question 3:
- Let the dimension of a shape be x and the dimension of its enlarged or reduced image be y.
- The linear scale factor will be:
[tex]sf_L=\frac{y}{x}[/tex]- If the area of the original shape is Ax and the Area of the enlarged or reduced image is Ay, then, the Area scale factor is:
[tex]sf_A=\frac{A_y}{A_x}=\frac{y^2}{x^2}[/tex]- We have been given the area of the big shape to be 99un² and the dimensions of the big and small shapes are 6 and 2 respectively.
- Based on the explanation given above, we can conclude that:
[tex]\begin{gathered} \text{ If we choose }x\text{ to be 6, then }y\text{ will be 2. And if we choose }x\text{ to be 2, then }y\text{ will be 6} \\ \text{ So we can choose any one.} \\ \\ \text{ For this solution, we will use }x=6,y=2 \end{gathered}[/tex]- Now, solve the question as follows:
[tex]\begin{gathered} \text{ Linear Scale factor:} \\ sf_L=\frac{y}{x}=\frac{2}{6}=\frac{1}{3} \\ \\ \text{ Area Scale factor:} \\ sf_A=\frac{y^2}{x^2}=\frac{2^2}{6^2}=\frac{1}{9} \\ \\ \text{ Also, we know that:} \\ sf_A=\frac{A_y}{A_x}=\frac{y^2}{x^2} \\ \\ \text{ We already know that }\frac{y^2}{x^2}=\frac{1}{9} \\ \\ \therefore\frac{A_y}{A_x}=\frac{1}{9} \\ \\ A_x=99 \\ \\ \frac{A_y}{99}=\frac{1}{9} \\ \\ \therefore A_y=\frac{99}{9} \\ \\ A_y=11un^2 \end{gathered}[/tex]Final Answer
The answers are:
[tex]\begin{gathered} \text{ Linear Scale Factor:} \\ \frac{1}{3} \\ \\ \text{ Area Scale Factor:} \\ \frac{1}{9} \\ \\ \text{ Area of smaller shape:} \\ 11un^2 \end{gathered}[/tex]