Answer:
[tex]\large \boxed{\text{A. }\dfrac{13}{7}\text{; B. }\dfrac{17}{14} \text{; C. 507 in}^{2}}[/tex]
Step-by-step explanation:
A. Scale factor
When you dilate an object by a scale factor, you multiply its line lengths by the same number.
If EF/AB = 13/7, the scale factor is 13/7.
B. Length of EF
[tex]\begin{array}{rcl}\dfrac{EF}{AB} & = & \dfrac{13}{7}\\\\\dfrac{EF}{\frac{17}{26}} & = & \dfrac{13}{7}\\\\EF & = & \dfrac{13}{7}\times\dfrac{17}{26}\\\\ & = &\dfrac{1}{7}\times\dfrac{17}{2}\\\\ & = & \mathbf{\dfrac{17}{14}}\\\end{array}\\\text{The length of EF is $\large \boxed{\mathbf{ \dfrac{17}{14}}}$}[/tex]
C. Area of EFGH
If the lengths in a shape are all multiplied by a scale factor, then the areas will be multiplied by the scale factor squared.
ABCD is dilated by a scale factor of 13/7, so its area is dilated by a scale factor of
[tex]\left(\dfrac{13}{7} \right)^{2} = \dfrac{169}{49}[/tex]
The area of its dilated image EFGH is
[tex]\text{Area of EFGH} = \text{147 in}^{2} \times \dfrac{\text{169}}{\text{49}} = 3 \times 169\text{ in}^{2} = 507 \text{ in}^{2}\\\\\text{The area of EFGH is $\large \boxed{\textbf{507 in}^{\mathbf{2}}}$}[/tex]
