Question

△abc is the image of △abc under a dilation whose center is a and scale factor is $\frac{3}{4}$. which figure correctly shows △abc using the solid line? choose 1 answer:

Explanation:

Dilation with center \( A \) and scale factor \( \frac{3}{4} \) means \( A' \) coincides with \( A \) (since center of dilation is \( A \)), and \( B' \) and \( C' \) lie on segments \( AB \) and \( AC \) respectively, with \( \frac{AA'}{AB}=\frac{3}{4} \) (but \( A' = A \), so \( AB'=\frac{3}{4}AB \)) and \( AC'=\frac{3}{4}AC \). So the image \( \triangle A'B'C' \) should have \( A' = A \), and \( B', C' \) closer to \( A \) than \( B, C \) (since scale factor \( \frac{3}{4}<1 \), it's a reduction). Looking at the options, option B shows \( A' = A \) and \( B', C' \) inside \( \triangle ABC \) (closer to \( A \)), matching the dilation with center \( A \) and scale factor \( \frac{3}{4} \). Option A has \( A' \) not at \( A \), option C also misplaces \( A' \).

Answer:

B. The figure where \( A' \) coincides with \( A \) and \( B', C' \) are on \( AB, AC \) (closer to \( A \)) forming \( \triangle A'B'C' \) with solid lines.