Given:
BC = 3 cm, FG = 4 cm
Required: bLength of EF and ratio of areas
Explanation:
(a) Since the rectangles ABCD and EFGH are similar, the correponding angles are proportional. Hence
[tex]\frac{AB}{EF}=\frac{BC}{FG}[/tex]
Plug the given values.
[tex]\frac{AB}{EF}=\frac{3}{4}[/tex]
If AB = a cm, then
[tex]\begin{gathered} \frac{a}{EF}=\frac{3}{4} \\ EF=\frac{4a}{3} \end{gathered}[/tex]
(b) Ara of ABCD
[tex]\begin{gathered} =\text{ Length}\times\text{ Width} \\ =3a\text{ cm}^2 \end{gathered}[/tex]
Area of EFGH
[tex]\begin{gathered} =\text{ Length}\times\text{ Width} \\ =4\times\frac{4a}{3} \\ =\frac{16a}{3}\text{ cm}^2 \end{gathered}[/tex]
Ratio of areas
[tex]\begin{gathered} =3a:\frac{16a}{3} \\ =9:16 \end{gathered}[/tex]
Final Answer: The ratio of areas of ABCD to EFGH is 916.
