Three rectangles

Some wire is used to make three rectangles: A, B, and C. Rectangle B's dimensions are 3/5 cm larger than Rectangle A's dimensions, and Rectangle C's dimensions are again 3/5 cm larger than Rectangle B's dimensions. Rectangle A is 2 cm by 3 1/5 cm.

What is the total area of all three rectangles?

Final Answer:

S =  30 ⁹⁄₂₅ = 759/25 cm²

Step-by-step explanation:

$$\begin{gathered} a_1=2\text{cm} \\ {a}_2=3\frac{1}{5}=3+\frac{1}{5}=\frac{3\cdot 5+1}{5}=\frac{15+1}{5}=\frac{16}{5}=3.2\text{cm} \\ x=\frac{3}{5}=0.6\text{cm} \\ {b}_1=a_1+x=2+\frac{3}{5}=\frac{2\cdot 5}{5}+\frac{3}{5}=\frac{10}{5}+\frac{3}{5}=\frac{10+3}{5}=\frac{13}{5}=2.6\text{cm} \\ {b}_2=a_2+x=\frac{16}{5}+\frac{3}{5}=\frac{16+3}{5}=\frac{19}{5}=3.8\text{cm} \\ {c}_1=b_1+x=\frac{13}{5}+\frac{3}{5}=\frac{13+3}{5}=\frac{16}{5}=3.2\text{cm} \\ {c}_2=b_2+x=\frac{19}{5}+\frac{3}{5}=\frac{19+3}{5}=\frac{22}{5}=4.4\text{cm} \\ {S}_1=a_1\cdot a_2=2\cdot \frac{16}{5}=\frac{2\cdot 16}{5}=\frac{32}{5}=6.4{\text{cm}}^2 \\ {S}_2=b_1\cdot b_2=\frac{13}{5}\cdot \frac{19}{5}=\frac{13\cdot 19}{5\cdot 5}=\frac{247}{25}=9.88{\text{cm}}^2 \\ {S}_3=c_1\cdot c_2=\frac{16}{5}\cdot \frac{22}{5}=\frac{16\cdot 22}{5\cdot 5}=\frac{352}{25}=14.08{\text{cm}}^2 \\ S=S_1+S_2+S_3=\frac{32}{5}+\frac{247}{25}+\frac{352}{25}=\frac{160}{25}+\frac{247}{25}+\frac{352}{25}=\frac{160+247+352}{25}=\frac{759}{25}=\frac{759}{25}{\text{cm}}^2=30\frac{9}{25}{\text{cm}}^2=30.36{\text{cm}}^2 \end{gathered}$$

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