Solution:

( Diện tích của hình nào dưới đây gần bằng 1 ? )

The area of figure (A) is $\frac{1}{2}\times \frac{7}{3}\times \frac{3}{5}=\frac{7}{10}$, that is $\frac{3}{10}$ different with 1

The area of figure (B) is $\frac{1}{2}\times (\frac{1}{2}+2)\times 1=\frac{5}{4}$, that is $\frac{1}{4}$ different with 1

The area of figure (C) is $\frac{1}{2}\times \frac{3}{2}\times \frac{3}{2}=\frac{9}{8}$ , that is $\frac{1}{8}$ different with 1

The area of figure (D) is $\frac{11}{10}\times \frac{11}{10}=\frac{121}{100}$ that is $\frac{21}{100}$ different with 1

The area of figure (E) is $\frac{3}{2}\times \frac{3}{5}=\frac{9}{10}$ that is $\frac{1}{10}$ different with 1

Since $\frac{3}{10}=\frac{60}{200}>\frac{1}{4}=\frac{50}{200}>\frac{21}{100}=\frac{42}{200}>\frac{1}{8}=\frac{25}{200}>\frac{1}{10}=\frac{20}{200}$ , hence the closest is (E)