Question

Determine which of the following rectangles has the least area among the four.
Rectangle $A:Length=\frac{7}{4}$ cm, $Width=\frac{7}{5}$ cm
Rectangle $B:Length=\frac{4}{5}$ cm, $Width=\frac{5}{4}$ cm
Rectangle $C:Length=\frac{2}{3}$ cm, $Width=\frac{3}{2}$ cm
Rectangle $D:Length=\frac{5}{4}$ cm, $Width=\frac{2}{3}$ cm

• A. Rectangle C
• B. Rectangle D
• C. Rectangle A
• D. Rectangle B

Answer 1

The correct option is B Rectangle D

Rectangle A
$Length=\frac{7}{4}$ cm, $Width=\frac{7}{5}$ cm
$Area=Length\times Width$
$=\frac{7}{4}\times \frac{7}{5}=\frac{7\times 7}{4\times 5}=\frac{49}{20}$ sq cm

Rectangle B
$Length=\frac{4}{5}$ cm, $Width=\frac{7}{4}$ cm
$Area=Length\times Width$
$=\frac{4}{5}\times \frac{7}{4}=\frac{4\times 7}{5\times 4}=\frac{28}{20}$ sq cm

Rectangle C
$Length=\frac{2}{3}$ cm, $Width=\frac{3}{2}$ cm
$Area=Length\times Width$
$=\frac{2}{3}\times \frac{3}{2}=\frac{2\times 3}{3\times 2}=\frac{6}{6}=1$ sq cm

Rectangle D
$Length=\frac{5}{4}$ cm, $Width=\frac{2}{3}$ cm
$Area=Length\times Width$
$=\frac{5}{4}\times \frac{2}{3}=\frac{5\times 2}{4\times 3}=\frac{10}{12}$ sq cm


To compare the fractions, we have to make them have the same denominators.

Rectangle A
$\frac{49}{20}sqcm=\frac{49\times 3}{20\times 3}=\frac{147}{60}$ sq cm

Rectangle B
$\frac{28}{20}$ sq cm = $\frac{28\times 3}{20\times 3}=\frac{84}{60}$ sq cm


Rectangle C
$1$ sq cm $=\frac{1\times 60}{1\times 60}=\frac{60}{60}$ sq cm

Rectangle D
$\frac{10}{12}$ sq cm $=\frac{10\times 5}{12\times 5}=\frac{50}{60}$ sq cm

$\frac{147}{60}>\frac{84}{60}\frac{60}{60}>\frac{50}{60}$

Rectangle A > Rectangle B > Rectangle C > Rectangle D
So, Rectangle D has the leastest area.
Hence, option D is correct.