The area of a rectangular plate is $5\dfrac{5}{7}$ m² and its length is $3\dfrac{3}{4}$ m, find its breadth and its perimeter.

Area of a rectangular plate = $5\dfrac{5}{7}$ m² = $\dfrac{40}{7}$ m²

Length of a rectangular plate = $3\dfrac{3}{4}$ m = $\dfrac{15}{4}$ m

Let the breadth of the rectangular plate be b.

Area = length x breadth

$\dfrac{40}{7} = \dfrac{15}{4} \times b\\[1em] \Rightarrow b = \dfrac{40}{7} ÷ \dfrac{15}{4}\\[1em] \Rightarrow b = \dfrac{40}{7} \times \dfrac{4}{15}\\[1em] \Rightarrow b = \dfrac{40 \times 4}{7 \times 15}\\[1em] \Rightarrow b = \dfrac{160}{105}\\[1em] \Rightarrow b = \dfrac{32}{21}\\[1em] \Rightarrow b = 1\dfrac{11}{21}$

Perimeter = 2(length + breadth)

$= 2 \times \Big(\dfrac{15}{4} + \dfrac{32}{21}\Big)$

LCM of 4 and 21 is 2 x 2 x 3 x 7 = 84

$= 2 \times \Big(\dfrac{15 \times 21}{4 \times 21} + \dfrac{32 \times 4}{21 \times 4}\Big)\\[1em] = 2 \times \Big(\dfrac{315}{84} + \dfrac{128}{84}\Big)\\[1em] = 2 \times \Big(\dfrac{315 + 128}{84}\Big)\\[1em] = 2 \times \Big(\dfrac{443}{84}\Big)\\[1em] = \Big(\dfrac{443 \times 2}{84}\Big)\\[1em] = \Big(\dfrac{886}{84}\Big)\\[1em] = \Big(\dfrac{443}{42}\Big)\\[1em] = 10\dfrac{23}{42}$

Hence, breadth = $1\dfrac{11}{21}$ and perimeter = $10\dfrac{23}{42}$