The area of a piece of paper is $7\dfrac{3}{26}$ cm² and its breadth is $2\dfrac{9}{13}$ cm. Find its length and perimeter.

Area of a rectangular paper = $7\dfrac{3}{26}$ cm² = $\dfrac{185}{26}$ cm²

Breadth of a rectangular paper = $2\dfrac{9}{13}$ m = $\dfrac{35}{13}$ cm

Let the length of the piece of paper be l.

Area = length x breadth

$\dfrac{185}{26} = l \times \dfrac{35}{13}\\[1em] \Rightarrow l = \dfrac{185}{26} ÷ \dfrac{35}{13}\\[1em] \Rightarrow l = \dfrac{185}{26} \times \dfrac{13}{35}\\[1em] \Rightarrow l = \dfrac{185 \times 13}{26 \times 35}\\[1em] \Rightarrow l = \dfrac{2405}{910}\\[1em] \Rightarrow l = \dfrac{37}{14}\\[1em] \Rightarrow l = 2\dfrac{9}{14}$

Perimeter = 2(length + breadth)

$= 2 \times \Big(\dfrac{37}{14} + \dfrac{35}{13}\Big)$

LCM of 14 and 13 is 2 x 7 x 13 = 182

$= 2 \times \Big(\dfrac{37 \times 13}{14 \times 13} + \dfrac{35 \times 14}{13 \times 14}\Big)\\[1em] = 2 \times \Big(\dfrac{481}{182} + \dfrac{490}{182}\Big)\\[1em] = 2 \times \Big(\dfrac{481 + 490}{182}\Big)\\[1em] = 2 \times \Big(\dfrac{971}{182}\Big)\\[1em] = \Big(\dfrac{971 \times 2}{182}\Big)\\[1em] = \Big(\dfrac{1942}{182}\Big)\\[1em] = \Big(\dfrac{971}{91}\Big)\\[1em] = 10\dfrac{61}{91}$

Length = $2\dfrac{9}{14}$ and perimeter = $10\dfrac{61}{91}$