Question

the length of the rectangle is thrice than its breadth and area is 192sq.cm ,then find its breadth​

Answer 1

Answer :-

Given :-

• Length = 3 × Breadth
• Area = 192 cm²

To Find :-

• Breadth

Solution :-

Let the breadth be x

Length = 3x

We know that,

→ Area = Length × Breadth

Substituting the values in formula :-

→ 192 = x × 3x

→ 192 = 3x²

→ x² = 192 / 3

→ x² = 64

→ x = √64

→ x = 8

Breadth = 8 cm

Answer 2

Given :

• The length of the rectangle is thrice than its breadth.
• Area of Rectangle = 192cm²

To Find :

• Breadth of the Rectangle

Solution :

⟾ Let the Breadth of Rectangle be p

⟾ Then Length of Rectangle will be 3p

⠀

According to the Question :

⟶ Area of Rectangle = Length × Breadth

⟶ 192 = 3p × p

⟶ 192/3 = p × p

⟶ 64 = p × p

⟶ 64 = p²

⟶ √64 = p

⟶ 8cm = p

⠀

Therefore :

• Breadth of Rectangle = 8cm

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$\color{red}{\boxed { \begin{array}{ |c|c| } \hline \\ \sf \blue{Perimeter \: of \: rectangle}& \sf \color{lime}{2(length + breadth)} \\ \\ \hline \\\blue{ \sf Area \: of \: rectangle}& \sf \color{lime}{length \times breadth}\\ \\ \hline \\ \sf\blue {Length \: of \: rectangle}& \sf \color{lime}{\dfrac{Area }{ Breadth}} \\ \\ \hline \\ \sf \blue{Breadth \: of \: rectangle}& \sf \color{lime}{ \dfrac{Area}{Length} }\\ \\ \hline \\ \sf \blue{Diagonal \: of \: rectangle}& \sf \color{lime}{\sqrt{ {length}^{2} + {breadth}^{2} }} \\ \\ \end{array}}}$