Question

the length of a rectangle exceed its breadth by 9 CM . if length and breadth are each increased by 3 cm the area of new rectangle will be 84 CM square more than that of the given rectangle find the length and breadth of the given rectangle.

please help me​

Answer 1

Answer:

If Length and Breadth Are Each Increased by 3 Cm, the Area of the New Rectangle Will Be 84 Cm2 More than that of the Given Rectangle.

Answer 2

$\large{\underline{ \underline{ \sf{Given :}}}}$

• Length of a rectangle exceeds by 9 cm.
• Length & Breadth of rectangle are increased by 3 cm then the area of new rectangle will be 84 cm² more than that of given rectangle.

$\large{\underline{ \underline{ \sf{To \: Find :}}}}$

• The length and breadth of the rectangle.

$\large{\underline{ \underline{ \sf{Solution :}}}}$

Let the breadth be x cm and length be (x + 9) cm

We know,

Area of rectangle = length × breadth

$\mapsto \tt \: x(x + 9) \\ \mapsto \tt \: ( {x}^{2} + 9x) {cm}^{2}$

• New length :

$\mapsto \tt \: (x + 9) + 3 \: cm \\ \mapsto \tt \: x + 12 \: cm$

• New breadth :

$\mapsto \tt \: (x + 3) \: cm$

Now calculating the new area,

$\implies \tt \: (x + 12)(x + 3) {cm}^{2} \\ \implies \tt \: {x}^{2} + 3x + 12x + 36 {cm}^{2} \\ \implies \tt \: {x}^{2} + 15x + 36 {cm}^{2}$

A.T.Q :

New area - Area = 84 cm²

$\implies \tt \: ( {x}^{2} + 15x + 36) - ( {x}^{2} + 9x) = 84 \: {cm}^{2} \\ \implies \tt \cancel{ {x}^{2} } + 15x + 36 - \cancel{ {x}^{2} } - 9x = 84 \: cm \\ \implies \tt \: 6x + 36 = 84 \\ \implies \tt \: 6x = 84 - 36 \\ \implies \tt \: 6x = 48 \\ \implies \tt \: x = \cancel\frac{48}{6} \\ \implies \bf \: x = 8$

Therefore,

• Breadth = x = 8 cm
• Length = (x + 9) = 8 + 9 = 17 cm

$\star$ $\underline{ \boxed{ \bf{ \red{Length \: of \: the \: rectangle = 17 \: cm}}}}$

$\star$ $\underline{ \boxed{ \bf{ \green{Breadth \: of \: the \: rectangle = 8 \: cm}}}}$