Question

14. The length of a rectangle is greater than the breadth by 3 cm. If the length is
increased by 9 cm and the breadth is reduced by 5 cm, the area remains the same

Find the dimensions of the rectangle.​

Answer 1

Answer:

Answer. Let the breadth be x cm. Then the length will be (x+3) cm. Given that length is increased by 9cm = x + 3 + 9 = x + 12 cm.

Answer 2

Given:-

• Length of rectangle is greater than breadth by 3 cm.
• Area remains same , after increasing & decreasing length and breadth by 9 and 5 cm respectively.

To Find:-

• Dimensions of rectangle ?

Solution: -

• Let the breadth of rectangle be x cm.

$\qquad$☀It’s given ,length of rectangle is greater than breadth by 3 cm.That means Length will be = 3 more than x.

• Length = (x + 3) cm

$\qquad$☀As we know that,Area of Rectangle = Length × Breadth.

Therefore :-

$\qquad$⎘Area of Rectangle

$\qquad$= (x + 3) × x

$\qquad$= x² + 3x

$\qquad$☀It's given, area remains same , after increasing & decreasing length and breadth by 9 and 5 cm respectively.

$\qquad\small\underline{\pmb{\sf \:According \: to \: the \: question :-}}$

$\pink{\qquad\leadsto\quad \pmb {\mathfrak{ (Length × Breadth) = x² + 3x }}}$

$\qquad\leadsto\quad \pmb {\mathfrak{ (x + 12) (x – 5) = x² + 3x }}$

$\qquad\leadsto\quad \pmb {\mathfrak{ x(x – 5) + 12(x – 5) = x² + 3x }}$

$\qquad\leadsto\quad \pmb {\mathfrak{ x² – 5x + 12x – 60 = x² + 3x }}$

$\qquad\leadsto\quad \pmb {\mathfrak{ 7x – 3x = 60 }}$

$\qquad\leadsto\quad \pmb {\mathfrak{ 4x = 60 }}$

$\qquad\leadsto\quad \pmb {\mathfrak{ x = \dfrac{60}{4} }}$

$\pink{\qquad\leadsto\quad \pmb {\mathfrak{ x = 15 cm }}}$

$\qquad$ ⎘ Breadth is x = 15 cm

$\qquad$ ⎘ Length is (x + 3) = 15 + 3 = 18 cm

⠀⠀⠀⠀¯‎¯‎‎‎¯‎¯¯‎¯‎‎‎¯‎‎‎¯‎¯‎¯‎‎¯‎‎‎¯‎¯‎‎‎¯‎‎‎¯‎¯‎¯‎‎‎¯‎‎¯‎‎‎¯‎‎¯‎¯‎‎‎¯‎‎‎¯‎¯‎¯‎‎¯‎¯‎‎¯‎¯¯‎¯‎¯‎‎‎¯‎¯‎‎‎¯‎‎‎¯‎‎‎¯‎‎‎¯‎