Question

the length of the rectangle exceeds its breadth by 3 cm . if the length and breadth are each increased by 2 cm ,then the area of the new rectangle will be 70 sq cm more than that of given rectangle.Find the length and breadth of the given rectangle.

Answer 1

$Let\ the\ breadth\ and\ the\ length\ of\ the\ rectangle\ be\ x\ and\ x + 3 \\ respectively. \\ \\ The\ breadth\ and\ the\ length\ of\ the\ new\ rectangle\ will\ be\ x + 2 \\ and\ x + 5\ respectively. \\ \\ \\ (x + 2)(x + 5) - x(x + 3) = 70 \\ \\ = (x^2 + 7x + 10) - (x^2 + 3x) = 70 \\ \\ = x^2 + 7x + 10 - x^2 - 3x = 70 \\ \\ = 4x + 10 = 70 \\ \\ 4x = 70 - 10 = 60 \\ \\ x = \frac{60}{4} = \bold{15}$

$\\ \\ \\ Breadth\ of\ the\ rectangle = x = \bold{15\ cm} \\ \\ Length\ of\ the\ rectangle = x + 3 = 15 + 3 = \bold{18\ cm} \\ \\ \\ Breadth\ of\ new\ rectangle = x + 2 = 15 + 2 = \bold{17\ cm} \\ \\ Length\ of\ new\ rectangle = x + 5 = 15 + 5 = \bold{20\ cm}$

$\\ \\ \\ (17 \times 20) - (15 \times 18) \\ \\ = 340 - 270 \\ \\ = 70$

$\\ \\ \\ Thank\ you.\ Have\ a\ nice\ day. \\ \\ \\ \#adithyasajeevan$

Answer 2

Answer:

Let be the breadth x

length =x+3

according to the question if we compare the rectangle's area to the new rectangle's area the difference will be 70sq cm

area of the first rectangle =

x ( x + 3 )

= $x^{2}$ + 3x

after adding 2cm to length and breadth will x + 3 + 2 and x + 2

area of the new rectangle =

( x + 3 + 2 ) ( x + 2 )

( x + 5) ( x + 2 )

x ( x + 2 ) + 5 ( x + 2 )

$x^{2}$ + 2x + 5x + 10

= $x^{2}$ + 7x + 10

according to the question

( $x^{2}$ + 7x + 10 ) - ( $x^{2}$ + 3x ) = 70

$x^{2}$ + 7x + 10 - $x^{2}$ - 3x = 70

$x^{2}$ - $x^{2}$  + 7x - 3x + 10 = 70

4x + 10 = 70

4x = 70 - 10

4x = 60

x = 15

length of rectangle = 18 cm

breadth of rectangle = 15 cm

Step-by-step explanation: