Question

6 The length of a rectangle is 4 cm more than its breadth. If the length is increased by 4 cm and breadth
is decreased by 2 cm, the area remains the same as that of original rectangle. Find the length and
breadth of the rectangle.
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Answer 1

Answer :

Let, Length = l

breadth = b

Formula of Area of the rectangle = length × breadth

Given ...

length = breadth + 4

here..Length is increased by 4 cm .

and

Breadth is decreased by 2 cm .

New length = l + 4

New breadth = b - 2

$\sf( l + 4 )( b - 2 ) = lb ( Area \: given. ) \\ \sf= \cancel{lb} - 2l + 4b - 8 = \cancel{lb} \\ \sf = - 2l + 4b = 8 \\ \sf = - l + 4b = \frac{ \cancel8}{ \cancel2} = 4 \\ \sf = - l + 4b = 4 \\ \sf = - b - 4 + 2b = 4 \\(\sf l = b + 4 \: \: \therefore \: - l = - (b + 4) = - b - 4)$

Here

b = 8 cm

l = 4 + 8 = 12 cm .

Answer 2

Step-by-step explanation:

$let \: the \: breadth \: of \: rectangle \: be \: x.$

$and \: let \: the \: length \: of \: rectangle \: be \: x + 4$

$area \: of \: rectangle \: = \: l \times b \\ = x(x + 4) \\ = x^{2} + 4x$

$let \: the \: length \: be \: (x + 4) + 4.$

$let \: the \: breadth \: be \: x - 2$

${x}^{2} + 4x = (x + 4 + 4)(x - 2) \\ {x}^{2} + 4x = (x + 8)(x - 2) \\ {x}^{2} + 4x = {x}^{2} - 2x + 8x - 16 \\ {x}^{2} + 4x = {x}^{2} + 6x - 16 \\ 4x - 6x = - 16 \\ - 2x = - 16 \\ x = \frac{ - 16}{ - 2} \\ x = 8$

$therefore \: length \: = (x + 4) = 8 + 4 = 12 \\ breadth \: = x = 8$

HOPE IT HELPS!!