Answer:
[tex]\qquad \:\boxed{\begin{aligned}& \qquad \:\sf \: Breadth \: of \: rectangle=9 \: cm \qquad \: \\ \\& \qquad \:\sf \: Length \: of \: rectangle=16 \: cm \end{aligned}} \qquad \\ \\ [/tex]
Step-by-step explanation:
Given that, length of a rectangle exceeds its breadth by 7 cm.
Let assume that
Breadth of a rectangle = x cm
So,
Length of a rectangle = x + 7 cm
We know,
[tex]\sf \:Area_{(Rectangle)} = Length \times Breadth \\ \\ [/tex]
So, on substituting the values, we get
[tex]\sf\implies \sf \:Area_{(Rectangle)} = (x + 7)x - - - (1) \\ \\ [/tex]
Further given that, If the length is decreased by 4 cm and the breadth is increased by 3 cm, the area of the new rectangle is the same as the area of the original rectangle.
Breadth of a rectangle = x + 3 cm
Length of a rectangle = x + 7 - 4 = x + 3 cm
Now,
[tex]\sf\implies \sf \:Area_{(Rectangle)} = (x + 3)(x + 3) - - - (2) \\ \\ [/tex]
So, According to statement,
[tex]\sf \:x(x + 7) = (x + 3)(x + 3) \\ \\ [/tex]
[tex]\sf \: {x}^{2} + 7x = x(x + 3) + 3(x + 3) \\ \\ [/tex]
[tex]\sf \: {x}^{2} + 7x = {x}^{2} + 3x + 3x + 9 \\ \\ [/tex]
[tex]\sf \: 7x = 6x + 9 \\ \\ [/tex]
[tex]\sf \: 7x - 6x = 9 \\ \\ [/tex]
[tex]\sf\implies \: x = 9 \\ \\ [/tex]
Hence,
[tex]\sf\implies \:\boxed{\begin{aligned}& \qquad \:\sf \: Breadth \: of \: rectangle=9 \: cm \qquad \: \\ \\& \qquad \:\sf \: Length \: of \: rectangle=16 \: cm \end{aligned}} \qquad \\ \\ [/tex]
