Question

The perimeter of a rectangle and square are equal,but the area of the former is 225 sq.M less than the later.Find the length and breadth of the rectangle.

Answer 1

Answer:

$\boxed{a \geqslant 15}$

$\huge\underline\textsf{Explantion:- }$

Let,L = length of the rectangle

B = breadth

A = side of Square

Now, given that

perimeter of rectangle =perimeter of square

$\large\underline\textsf{ Formula Used }$

=2(l+b)=4a

=l+b=2a

Now, we have given the

Area of rectangle =Are of square-225

LB= a^2

a^2-15^2

$\large\underline\textsf{Identity used:- }$

$\boxed{(a {}^{2} - b {}^{2} ) = (a + b)(a - b)}$

Now,

LB =(a+15)(a-15)

Now ,as length is always greater than than the breadth of a rectangle.

$\therefore$ L=a+15 and B=a-15

And ,as the length of any side is always positive so,

$\boxed{a - 15 \geqslant 0 = > a \geqslant 15}$

L=a +15

& B=a-15

$\boxed{a \geqslant 15}$

Answer 2

$\huge{ \underline{ \overline{ \mid{ \mathfrak{ \purple{ \: \: SOLUTION \: \: } \mid}}}}}$

$\underline{ \mathfrak{ \: \: Given, \: \: }} \\ \\ \text{Perimeter of rectangle = Perimeter of } \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \text{square} \\ \\ \text{Area of rectangle is 225 m less than the} \\ \text{ area of square.} \\ \\ \\ \underline{ \mathfrak{\: \: Suppose, \: \: }} \\ \\ : \rightsquigarrow \text{ \: Length of the rectangle = l} \\ \\ : \rightsquigarrow \text{ \: Breadth of the rectangle = b} \\ \\ : \rightsquigarrow \: \text{ \: Side of Square = a }$

$\underline{ \text{ \: According to the first condition, \: }} \\ \\ \: \: \: \: \: \: \: \: \: \sf{2(l+b) = 4a} \\ \\ \sf{ \implies \: l+b=2a}$

$\underline{ \text{ \: According to the second \: condition, \: }} \\ \\ \: \: \: \: \: \: \: \sf{l \times b = a {}^{2} - 225} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf{ = a {}^{2} - (15) {}^{2} } \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf{ =(a + 15)(a - 15)} \\ \\ \sf{ \therefore \: \: l \times b = (a + 15)(a - 15) \: \: - - - > (1)}$

$\underline{ \mathfrak{ \: \: We \: know \: \: that, \: \: }} \\ \\ \: \: \: \: \: \text{Length is always greater than the breadth} \\ \text{ of a rectangle.}$

So, comparing L.H.S. and R.H.S. of the eq. (1), we get

$: \rightsquigarrow \: \: \: \: \: \tt{ \red{ l = a +15 } \: \: \: and\: \: \: \: \red{ b=a-15}}$

$\underline{ \mathfrak{ \: \: Again, \: \: }} \\ \\ \text{The length or breadth of any dimension} \\ \text{ is always positive.}$

$\tt{ \therefore \: \: \: a + 15 \geqslant 0 \: \: \: \: \: \: \: \: and \: \: \: \: \: \: \: a - 15\geqslant 0} \\ \\ \tt{ \implies \: a \geqslant 15 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \implies \: a \geqslant 15}$

$\: \: \: \tt{ \therefore\: \: \red{Length, l = a + 15 }\: \: \: \: \: \: \: \forall \: \: \: \red{a \geqslant 15}} \\ \\ \tt{And,}\\ \\ \tt{ \: \: \: \: \: \: \red{Breadth,b = a - 15 }\: \: \: \: \: \: \: \forall \: \: \: \: \red{a \geqslant 15}}$