Question

The perimeter of a rectangle is 42 metres and it's diagonal is15 metres.what are the length s of it's sides?​

Answer 1

Answer:

length of the rectangle , l = 21 - b

Answer 2

S O L U T I O N :

$\underline{\bf{Given\::}}$

The perimeter of a rectangle is 42 m & it's diagonal is 15 m.

$\underline{\bf{Explanation\::}}$

$\underbrace{\bf{1^{st}\:Case\::}}$

As we know that formula of the diagonal of rectangle;

$\boxed{\bf{(D)^{2} = (l)^{2} + (b)^{2}}}$

A/q

$\mapsto\tt{(15)^{2} = (l)^{2} + (b)^{2}............(1)}$

$\underbrace{\bf{2^{nd}\:Case\::}}$

As we know that formula of the perimeter of rectangle;

$\boxed{\bf{Perimeter = 2(l+b)}}$

So,

$\mapsto\tt{2(l+b) = 42}$

$\mapsto\tt{(l+b) = \cancel{42/2}}$

$\mapsto\tt{l+b = 21}$

$\mapsto\tt{b = 21-l..............(2)}$

∴ Putting the value of b in equation (1),we get;

$\mapsto\tt{(15)^{2} = (l)^{2} + (21-l)^{2} }$

$\mapsto\tt{225 = (l)^{2} + (21)^{2} + (l)^{2} -2 \times 21 \times l \:\:[\therefore using \:(a-b)^{2}]}$

$\mapsto\tt{225 = l^{2} + 441 + l^{2} - 42l}$

$\mapsto\tt{225 - 441= 2l^{2} - 42l}$

$\mapsto\tt{-216= 2l^{2} - 42l}$

$\mapsto\tt{ 2l^{2} - 42l+216 = 0}$

$\mapsto\tt{ 2(l^{2} - 21l+108) = 0}$

$\mapsto\tt{ l^{2} - 21l+108 = 0/2}$

$\mapsto\tt{ l^{2} - 21l+108 =0}$

$\mapsto\tt{ l^{2} -12l - 9l+108 =0}$

$\mapsto\tt{ l(l - 12) -9(l-12)=0}$

$\mapsto\tt{ (l - 12) (l-9)=0}$

$\mapsto\tt{l-12 = 0\:\:Or\:\:l - 9 = 0}$

$\mapsto\bf{l=12\:m\:\:Or\:\:l =9\:m}$

∴ Putting the value of l = 12 m in equation (2),we get;

→ b = 21 m - 12 m

→ b = 9 m

∴ Putting the value of l = 9 m in equation (2),we get;

→ b = 21 m - 9 m

→ b = 12 m

Thus,

The length of the rectangle will be 12 m & 9 m .