Question

Sum of areas of two squares is 468 m2

. If, the difference of their perimeters is 24 metres, find

the sides of the two squares.

Answer 1

Solution :

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

Sum of areas of two squares is 468 m². If the difference of their perimeter is 24 m.

$\bf{\red{\underline{\bf{To\:find\::}}}}$

The sides of the two square.

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

Let the side of two square be r m and m m respectively.

$\underbrace{\tt{FIRST\:CASE\::}}}}}$

A/q

$\boxed{\bf{Perimeter\:of\:square=4\times Side}}}}}$

$\longrightarrow\sf{4r-4m=24}\\\\\longrightarrow\sf{4(r-m)=24}\\\\\longrightarrow\sf{r-m=\cancel{{\dfrac{24}{4}}}}\\\\\longrightarrow\sf{r-m=6}\\\\\longrightarrow\bf{r=6+m..................(1)}}}$

$\underbrace{\tt{SECOND\:CASE\::}}}}}$

$\boxed{\bf{Area\:of\:square=Side\times Side}}}}}$

$\longrightarrow\sf{(r)^{2} +(m)^{2} =468}\\\\\longrightarrow\sf{(6+m)^{2} +m^{2} =468\:\:\:\:\:[from(1)]}\\\\\longrightarrow\sf{6^{2} +m^{2} +2*6*m+m^{2} =468}\\\\\longrightarrow\sf{36+m^{2} +12m+m^{2} =468}\\\\\longrightarrow\sf{2m^{2} +12m+36-468=0}\\\\\longrightarrow\sf{2m^{2} +12m-432=0}\\\\\longrightarrow\sf{2(m^{2} +6m-216)=0}\\\\\longrightarrow\sf{m^{2} +6m-216=\cancel{\dfrac{0}{2} }}\\\\\longrightarrow\sf{\pink{m^{2} +6m-216=0}}$

$\boxed{\bf{Using\:Quadratic\:Formula\::}}}}}$

As we can compared the given equation with ax² + bx + c = 0

• a = 1
• b = 6
• c = -216

So;

$\mapsto\sf{x=\dfrac{-b\pm\sqrt{b^{2} -4ac} }{2a} }\\\\\\\mapsto\sf{x=\dfrac{-6\pm\sqrt{(6)^{2}-4*1*(-216) } }{2*1} }\\\\\\\mapsto\sf{x=\dfrac{-6\pm\sqrt{36+864} }{2} }\\\\\\\mapsto\sf{x=\dfrac{-6\pm\sqrt{900} }{2}} \\\\\\\mapsto\sf{x=\dfrac{-6\pm30}{2} }\\\\\\\mapsto\sf{x=\dfrac{-6+30}{2} \:\:Or\:\:x=\dfrac{-6-30}{2} }\\\\\\\mapsto\sf{x=\cancel{\dfrac{24}{2} }\:\:Or\:\:x=\cancel{\dfrac{-36}{2} }}\\\\\\\mapsto\sf{\red{x=12\:m\:\:Or\:\:x\neq -18}}$

We know that negative value isn't acceptable.

Putting the value of m = 12 in equation (1),we get;

$\longrightarrow\sf{r=6+12}\\\\\longrightarrow\sf{\pink{r=18\:m}}$

Thus;

$\underline{\large{\sf{The\:sides\:of\:the\:two\:square\:is\:r=18\:m\:\:\&\:m=12\:m.}}}}$

Answer 2

$\tt{\huge{\orange {Hello!!! }}} M.V$

QUESTION :

Sum of areas of two squares is 468 m2

Sum of areas of two squares is 468 m2. If, the difference of their perimeters is 24 metres, find

Sum of areas of two squares is 468 m2. If, the difference of their perimeters is 24 metres, findthe sides of the two squares.

SOLUTION :

Let the sides of the squares be A and B

Area 1 = A^2

Area 2 = B^2

Perimeter 1 = 4 A

Perimeter 2 = 4 B

4 ( A - B ) = 24

=> A - B = 6.........(1)

=> A = B + 6

( B + 6 ) ^2 + B^2 = 462

=> 2 B^2 + 12 B + 36 = 462

=> 2B^2 + 12 B - 432 = 0

=> B^2 + 6 B - 216 = 0

=> B^2 + 18 B - 12 B - 216 = 0

=> B ( B + 18 ) - 12 ( B + 18 ) = 0

=> ( B - 12 ) ( B + 18 ) = 0

=> B is 12 m.

A = B + 6

=> A = 16 m.

Hence the sides of the squares are 12 m and 16 m respectively.