Question

If the sum of the areas of two squares is 468 m2 and the difference of their perimeters is24m .find the meserments of their sides ​????

Answer 1

Answer:

$\boxed{\sf The\:measurements\:of\:both\:of\:the\:sides\:is\:18m\:and\:12m\:respectively.}$

Step-by-step explanation:

→ Let the side of the first square be x

→ Let the side of the second square be y

Their squares,

→ Area of first square(x) = x²

→ Area of second square(y) = y²

Let their perimeters be 4x and 4y

★ According To The Question :

→ 4x - 4y = 24

= x - y = 6

= x = 6 + y

★ Equation :

→ (y + 6)² + y² = 468 m²

= 2y² + 12y + 36 = 468 m²

= 2y² + 12y + 36 - 468 = 0

= 2y² + 12y - 432 = 0

= y² + 6y - 216 = 0

= y² + 18y - 12y - 216 = 0

= y(y + 18) - 12(y + 18) = 0

= (y - 12) (y + 18) = 0

→ So, we can say y = 12 or y = -18

→ We know that, sides can not be negative, thus y = 12 and x = y + 6 = 12 + 16 = 18.

__________________________

Answer 2

Given,

• The sum of the areas of two square is 468 m².
• The Difference of their perimeters is 24 m .

To Find,

• The measurement of their sides .

Solution,

☯$\sf{Suppose \ the \ sides \ of \ two \ squares \ be \ "a" \ and \ "b"}$

We know that :

• Perimeter of Square = 4 x side
• Area of square = (side)²

$\boxed{\sf{\red{Therefore,}}}$

• First Square Perimeter = 4a
• Second Square Perimeter = 4b
• First Square Area = a²
• Second Square Area = b²

According to the Question :-

$\boxed{\sf{\purple{Case - (i)}}}$

• The Difference of their perimeter is 24 m.

$\implies \sf{4a - 4b = 24 \ (Difference \ of \ Perimeter)}$

Or,

$\implies \sf{a - b = 6}$

$\implies \boxed{\sf{a = b + 6}}$

$\boxed{\sf{\purple{Case - (ii)}}}$

• The sum of the area of two square is 468 m².

$\implies \sf{a^{2} + b^{2} = 468 }$

$\implies \sf{(b + 6)^{2} + b^{2} = 468}$

$\implies \sf{36 + b^{2} + 12b + y^{2} = 468}$

$\bold{[ Use \ Formula :- (a + b )^{2} = a^{2} + b^{2} + 2ab]}$

$\implies \sf{2b^{2} + 12b + y^{2} = 468 - 36}$

$\implies \sf{2b^{2} + 12b = 432 }$

$\implies \sf{2b^{2} + 12b - 432 = 0}$

$\implies \sf{2(b^{2} + 6b - 216) = 0}$

Or,

$\implies \sf{b^{2} + 6b -216 = 0}$

$\implies \sf{b^{2} + 18b - 12b - 216 = 0 \ (Factorise)}$

$\implies \sf{b(b + 18)-12(b +18) = 0}$

$\implies \sf{(b - 12)(b + 18) = 0}$

Therefore,

→b = 12 or b = -18

(So side cannot be negative, then b = 12)

Now Put the value of b in First Case to find value of a :-

$\implies \sf{a = 6 + b}$

$\implies \sf{a = 6 + 12}$

$\implies \boxed{\sf{a = 18}}$

Now Find Measurement sides of Square

$\boxed{\bold{\purple{Side \ of \ First \ Square = a = 18 \ m}}}$

$\boxed{\bold{\purple{Side \ of \ Second \ Square = b = 12 \ m}}}$

$\rule{200}2$