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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.(Use quadratic formula)
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Answer 1

Question:-

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

Given:-

• sum of area of two squares = 468 m ²
• difference of their perimeter = 24 m

To find :-

• the sides of the 2 squares = ?

Solution:-

→Let the side of 1st square be "a" m

→the side of 2nd square be "A" m

→Area of 1st square = a² sq. m.

→Area of 2nd square = A² sq. m.

⇒so , their perimeter would be 4a and 4A

→Given that , 4A -4a = 24

→A - a = 6

→A² + a² = 468

⇒From (1), a = A + 6

Now, substituting the value of equation (1) in equation (2), we get:

= (A+6)² + A² = 468

= A² + 12A + 36 +A² = 468

= 2A² + 6A + 18 = 264

= A² + 6A + 18 = 264

= A² +  6A + 18 -264 = 0

= A² + 6A - 216 = 0

The equation A² + 6A - 216 = 0 where:-

a = co-efficient of A² = 1

b = co-efficient of A = 6

c = constant term = -216

and , by using quadratic formula,

here ,  D =  b² - 4ac

= 6² - 4 * 1 * (-216)

= 36 + 864

=900

$now , y=\frac{-b+\sqrt{D} }{2a} = \frac{-6+\sqrt{900} }{2*1} \\\\= \frac{-6+30}{2} = \frac{24}{2} , \frac{-36}{2} = 12 , -18$

Now,  substituting the value of 'A' in equation (1),  we get :-

a = 6 + A

a = 6 + 12

a = 18

∴ the side of 1st square is 12m and the side of 2nd square is 18m.

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hope it helps :D

Answer 2

Pre-requisite knowledge:

The area of a square is defined as the product of the two sides of the square. It is also known as squares of sides. The unit of the area is in square units.

${\;Area = (side)^2 = side \times side}$

The perimeter of a square is equal to the sum of all sides. As we know, all sides of a square are equal so the perimeter is four times the length of any side of a square.

${\;Perimeter = 4 \times side}$

A quadratic equation in a variable $x$ is an equation which is of the form $ax^2 + bx + c = 0$ where constants $a$, $b$ and $c$ are all real numbers and $a \neq 0$.

There are three ways to solve a quadratic equation:

• Factoring
• Complete the Square
• Quadratic Formula

Factoring uses the logic that the product of any number and zero is zero.

Complete the square method uses square root, and the quadratic formula is the simpler method of it.

Quadratic Formula, $x=\dfrac{-b\pm\sqrt{b^2-4ac} }{2a}$

In this problem we'll use the quadratic formula to solve the required problem.

$\rule{300}{2}$

Solution:

Let the first side of the square be $a$ m and second side of the square be $A$ m.

We know that, the area of a square is the squares of two sides, $a^2$.

So, the area of the first square will be $a^2$ the perimeter of the first square will be $4a$ respectively.

Similarly, the area of the second square will be $A^2$ the perimeter of the second square will be $4A$ respectively.

As it is given that, the difference of their perimeters is 24 metres. Therefore our equation becomes:

$\implies 4a - 4A = 24$

$\implies a - A = 6$

$\implies a = 6 + A \qquad \textsf{------ (1)}$

Also, it is given that, Sum of areas of two squares is 468m². Therefore our equation becomes:

$\implies a^2 - A^2 = 468 \qquad \textsf{------ (2)}$

Now, substituting the value of equation (1) in equation (2), we get:

$\implies (A + 6)^2 + A^2 = 468$

$\implies A^2 + 12A + 36 + A^2 = 468$

$\implies 2A^2 + 12A + 36 = 468$

$\implies A^2 + 6A + 18 = 264$

$\implies A^2 + 6A + 18 - 264 = 0$

$\implies A^2 + 6A - 216 = 0$

Now, we can see that the above equation came in the form of a quadratic equation.

The equation ${A^2 + 6A - 216 = 0}$ where;

• a = co-efficient of A² = 1
• b = co-efficient of A = 6
• c = constant term = -216

Now we can use the quadratic formula to find the sides of the two squares.

The quadratic formula is given by,

$\;x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

By substituting the known values in the formula, we get the following results:

$\implies x=\dfrac{-6\pm\sqrt{6^2-4(1)(-216)}}{2(1)}$

$\implies A=\dfrac{-6\pm\sqrt{36-4(1)(-216)}}{2}$

$\implies A=\dfrac{-6\pm\sqrt{36+864}}{2}$

$\implies A=\dfrac{-6\pm\sqrt{900}}{2}$

$\implies A=\dfrac{-6\pm30}{2}$

$\implies A=\dfrac{-6+30}{2} \quad \bigg| \quad A=\dfrac{-6-30}{2}$

$\implies A=\dfrac{24}{2} \quad \bigg| \quad A= \dfrac{-36}{2}$

$\implies A=12 \quad \bigg| \quad A=-18$

Since, length cannot be negetive. So, taking $A=12$.

Now substituting the value of 'A' in equation (1), we get:

$\implies a = 6 + A$

$\implies a = 6 + 12$

$\implies a = 18$

Hence, the side of first square is 12m and the side of second square is 18m respectively.