Prove the following using proof by contrapositive, proof by contradiction or an existence proof (a) if the gcd(a, b)1, then at most one of a or b is even (b) there are no positive real roots of 2 4x +5 (c) if r and s are both positive real numbers, then + st vf+ vs. (d) the only three consecutive odd natural numbers that are all prime are 3, 5, and (e) v5 is not a rational number (0) if r is a rectangle of area a, then there exists a square with the same area.

The required value of xx is \boxed{\bf x=6.2}

x=6.2

Step-by-step explanation:

Consider the length and breadth of the rectangle as LL and BB respectively.

The expression which represent the length of the rectangle is as follows:

L=\left(x-3\right)\text{m}L=(x−3)m

The expression which represent the breadth of the rectangle is as follows:

B=\left(x-6\right)\text{m}B=(x−6)m

The expression which represent the side of the square is as follows:

\text{Side}=\left(x-7\right)Side=(x−7)

Area of a rectangle is calculated as shown below:

\boxed{\text{Area}=\text{Length}\times \text{Breadth}}

Area=Length×Breadth

To obtain the expression for the area of the rectangle substitute the expression of length and breadth is the above equation as shown below:

\begin{gathered}\begin{aligned}\text{Area of rectangle}&=\left(x-6\right)\times \left(x-3\right)\\&=x^{2}-9x+18\end{aligned}\end{gathered}

Area of rectangle

=(x−6)×(x−3)

=x

2

−9x+18

The formula to calculate the area of a square is as follows:

\boxed{\text{Area of square}=\left(\text{Side}\right)^{2}}

Area of square=(Side)

2

I hope that this'll help you!

Calculate the area of the square as shown below:

\begin{gathered}\begin{aligned}\text{Area of square}&=\left(x-7\right)^{2}\\&=x^{2}-14x+49\end{aligned}\end{gathered}

Area of square

=(x−7)

2

=x

2

−14x+49

It is given that the area of the rectangle is 24\text{m}^{2}24m

2

greater than the area of the square.

The equation formed as per the above statement is as follows:

\begin{gathered}\begin{aligned}\text{Area of rectangle}&=\text{Area of square}+24\\x^{2}-9x+18&=x^{2}-14x+49\\14x-9x&=49-18\\5x&=31\\x&=\dfrac{31}{5}\\x&=6.2\end{aligned}\end{gathered}

Area of rectangle

x

2

−9x+18

14x−9x

5x

x

x

=Area of square+24

=x

2

−14x+49

=49−18

=31

=

5

31

=6.2

Therefore, the required value of xx is \boxed{\bf x=6.2}

x=6.2

.