BC = 18.385
x = 7
The figure given is a square.
In a square, all the sides are equal i.e.
AB = BC = DC = AD
The diagonals of a square also equal. i.e.
AC = BD
We have been given AC = 26 in the question, therefore,
AC = BD = 26
The diagonals of the square divide the square up into right-angled triangles. The triangle we shall consider is BDC because it contains the side we will like to calculate (BC)
Since BDC is a right-angled triangle, we can use the Pythagoras theorem to solve for
BC.
Pythagoras theorem for this triangle is:
[tex]\begin{gathered} BD^2=BC^2+CD^2 \\ \text{but we know that BC = CD = x (since it is a square)} \\ BD=26\text{ (Since it is a diagonal)} \\ \\ 26^2=x^2+x^2 \\ 676=2x^2 \\ \text{Divide both sides by 2} \\ \\ 338=x^2 \\ Get\text{ the square root of both sides} \\ \\ \sqrt[]{x^2}=\sqrt[]{338} \\ \therefore x=18.385 \end{gathered}[/tex]
Thus, if x = 18.385, then,
BC = 18.385 (From the diagram drawn above)
Now to get x:
The diagonals of a Square split the angle at the square into two equal parts.
The angles at the vertices of a square are 90 degrees each
Since the angles are 90 degrees each and the diagonals bisect the angles, then:
[tex]<\text{ACB}=\frac{90^0}{2}=45^0\text{ (As se}en\text{ on the diagram)}[/tex]
The question states that:
And we just figured out that
[tex]\begin{gathered} 45^0=11x-32 \\ \text{add 32 to both sides} \\ 45+32=11x-32+32 \\ 77=11x \\ \\ \text{divide both sides by 11} \\ x=\frac{77}{11}=7 \end{gathered}[/tex]
Therefore, x = 7
