An arch is in the shape of a parabola whose axis is vertically downwords and measures 80 mts across its bottom on the ground. Its highest point is 24 mts. The measure of the horizontal beam across its cross section at a height of 18 mts is

Let equation of the given vertically downward parabola be $x^2$=−4ay→(1)
Given the measure of bottom on ground is 80m
And the highest point is 24m
So coordinates of A and B are A≡(−40,−24)&B≡(40,−24)

Since both the points lie on parabola then

$$\begin{gathered} {(-40)}^2=-4a(-24) \\ \Rightarrow 1600=96a \\ \Rightarrow a=\frac{1600}{96} \\ \Rightarrow a=\frac{50}{3} \\ \left(1\right) becomes x^2=-\frac{200}{3}y \\ we need to find the value of x when y=-6 \\ \Rightarrow x^2=-\frac{200}{3}(-6) \\ \Rightarrow x^2=400 \\ \Rightarrow x=\pm 20 \\ coordinates of C and D are \\ C=(-20,-6) and D=(20,-6) \\ Now CD=\sqrt{{(-20-20)}^2+0^2} \\ =\sqrt{1600} \\ =40 \end{gathered}$$