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# An arch is in the shape of a parabola whose axis is vertically downward and measures $80\,m$ across its bottom ground. Its height is $24\,m$. The measure of the horizontal beam across its cross section at a height of $18\,m$is:$a)\,50 \\ b)\,40 \\ c)\,45 \\ d)\,60 \\$

Last updated date: 03rd Aug 2024
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Hint:In this question we will try to make a figure according to the given information and determine the points on parabola at height $18\,m$ where the cross section of beam measures .Then we will write the equation of the parabola. And with the help of equations we will find the answer.

Let equation of the given vertically downward parabola be ${x^2} = - 4ay\,\,\,\,\,\,\,\,\,\,\,\,\, \to (1)$
Given the measure of bottom on ground is $80\,m$
And the highest point is $24\,m$
So we can clearly now find the points A and B‘s coordinates.
$A \equiv ( - 40, - 24)\,\,\,\,\& \,\,\,B \equiv (40, - 24)$
And both of these points lie on parabola so A must satisfy (1)
${(40)^2} = - 4a( - 24) \\ 1600 = 96a \\ a = \dfrac{{50}}{3}\,\,\,\,\,\,\,\,\,\,\, \to (2) \\$
Now putting this value in (1)
${x^2} = - 4\left( {\dfrac{{50}}{3}} \right)y \\ 3{x^2} = - 200y\,\,\,\,\,\,\,\, \to (3) \\$
Now we need to find the value of $x$ at $y = - 6$ by putting it in (3)
$3{x^2} = - 200( - 6) \\ 3{x^2} = 1200 \\ {x^2} = 400 \\ x = \pm 20 \\$
Now coordinates of C becomes $( - 20, - 6)$ and D becomes $(20, - 6)$.
Now distance between C and D is,
$CD = \sqrt {{{(20 - ( - 20))}^2} + {{( - 6 - ( - 6))}^2}} \\ CD = \sqrt {{{40}^2} + {0^2}} \\ CD = 40 \\$
So the horizontal beam across its cross section at a height of $18\,m$ is of measure $40\,m$.

So, the correct answer is “Option B”.

Note:The tricky part in the question is that we need to find the measure of height of the horizontal beam across its cross section at a height of $18\,m$. Generally we take $18\,m$ as the ordinate but actually by seeing the figure we can clearly see that $18\,m$ is not ordinate and we need to subtract $24\,m$ from it.