Question

An arch is in the form of semi-ellipse whose span is 48m wide. The height of the arch is 20m. Then the width of the arch at a

height of 10m above the base is​

Answer 1

Answer:

An arch is in the form of a semi ellipse. It is 50 metres wide at the base and has a height of 20 metres. How wide is the arch at the height of 18 metres above the base?

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x^2/25^2 + y^2/20^2 = 1

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x^2/625 + y^2/400 = 1

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Solve for "x" if y = 18

x^2/625 + 18^2/400 = 1

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x^2/625 = 1-0.81

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x^2/625 = 0.19

x^2 = 118.75

x = 10.89 ft.

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Answer 2

Given,

The length of the span in meters$\[ = 48\]$

The height of the arch in meters$\[ = 20\]$

Solution,

Assume that the arch is a semi-ellipse.

Assume that the width is y meters.

So,

$\[\begin{array}{l} \Rightarrow \frac{{{x^2}}}{{{{\left( {25} \right)}^2}}} + \frac{{{y^2}}}{{{{\left( {20} \right)}^2}}} = 1\\ \Rightarrow \frac{{{x^2}}}{{625}} + \frac{{{y^2}}}{{400}} = 1\end{array}\]$

Apply $\[y = 18m.\]$

$\[\begin{array}{l} \Rightarrow \frac{{{x^2}}}{{625}} + \frac{{{{\left( {18} \right)}^2}}}{{400}} = 1\\ \Rightarrow \frac{{{x^2}}}{{625}} + \frac{{324}}{{400}} = 1\\ \Rightarrow \frac{{{x^2}}}{{625}} = 1 - \frac{{324}}{{400}}\end{array}\]$

Solve further,

$\[\begin{array}{l} \Rightarrow \frac{{{x^2}}}{{625}} = \frac{{76}}{{400}}\\ \Rightarrow x = \sqrt {\frac{{76 \times 625}}{{400}}} \\ \Rightarrow x = 10.9\end{array}\]$

Hence, the width is $\[10.9\]$ meters.