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2 years ago

Find the area of an equilateral triangle with 12-inch altitudes

1 answer:

4 0

$\textit{height of an equilateral triangle}\\\\ h=\cfrac{s\sqrt{3}}{2}~~ \begin{cases} s=\stackrel{side's}{length}\\[-0.5em] \hrulefill\\ h=12 \end{cases}\implies 12=\cfrac{s\sqrt{3}}{2}\implies 24=s\sqrt{3}\implies \cfrac{24}{\sqrt{3}}=s \\\\[-0.35em] ~\dotfill$

$\textit{area of an equilateral triangle}\\\\ A=\cfrac{s^2\sqrt{3}}{4}\qquad \qquad A=\cfrac{~~ \left( \frac{24}{\sqrt{3}} \right)^2 \sqrt{3}~~}{2}\implies A=\cfrac{~~ \frac{24^2}{3} \sqrt{3}~~}{2} \\\\\\ A=\cfrac{192\sqrt{3}}{2}\implies A=96\sqrt{3}\implies A\approx 166.28$

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3 years ago

Read 2 more answers

Texas University Quarterback, Michael Daniels, is standing on his own 10-yard line. He throws a pass tword the opposite goal lin

dimulka [17.4K]

Answer: The function is $y=-\frac{28}{625}(x-35)^2+30$ .

Explanation:

It is given that Michael Daniels, is standing on his own 10-yard line. He throws a pass toward the opposite goal line. The football is 2 yards above the ground when the quarterback lets it go.

It follows a parabolic path. Reaching its highest point, 30 yards above the ground. It is caught 50 yards downfield at a point 2 yards above the ground.

So, the initial point is (10,2) and the other point is (60,2).

The height function of a football represents a downward parabola. The maximum point of the function is called vertex. So the vertex is (h,30).

The two point (10,2) and (60,2) have same y-coordinate, therefore the function is maximum at the midpoint of both points.

$midpoint=(\frac{10+60}{2}, \frac{2+2}{2})=(35,2)$

So, the function is maximum at x=35. Hence the vertex is (35,30)

The standard form of the parabola is,

$y=a(x-h)^2+k$

Where (h,k) is vertex and a is scale factor.

Since vertex is (35,30).

$y=a(x-35)^2+30$

The initial point is (10,2).

$2=a(10-35)^2+30$

$-28=625a$

$a=-\frac{28}{625}$

So the function of height is.

$y=-\frac{28}{625}(x-35)^2+30$

The graph of the function is shown below.

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3 years ago

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