What is the area of a triangle with a base of 4v^3 and a height of vu^2?

1 answer:

3 0

Answer: The area of the triangle is: " 2v⁴u² square units" ;
or, write as: " 2v⁴u² units² " .
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<u>Note</u>: The formula/equation for the area, "A", of a triangle is:

A = (1/2) * b * h ; or: A = (b * h) / 2 ;
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in which: "A" = "area" ;
"b" = "base length" ;
"h" = "[perpendicular] height" ;
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We are asked to solve for the area, "A" .

We are given: "b = 4v³ " ; "h = vu² " ;

So; to find the area, "A", of the triangle, we plug these given values into the formula; and solve/ simplify:
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A = (1/2) * b * h ;

A = (1/2) * (4v³) * (vu²) ;

A = 2v³ * vu² ;

A = 2v⁴u² square units.

Answer: The area of the triangle is: " 2v⁴u² square units" ;
or, write as: " 2v⁴u² units² " .
___________________________________________________

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$\bf \textit{parabola vertex form with focus point distance}\\\\
\begin{array}{llll}
4p(x- h)=(y- k)^2
\\\\
\boxed{4p(y- k)=(x- h)^2}
\end{array}
\qquad 
\begin{array}{llll}
vertex\ ( h, k)\\\\
 p=\textit{distance from vertex to }\\
\qquad \textit{ focus or directrix}
\end{array}\\\\
-------------------------------\\\\
\begin{cases}
h=1\\
k=2\\
p=1
\end{cases}\implies 4(1)(y-2)=(x-1)^2\implies 4(y-2)=(x-1)^2
\\\\\\
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