An isosceles triangle has its base along the x-axis with one base vertex at the origin and its vertex in the first quadrant on the graph of $y = 6-x^2$ . How do you write the area of the triangle as a function of length of the base?

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Answer 1 · 2017-01-16T16:39:36

$"Area, expressed as the function of base-length "l," is "1/8l(24-l^2)$ .

We consider an Isosceles Triangle $Delta ABC$ with base vertices

$B(0,0) and C(l,0), (l >0) in" the X-axis="{(x,0)| x in RR}$

and the third vertex

$A in G={(x,y)|y=6-x^2; x,y in RR} sub Q_I......(star)$ .

Obviously, the length $BC$ of the base is $l$ .

Let $M$ be the mid-point of the base $BC. :. M(l/2,0)$ .

$Delta ABC" is isosceles, ":. AM bot BC.$ So, if $h$ is the height of

$Delta ABC," then," because, BC$ is the X-Axis, $AM=h.$

Clearly, $A=A(l/2,h).$

Now, the Area of $Delta ABC=1/2lh...................(ast)$

But, $A(l/2,h) in G:. (star) rArr h=6-l^2/4$

Therefore, the Area of $Delta ABC=1/2l(6-l^2/4)...[because, (ast)],$ or,

$"Area, expressed as the function of base-length "l," is "1/8l(24-l^2)$ .

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An isosceles triangle has its base along the x-axis with one base vertex at the origin and its vertex in the first quadrant on the graph of y = 6-x^2y = 6-x^2. How do you write the area of the triangle as a function of length of the base?