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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