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?

Question

Julia Adams · Accepted Answer

#"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)#. Spread the Joy of Maths.!

Isaiah Anthony · Answer

undefined The area of the isosceles triangle as a function of the length of the base $ b $ is given by: 
$$ A(b) = \frac{1}{2} \times b \times \sqrt{6 - \left(\frac{b}{2}\right)^2} $$