Question

The base of an isosceles triangle is parallel to x-axis and its end-points belong to the parabola, y=x(10−x) . The vertex of the triangle belongs to the x-axis. Find area of the triangle if the length of its base is 8 units.

Answer 1

36

Explanation:

that is the right answer

Answer 2

so, the triangle is an isosceles, two sides are equal, parallel to the x-axis simply means one side is horizontal, say the base, and a vertex of it touches the x-axis. Check the picture below, the vertex is much higher btw, is at 5, 25, that's just a sketch, and since the function is y = x(10-x), if you zero it out, you get the x-intercepts of 0 and 10.

so.. hmmm if the base is 8 units long, that means, the first point is say at "x", and the second point will then be 8 units later or "x + 8".

Now, when at those x-coordinates, since the base is horizontal, then the value of "y"is the same, so is the same exact value at f(x) as it'll be at f(x+8), namely the "rise" is 0.

so, we know what f(x) is, hmm let's check what f(x+8) is then.


`\bf \stackrel{y}{f(x)}=x(10-x)\implies \boxed{f(x)=-x^2+10x}\\\\ -------------------------------\\\\ f(x+8)=-(x^2+16x+64)+10x^2+80x \\\\\\ f(x+8)=-x^2-16x-64+10x^2+80x \\\\\\ \boxed{f(x+8)=9x^2+64x-64}\\\\ -------------------------------\\\\ f(x)=f(x+8)\qquad thus\qquad -x^2+10x~=~9x^2+64x-64 \\\\\\ 0=10x^2+54x-64\implies 0=5x^2+27x-32`


`\bf \textit{using the quadratic formula} \\\\\\ x=\cfrac{-27\pm√(27^2-4(5)(-32))}{2(5)}\implies x=\cfrac{-27\pm√(729+640)}{10} \\\\\\ x=\cfrac{-27\pm 37}{10}\implies x= \begin{cases} \boxed{1}\\\\ -(32)/(5) \end{cases}`

since we're on the 1st quadrant, it can't be a negative value, so x = 1, well, what's f(1) then? well is 9.


`\bf A=\cfrac{1}{2}(base)(height)\quad \begin{cases} base=8\\ height=9 \end{cases}\implies A=\cfrac{1}{2}(8)(9)\implies A=36`