How do i find the surface of the triangle?(with a graphic and a x,y coordinate table)

Question

How do i find the surface of the triangle ASB?(with a graphic and a x,y coordinate table)

Answer 1

From the tabled coordinates, the equation of the parabola is obtained as

$$y=-(x-3)^2+4$$

Then, the vertexes are $A(1,0)$, $B(4,3)$ and $S(3,4)$. Apply the triangle area formula with coordinates below

$$\begin{align} Area &= \frac12|x_A(y_B-y_S) + x_B(y_S-y_A) + x_S(y_A-y_B)|\\ &= \frac12|1(3-4) + 4(4-0) + 3(0-3)| =3 \end{align}$$

Answer 2

First you have to determine the equation of the parabola, and based on that you determine the points A = (a, 0) and B = (4, b). I suppose that S = (3,4).

Answer 3

1. Rule of the function: $$f(x)=-x^2+6x-5$$

2. Find coordinates of A, B, S:

For S:

$$f'(x)=-2x+6$$ $$0=-2x+6$$ $$x=3$$

By replacing to get y, we get: $$y=4$$

so, $S=(3,4)$

For A:

By plugging in $y=0$:

$$0=-x^2+6x-5$$ $$x_1=1, x_2=5$$

We choose x=1 in this case because of A's position on the graph

so, $A=(1,0)$

For B:

We know x=4, so $$f(4)=-16+24-5$$ $$y=3$$

so, $B=(4,3)$

3. Find the measurement of the segments:

$$(AB)^2=3^2+3^2$$ $$AB=3\sqrt2$$

$$ $$ $$(SB)^2=1^2+1^2$$ $$SB=\sqrt2$$

$$ $$ $$(SA)^2=2^2+4^2$$ $$SA=2\sqrt5$$

4. Using Heron's law:

$$A=\sqrt{s(s-SA)(s-SB)(s-AB)}$$

where s = half the perimeter of the triangle

$$A=\sqrt{5(5-2\sqrt5)(5-\sqrt2)(5-3\sqrt2)}$$

so, $$A=2.7u^2$$