How can I prove that this expression defines the area of the quadrilateral?

Question

(I describe the problem below)

So, let's consider a quadratic function: $f(x)=-\frac{1}{2}x^{2}+x+4$ .

• Point $A$ is the intersection of the function with the $x$ axis and it has coordinates $(0,4)$ ;
• Point $B$ is the intersection of the function with the $y$ axis and it's coordinates are $(4,0)$ ;
• Point $P$ is a point in the parabola of $f(x)$ with coordinates $(x_{P}, y_{P})$ $x_{_{P}}\in \,]0, a[$ ($a$ is the $x$ coordinate of point $A$) .

Let's consider a quadrilateral with vertices $A$, $O$, $B$ and $P$ :

Let $g$ be the function that relates $x_{_{P}}$ with the area of the quadrilateral.

Prove that $g(x)=-x^{2}+4x+8$

Why is this the expression that gives us the area of the quadrilateral $[AOBP]$ ? How can I prove that it is? I've been trying to but I can't seem to be able to do it.

Can someone explain it to me please?

(I'm not supposed to use calculus to solve this. This is a 10th grade problem) (I'm freaking out)

Answer 1

We know that $x_p = (x,-\frac{1}{2}x^2 + x +4)$, so we divide the area in three polygons which area is evaluable

So for the rectangle we have $$A_1(x) = x \cdot \bigg( -\frac{1}{2}x^2 + x +4\bigg) = -\frac{1}{2}x^3 + x^2 +4x$$

The upper triangle $$A_2(x) = \frac{1}{2}x \cdot \bigg( 4 - \bigg[-\frac{1}{2}x^2 + x +4\bigg]\bigg) = \frac{1}{4}x^3 -\frac{1}{2}x^2$$

Side triangle $$A_3(x) = \frac{1}{2}(4-x) \cdot \bigg(-\frac{1}{2}x^2 + x +4\bigg) = \frac{1}{4}x^3 - \frac{3}{2}x^2 + 8$$ Total area is the sum of the three areas evaluated above so should result

$$A_{\text{TOT}}(x) = \sum_{i=1}^3A_i(x) = - x^2 + 4x +8$$

As desired!

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Worth mentioning as pointed out in the comment that my solution is valid for $y_p \leq 4$. In the case $y_p > 4 $, point $P$ is above $B$ so we still form a triangle which height is $$h(x) = \bigg(-\frac{1}{2}x^2 + x + 4\bigg) - 4 = -\frac{1}{2}x^2 + x$$

the area of the upper triangle is $$A'_2(x) = \frac{1}{2}x \cdot h(x) = -\frac{1}{4}x^3 + \frac{1}{2}x^2$$

while the rectangle its height is fixed to 4 and its base is $x$, summing everything as before

$$\begin{align*} A'_{\text{TOT}}(x) &= 4x + \bigg (\frac{1}{4}x^3 - \frac{3}{2}x^2 + 8\bigg)+ \bigg( -\frac{1}{4}x^3 + \frac{1}{2}x^2\bigg) \\ &= -x^2 +4x +8 \end{align*}$$

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P.S. Same result in the end, I should be more careful or at least justify as in the comments why solution works as fine!

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Even without the reasoning above the formula still holds in every case. Looking at the following picture should clear this

Here upper triangle's area should be subtracted instead of summed so the expression for $g(x)$ still holds. Hope this helps!

Answer 2

You are asking the result for $x\in]0; a[$.

As your notations are inconsistent with the schema I precise here the coordinates of the points that I shall use: $O(0,0), A(4, 0), B(0, 4), P(x, f(x)$. I shall add $H$ the projection of $P$ on x-axis: $H(x, 0)$.

The surface of $AOBP$ is the sum of the surfaces of $HOBP$ which is a right trapezoid, and $AHP$ which is a right triangle. It gives:

$$\begin{aligned} g(x)&=x\times\frac{f(0)+f(x)} 2+ \frac{(a-x)\times f(x)}2\\ &=\frac 1 2\left((4x+xf(x))+(4f(x)-xf(x))\right) = 2(f(x) + x)\\ &=2(-\frac{1}{2}x^{2}+x+4 + x)\\ &= -x^2+4x+8 \end{aligned}$$

Answer 3

We can use a tenth grade formula of area of a triangle here $$\Delta=\frac{1}{2}(x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2))$$ Divide your quadrilateral into $\triangle OBP$ and $\triangle OAP$ and then simply add the areas.

But your diagram and question is totally flawed. The function doesn't have real roots. Please make the necessary changes.

Answer 4

Firstly, The parabola shown neither represents the graph of $f(x)=1/2x^2+x+4$ or $g(x)=-x^2+4x+8$. Maybe I am missing something in your explanation.

To find the area of a quadrilateral like the one shown above, choose 2 vertices opposite each other and draw a line between them creating 2 triangles. Solve for the area of both triangles using the formula $A=1/2*b*h$. The sum of the area of both triangles will equal the area of the quadrilateral.

Drawing a line between point B & point C gives a right triangle, which means we can use the Pythagorean theorem to solve for $a$ (opposite of angle $A$). All that is left after doing this is measuring the distance from $b$ to $p$, & from $p$ to $c$ to have all side lengths of the triangles.