What is the Area of a Quadrilateral with Given Coordinates?

[Ilikebugs]

View attachment 6570 I think I got 4 points of (1,0) (3,5) (9,8) (13,0) but I don't know how to get the area

 

[MarkFL]

Using the vertex form for a quadratic, we may state:

\(\displaystyle y=-a(x-7)^2+9\)

To determine $a$, we may use the other known point:

\(\displaystyle 8=-a(9-7)^2+9\implies a=\frac{1}{4}\)

Hence:

\(\displaystyle y=-\frac{1}{4}(x-7)^2+9\)

To determine the $x$-coordinates of points $C$ and $D$, we solve:

\(\displaystyle 0=-\frac{1}{4}(x-7)^2+9\)

\(\displaystyle (x-7)^2=6^2\)

\(\displaystyle x-7=\pm6\)

\(\displaystyle x=7\pm6\)

To find $b$, we solve:

\(\displaystyle 5=-\frac{1}{4}(x-7)^2+9\)

\(\displaystyle (x-7)^2=4^2\)

Take the smaller root:

\(\displaystyle x=7-4=3\)

So, I agree with all of the coordinates you found. Now, how about we find the areas of \(\displaystyle \triangle{BCD}\) and \(\displaystyle \triangle{ABD}\) and add them together.

Or, you can drop vertical lines down from points $A$ and $B$ to the $x$-axis, and draw the segment $\overline{AB}$, and you have two right triangles and a trapezoid.

Can you proceed?

 

[Ilikebugs]

(12*5)/2=30

()/2=30

30+30=60?

 

[MarkFL]

(12*5)/2=30

()/2=30

30+30=60?

That's different than the value I got using the second method I suggested (which I found simpler to do).

 

[Ilikebugs]

T1=13.5
T2=16
Trapezoid=height 8, b1= 6 b2=sqr(45) 1/2(6+sqr(45))*8= 24+4(sqr(45))

43.5+4(sqr(45))?

I got 60 through View attachment 6571

 

[MarkFL]

Yeah...I used a wrong value in one of the computations. It is 60.