MHB What is the Area of a Quadrilateral with Given Coordinates?

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The discussion focuses on calculating the area of a quadrilateral defined by the coordinates (1,0), (3,5), (9,8), and (13,0). Participants use the vertex form of a quadratic equation to derive necessary points and then explore different methods to compute the area. They suggest breaking the quadrilateral into triangles and trapezoids for easier calculation. The area calculations yield a total of 60, although discrepancies arise in intermediate steps. Ultimately, the correct area of the quadrilateral is confirmed to be 60.
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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
 

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

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

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

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

Hence:

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

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

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

$$(x-7)^2=6^2$$

$$x-7=\pm6$$

$$x=7\pm6$$

To find $b$, we solve:

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

$$(x-7)^2=4^2$$

Take the smaller root:

$$x=7-4=3$$

So, I agree with all of the coordinates you found. Now, how about we find the areas of $$\triangle{BCD}$$ and $$\triangle{ABD}$$ 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?
 
(12*5)/2=30

()/2=30

30+30=60?
 
Ilikebugs said:
(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).
 
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
 

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