- #1
Ilikebugs
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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
What is the Area of a Quadrilateral with Given Coordinates?
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In summary, the conversation discussed finding the area of two triangles and a trapezoid using the vertex form of a quadratic equation. The coordinates (1,0), (3,5), (9,8), and (13,0) were used to determine the equation y=-1/4(x-7)^2+9. The areas of the triangles and trapezoid were then calculated and found to be 60.
- #2
MarkFL
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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?
\(\displaystyle y=-a(x-7)^2+9\)
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?
- #3
Ilikebugs
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(12*5)/2=30
()/2=30
30+30=60?
()/2=30
30+30=60?
- #4
MarkFL
Gold Member
MHB
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Ilikebugs said:
That's different than the value I got using the second method I suggested (which I found simpler to do).
- #5
Ilikebugs
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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
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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- #6
MarkFL
Gold Member
MHB
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Yeah...I used a wrong value in one of the computations. It is 60.
FAQ: What is the Area of a Quadrilateral with Given Coordinates?
What is the formula for finding the area of a quadrilateral?
The formula for finding the area of a quadrilateral varies depending on the type of quadrilateral. For a rectangle, the formula is length x width. For a square, it is side length squared. For a parallelogram, it is base x height. For a trapezoid, it is 1/2 x (sum of parallel sides) x height. For a general quadrilateral, the formula is (1/2 x (product of diagonals) x sin(theta)), where theta is the angle formed by the diagonals.
Can you use the same formula to find the area of any quadrilateral?
No, the formula for finding the area of a quadrilateral varies depending on the type of quadrilateral. For example, a trapezoid has a different formula than a square. It is important to identify the type of quadrilateral before using the appropriate formula to find its area.
What is the difference between a convex and a concave quadrilateral?
A convex quadrilateral has all angles less than 180 degrees, while a concave quadrilateral has at least one angle greater than 180 degrees. In other words, a convex quadrilateral has no indentations or "dips" in its shape, while a concave quadrilateral does.
How do you find the area of an irregular quadrilateral?
An irregular quadrilateral is a quadrilateral with no specific shape or symmetry. To find its area, you can divide it into smaller shapes that you know how to find the area of, such as triangles or rectangles. Then, you can find the sum of the areas of these smaller shapes to find the area of the irregular quadrilateral. Alternatively, you can use the shoelace formula, which involves plotting the coordinates of the vertices and using a specific formula to find the area.
What are some real-world applications of finding the area of a quadrilateral?
Finding the area of a quadrilateral is useful in many fields, such as engineering, architecture, and construction. The area of a quadrilateral can also be used to determine the amount of material needed to cover a specific surface or to calculate the cost of a project. In addition, understanding the area of a quadrilateral is important in fields such as surveying and geography, where accurate measurements of land and maps are necessary.