The Solution: In solving this problem, we first drew all of the shapes mentioned above. We then examined the accepted formulas for their areas. At first, we looked at methods of standardizing the shapes to fit a certain equation, but soon realized that that was not necessary at all - that indeed there was one formula that all of us already knew that would work for each of the shapes.
We developed a theory that the formula A=1/2h(b1+b2) would work for any of these shapes. We then went about testing our theory by first drawing the figures and giving measures to their sides then finding the area using A=1/2 h(b1+b2). Finally, we checked the area by finding it again using the commonly accepted formula. We found our theory to be true. These tests can be found at the end of the problem. The following is a shape-by-shape analysis of why this formula works and how.
We will start by examining the trapezoid, since its area formula is the one we have determined to be the formula that will find that area of any of the shapes. The area of a trapezoid can be determined by the formula A=1/2 h(b1+b2). To prove this we will follow the two column format. Please follow along with the supplementary figure 1 below.
Statements Reasons
1. ABCD is a trapezoid. 1. Given
2. AD || BC 2. Definition of a
trapezoid.
3. Let the height of ABCD be defined 3. Definition of height of
as the distance between AD and B a trapezoid.
4. Intro BD 4. Line Postulate
5. The h ABD = h BDC 5. Parallel lines are
everywhere equidistant.
6. Area ABD = 1/2 AD * h 6. Area of a triangle =
1/2 b*h
7. Area BDC = 1/2 BC * h 7. Area of a triangle =
1/2 b*h
8. Area ABCD = Area ABD + Area BDC 8. Area Addition Postulate
9. Area ABCD = (1/2 AD * h ) + 9. Substitution property
(1/2 BC * h)
10. Area ABCD = 1/2 * h (AD + BC) 10. Distributive Property
Since ABCD was created as a generic trapezoid with no specific restrictions and taking AD and BC as b1 and b2, we conclude that the area of any trapezoid=1/2 h(b1+b2).
This formula, Area=1/2 h(b1+b2), also applies to a parallelogram. For any parallelogram ABCD, AB || DC. Also, DC=AB and AD=BC. The distance between any two parallel segments, called the height, is everywhere equidistant. We know that the area of a parallelogram can be expressed as: Area=bh, since it is given that both bases are congruent. But, since the average of two equal lengths is another equal length, this can also be expressed as Area=1/2 h(b1+b2).
The formula A=1/2 h(b1+b2) will also work to find the area of a square or rectangle. The height can be any side and then the corresponding bases are used. The accepted formula for the area of these two figures is Area=bh. The new formula will work for the same reason it worked for the parallelogram - the adding of the second base compensated for the loss of dividing the height in half.
Using the formula Area=1/2 h(b1+b2), the area of a triangle can also be found. The accepted formula for the area of a triangle is Area=1/2 bh. To apply the new formula, the normal base and height are found, and then to find the second base, an imaginary line is drawn parallel to the first base and through the vertex of the triangle opposite the first base. The vertex along this new line is a point. We recall that a point has no height or length, and hence, no area, so the measure of the second base is taken to be zero. Therefore, because one of the bases is zero, the new formula becomes essentially the same as the accepted Area=1/2 bh, and the formula will work.
In conclusion, we have explained how and why the formula Area=1/2 h(b1+b2) will work to find the area of a trapezoid, a parallelogram, a square, a rectangle, and a triangle. Further proof of this can be found in the examples that follow.
In conclusion, the area of any of the mentioned polygons (parallelogram, trapezoid, triangle, rectangle, square) can be determined by the formula: A= 1/2h (b1+b2).
These Scholary Conclusions were drawn by : Meghan Von Geis, Sarah Clark, and Liz Federowicz of Shaler Area High School.