I have an isosceles trapezoid, with a semicircle in the middle. I need to know the difference in area of the two shapes. Radius of the semicircle is $6$cm, and the longest base is $14$cm.
[Math] Isosceles trapezoid with semicircle
geometry
Related Solutions
Your claim: $a=b+c$ is not true. Since your trapezoid is isosceles, the two lower angles are the same. Therefore, the two diagonals are equal because the lower triangle each creates when dividing the trapezoid are congruent (by SAS). Then, by the Pythagorean theorem, as long as the two perpendiculars are not collinear, $a>b+c$:
Only when the two perpendiculars are collinear will $a=b+c$, but then your isosceles trapezoid will be a square. (As noted by Issac, the two perpendiculars can be collinear with out the isosceles trapezoid being a square.)
Clearly the area of the colored region is 3 times the area of one of the curved triangles.
Draw a regular hexagon around one of the circles, in such a way that all of its sides are tangent to the circle (i.e., the circle is inscribed in the hexagon). There are 6 regions of the hexagon that are not contained in the circle, one at each vertex of the hexagon:
Note that any one of the curved triangles in your picture consists of 3 of these hexagon corners.
Let $A$ be the area of a hexagon with inradius 4. Let $B$ be the area of a circle with radius 4. The total area of 6 hexagon corner regions is $A-B$. The region we want to find is made of 3 curved triangles, each of which is made of 3 hexagon corners, for a total of 9 hexagon corners. Thus, the area of the colored region is $\frac{3}{2}(A-B)$.
Best Answer
Assuming that the base of the circle lies along the longest base, here is a way to find the length of the shorter base (which along with the length of the longer base, 14, and the height of the trapezoid, 6, readily yields its area):
Label the vertices of the trapezoid ABCD, with AD being the longer base of length 14. Let the smaller of the bases have length $x$, and let the center of the circle be at $O$.
Area of trapezoid= $$\frac{6}{2}(x+14)=3x+42$$ by the area of a trapezoid formula. We can calculate this in a different manner by doing area=
$$A_{OAB}+A_{OBC}+A_{OCA}=\frac{1}{2}(6AB+6*x+6*CD)=6AB+3x$$
Therefore, $$3x+42=6AB+3x\Leftrightarrow AB=7$$
Drop the perpendicular Q from B to AD and applying Pythagorean Theorem, we get $$AB^2=AQ^2+BQ^2 \Leftrightarrow 49=AQ^2+36 \Leftrightarrow AQ=\sqrt{13}$$
By symmetry, $$AD=2AQ+BC$$, which gives $$BC=14-2\sqrt{13}$$
Cheers,
Rofler