# I need help in understanding the alternative solution provided to solve this geometry question of calculating area of quadrilateral

Question:

Solution provided:

I understand this part that equal chords of a circle subtend equal angles at the center, but after this the faculty transformed this whole diagram to one shown below in the second picture.

I am not able to understand how did the two sides get interchanged here ?

Snapshot was taken at the end of online classes , where we do not know :

• what words the faculty spoke
• what order the red text was written in
• what additional text was written & summarized & removed Hence we can not make out all the Details of the Alternate Solution given.

"how did the two sides get interchanged here ?" : that can be answered easily.

Consider this Circle :

No matter where we make the angle $$\theta$$ , we can see that the Blue Areas will be Same , the Purple Areas will be Same & the Combined Blue+Purple Areas [ Sectors ] will be Same.

In Exercise , $$DOC$$ makes angle $$\alpha$$ : we can rotate that angle downwards such that $$DO$$ lines on the Original $$AO$$. Naturally , $$CO$$ would have to rotate too , while the angle $$\alpha$$ would not change.
In the New Position , the Area must be Same as before.
Like-wise , we can rotate angle $$\beta$$ [ $$AOD$$ ] upwards such that $$DO$$ lies on the Original $$CO$$ , where $$AO$$ would have to rotate too , while angle $$\beta$$ would not change.
In the New Position , the Area must be Same as before.
Now it will look like the Alternate Solution Image given.
That is due to $$\beta+\alpha+\alpha$$ (Original layout) matching $$\alpha+\beta+\alpha$$ (New layout) , with no change in Area.

Technically , that Image is wrong , because it is re-using the Letters $$ABCD$$
It should use new letters. At minimum , the Diagram should mark the Quadrilateral like $$ABDC$$ [ not $$ABCD$$ ] to indicate the rotational changes , which is confusing OP.
[[ Possibility is there that faculty might have spoken some words about this , though that is lost now & not included in the Diagram ]]

Let $$r$$ denote the radius. The area of the original quadrilateral in your picture is

$$A = \frac {1}{2} {r}^{2} \sin \left( \beta \right) + \frac {1}{2} {r}^{2} \sin \left( \alpha \right) + \frac {1}{2} {r}^{2} \sin \left( \alpha \right).$$

The area of the new quadrilateral is

$$A' = \frac {1}{2} {r}^{2} \sin \left( \alpha \right) + \frac {1}{2} {r}^{2} \sin \left( \beta \right) + \frac {1}{2} {r}^{2} \sin \left( \alpha \right).$$

Obviously, $$A = A'$$. I hope this helps.

So here's what happened. The first diagram tells us that the quadrilateral may be dissected into three isosceles triangles, two of which are congruent because they share a base length $$2 \sqrt{13}$$, and all of the legs are equal to the semicircle's radius, $$r$$. I believe this part should be fairly clear.

Where things get interesting is the second image, where it seems like a totally different quadrilateral is drawn. The missing insight is that the solution recognizes that the triangles in the dissection can be rearranged to form a trapezoid by interchanging $$\triangle AOD$$ and $$\triangle DOC$$. This doesn't change the total area of the quadrilateral, but it does change the location of point $$D$$ on the circumference. Unlike the lecturer, I will instead label this new point $$D'$$ so as to avoid confusion. Since the positions of $$A$$, $$B$$, $$C$$, and $$O$$ did not change, we can keep those.

Next, we can see that $$D'C \parallel AB$$ because $$AD' = BC$$ and $$\angle OAD' \cong \angle OBC$$. So $$ABCD'$$ is an isosceles trapezoid; as such, its area can be found if we know the height $$h$$. To obtain this, the solution employs two facts:

1. The Pythagorean theorem: $$(AE)^2 + (D'E)^2 = (AD')^2$$, where $$E$$ is the foot of the perpendicular from $$D'$$ to diameter $$AB$$.
2. The geometric mean in a semicircle: For any point $$D'$$ on the semicircle, the foot of the perpendicular $$E$$ divides diameter $$AB$$ into two segments $$AE$$, $$EB$$ such that $$D'E$$ is their geometric mean; i.e., $$D'E = \sqrt{(AE)(EB)}.$$

After labeling $$AE = x$$ and $$D'E = h$$, we have from the first fact $$x^2 + h^2 = (2 \sqrt{13})^2 = 52. \tag{1}$$ The second fact requires one more step to use, which is to observe that because $$ABCD$$ is isosceles, the corresponding foot $$F$$ of the perpendicular from $$C$$ divides diameter $$AB$$ in the same way as $$E$$ does, just mirrored. That is to say, $$BF = AE = x$$; moreover, $$EF = D'C = 5$$, because drawing these altitudes $$D'E$$ and $$CF$$ in an isosceles trapezoid creates a rectangle $$D'CFE$$. Consequently, if $$AE = x$$, then $$BE = 5 + x$$, and now we can use the second fact to assert $$h^2 = x(5+x). \tag{2}$$ Equations $$(1)$$ and $$(2)$$ comprise a system of two equations in two unknowns which we can solve via substitution: $$(2)$$ into $$(1)$$ gives $$x^2 + x(5+x) = 52, \tag{3}$$ and solving this quadratic gives the unique positive solution $$x = 4$$. Then substituting this back into, $$(2)$$ gives $$h^2 = 4(5+4) = 36, \tag{4}$$ hence $$h = 6$$, and the desired area is $$|ABCD| = |ABCD'| = \frac{1}{2}h(AB + CD) = \frac{1}{2}(6)(4+5+4+5) = 54.$$

The elegance in this solution lies in the fact that it does not require any knowledge of trigonometry, and that it leverages the rearrangement of a dissection to compute the area of an figure that has a simpler area formula than the original figure. As an exercise for the reader, can you generalize this method to find the area of a quadrilateral similarly inscribed, but with side lengths $$a, a, b$$?

Because that quadrilateral is cyclic (inscribed in a circle), you can use Brahmagupta's Formula:

Where a, b, c, and d are the side lengths of the cyclic quadrilateral.

• You only know three of the sides - you need to find the diameter (or radius) of the circle. Commented Jun 11 at 20:55