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Question:

Exercise

Solution provided:

Solution

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 ?

Alternate

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4 Answers 4

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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 :

Area

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 ]]

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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.

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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 $a, a, b$?

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

enter image description here

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

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

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