Proof: prove area of shaded lune equals area of triangle

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Faded-Maximus

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The right angle triangle has legs of length r and a right angle at O. A quarter of a circle is consturcted with the centre O and a radius of r. A semicircle is construted with a diameter of AB. Prove that the area of the shaded lune is equal to the area of triangle AOB.


Area of the triangle: (bh) / 2 = (r^2) / 2

Area of quarter circle: (pi r^2) / 4

Area of semi circle: (pi r^2) / 2

To get the area of the little lune you would subtract the area of the quarter circle by the triangle:

[(pi r^2) /4] - [(r^2) / 2]

. . .= [(pi r^2) / 4] - [(2r^2) / 4]

. . .= (pi - r^2) / 4

To get the area of the big lune (shaded area) you would subtract the area of the semicircle by the area of the little lune.

[(pir^2) / 2] - [(pi - r^2) / 4]

. . .= [2(pir^2) / 4] - [(pi - r^2) / 4]

. . .= (pi + 3r^2) / 4

The area of the trinagle does not equal the area of the shaded region. I have made a mistake somewhere, but am unsure where. Can somebody please help me out? Thanks.
 
I get the following
\(\displaystyle \L\frac{{r^2 \pi }}{4} - \left( {\frac{{r^2 \pi }}{4} - \frac{{r^2 }}{2}} \right).\)

That is the area of the dark region.
 
Can you please explain it? I don't understand how you got r^2 pi / 4 for the area of the semicircle.
 
Faded-Maximus said:
I don't understand how you got r^2 pi / 4 for the area of the semicircle.
Click to expand...
Which \(\displaystyle \frac{{r^2 \pi }}{4}\) don't you get?
There are two of them. I grouped the parts that go together.
 
Use the Pythagorean Theorem to find the length of the line segment AB. Then use this as the diameter of the half-circle.

Eliz.
 
The second one is one-quarter of the area of the circle.
The first one is the area of the semicircle.
 
I understand that the first one is suppose to be the area of a semicircle.
the area of a circle is pir^2
since it is a semi circle we would need to divide that by 2.
therefore it would be pir^2 / 2.

To find out the diameter you would use pythagorean.
By doing this you would have the diameter be sqrt 2r^2.
You would divide this by two to get the radius.

Something from here isn't clicking for me because I don't know how to make that equal to r^2pi / 4.
 
The radius of the semicircle is \(\displaystyle r\sqrt 2 .\)
 
I am so confused with this.

Would r sqrt 2 not be the diameter?

AO^2 + OB^2 = AB^2
AO = r and OB = r
thus, r^2 + r^2 = AB^2
2r^2 = AB^2
sqrt both sides
r sqrt2 = AB

then the radius would be 1/2 r sqrt 2

A = pi (1/2 r sqrt2)^2 / 2
A = pi 1/4 r^2 (2) / 2
A = 2pi r^2 / 8
A = pi r^2 / 4
is that right?
 
Relax, Maxim.

Diameter of semicircle = r sqrt(2); you got that correct; pka simply made a typo.

So radius of semicircle = r sqrt(2) / 2
(P = pi)
So area semicircle = P [r sqrt(2) / 2]^2 / 2 = P r^2 / 4 [1]

Area of quartercircle = P r^2 / 4
Area of triangle = r^2 / 2
So area of unshaded circle segment = P r^2 / 4 - r^2 / 2 = r^2(P - 2) / 4 [2]

So shaded area = [1] - [2] = r^2 / 2 = area of triangle

Can you get some sleep now :shock:
 
Hello, Faded-Maximus!

I found your error . . .

The right angle triangle has legs of length \(\displaystyle r\) and a right angle at \(\displaystyle O\).
A quarter of a circle is consturcted with the centre \(\displaystyle O\) and radius \(\displaystyle r\).
A semicircle is construted with a diameter \(\displaystyle AB\).
Prove that the area of the shaded lune is equal to the area of triangle AOB.
Click to expand...

You started okay . . .
Area of the triangle: \(\displaystyle \,\frac{1}{2}r^2\)

Area of quarter-circle: \(\displaystyle \,\frac{1}{4}\pi r^2\)

Area of semicircle: \(\displaystyle \,\frac{1}{2}\pi r^2\;\) . . . no
Click to expand...

The diameter of the semicircle \(\displaystyle (AB)\) is the hypotenuse of \(\displaystyle \Delta AOB.\)
. . Hence: \(\displaystyle \,AB \:=\sqrt{2}r\)
The radius of the semicircle is: \(\displaystyle \,\frac{1}{2}AB \:=\: \frac{\sqrt{2}}{2}r\)
. . The area of the semicircle is: \(\displaystyle \,\frac{1}{2}\pi\left(\frac{\sqrt{2}}{2}r\right)^2\:=\:\frac{1}{4}\pi r^2\)

We have shown that: \(\displaystyle \,\text{(quarter-circle)}\:=\:\text{(semicircle)}\;\) [1]


In the diagram, we see that: .\(\displaystyle \begin{arra}{cc}\text{(Lune)} & \;=\; & \text{(semicircle)} & - & \text{(segment)} \\ \text{(Triangle)} & \;=\; & \text{(quarter-circle)} & - & \text{(segment)} \end{array}\)

Therefore, from [1], we have: \(\displaystyle \:\text{(Lune)} \:=\:\text{(Triangle)}\)

 
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