Plz help me find the shaded area... Circle, quadrants n triangles...looks easy but it

The altitude of the triangle is \(\displaystyle \dfrac{\sqrt{3}}{2}b\) where \(\displaystyle b\) is the base. You should see that the base is one-half the length of the entire square in your drawing.


hi MarkFL,
this is my working n may I know if it is correct?

i divided the 1/4 of the picture to further 4 by 4 and this is my working...

area of square is 6*6=36
area of green triangle is 1/2 *4.5*6= 13.5
Area of "what it looks like 2 slices of pizzas = 1/2 * 6* 3 *2= 18
so.....

Area of tiny shaded part is 36-18-13.5=4.5...??
 
This is the 25th post and it seems to me that a lot of what is being done is not necessary. (Unless the question was changed in a post I didn't look at!)

In particular I see no reason to divide the figure into triangles. The entire figure is based on a square with each side of length 12 cm Each quarter, and particularly the upper right quarter, is a 6 by 6 square and so has area 36 square cm. The non-shaded region in that upper right quarter is one quarter of a circle of radius 6 cm which has area \(\displaystyle \pi(6^2)= 36\pi\) square cm and so the quarter has area \(\displaystyle 36\pi/4= 9\pi\) square cm. The shaded area in that quarter, then, has area \(\displaystyle 36- 9\pi\) square cm. The shaded area in the lower left quarter of the entire square is also \(\displaystyle 36- 9\pi\) square cm and, of course, the upper left quarter, which is completely shaded has area 36 squarte cm. The entire shaded region has area \(\displaystyle 36+ (36- 9\pi)+ (36- 9\pi)= 108- 18\pi\) square centimeters.
 
This is the 25th post and it seems to me that a lot of what is being done is not necessary. (Unless the question was changed in a post I didn't look at!)

In particular I see no reason to divide the figure into triangles. The entire figure is based on a square with each side of length 12 cm Each quarter, and particularly the upper right quarter, is a 6 by 6 square and so has area 36 square cm. The non-shaded region in that upper right quarter is one quarter of a circle of radius 6 cm which has area \(\displaystyle \pi(6^2)= 36\pi\) square cm and so the quarter has area \(\displaystyle 36\pi/4= 9\pi\) square cm. The shaded area in that quarter, then, has area \(\displaystyle 36- 9\pi\) square cm. The shaded area in the lower left quarter of the entire square is also \(\displaystyle 36- 9\pi\) square cm and, of course, the upper left quarter, which is completely shaded has area 36 squarte cm. The entire shaded region has area \(\displaystyle 36+ (36- 9\pi)+ (36- 9\pi)= 108- 18\pi\) square centimeters.


Hi there Hallsof Ivy,

I think you have overlooked the diagram I am attaching the picture again.
I have absolutely no problem finding the above mentioned answers it is the bottom shaded part that I am encountering difficulty on...
 

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Hi there Hallsof Ivy,

I think you have overlooked the diagram I am attaching the picture again.
I have absolutely no problem finding the above mentioned answers it is the bottom shaded part that I am encountering difficulty on...
Oh, Blast! That little teeny tiny part? Frankly, I would be inclined to believe that was error in the printing!

But, good luck convincing your teacher of that!
 
Oh, Blast! That little teeny tiny part? Frankly, I would be inclined to believe that was error in the printing!

But, good luck convincing your teacher of that!


Nahh... No printing error...
I just wish someone can tell me if the answer that I had found
is correct or incorrect..
Thanks for your time and patience hallsofivy :)
 
Last edited:
hi MarkFL,
this is my working n may I know if it is correct?

i divided the 1/4 of the picture to further 4 by 4 and this is my working...

area of square is 6*6=36
area of green triangle is 1/2 *4.5*6= 13.5
Area of "what it looks like 2 slices of pizzas = 1/2 * 6* 3 *2= 18
so.....

Area of tiny shaded part is 36-18-13.5=4.5...??

Yes, the area of the square is 36.

The area of the green triangle is \(\displaystyle \dfrac{1}{2}\cdot6\cdot\dfrac{6\sqrt{3}}{2}=9\sqrt{3}\)

The area of the two red circular sectors is \(\displaystyle 2\cdot\dfrac{1}{2}6^2\cdot\dfrac{\pi}{6}=6\pi\)

Thus, the area you want is:

\(\displaystyle A=36-\left(9\sqrt{3}+6\pi \right)\approx1.5619868103413452\)
 
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