# Math Help - quarter cirlces

1. ## quarter cirlces

HI
I would be most grateful if someone can help me in the below question

Thanks

The shaded figure below is formed by a semicircle and quarter circles. It lies within the rectangle PQRS in which RU = 1cm, UQ = 2cm and PT = 3cm.
(ans 6.57 square cm

2. I suggest you find the area of the 3 cm radius sector, then subtract this area from the 3 by 3 square to get the unshaded area.
Do the same for the 2 cm radius sector.

Then, find the area of the semicircle with radius 1 cm, subtract that from the square with sides 2 by 2 to get the unshaded area.

now, you only have to find the last unshaded area, at the top right corner. Find the area of the segment with 1 cm radius, subtract that from the 1 by 1 square.

Add all the unshaded areas together, then subtract that from the total area of the rectangle with area 4 by 3.

3. Originally Posted by kingman
HI
I would be most grateful if someone can help me in the below question

Thanks

The shaded figure below is formed by a semicircle and quarter circles. It lies within the rectangle PQRS in which RU = 1cm, UQ = 2cm and PT = 3cm.
(ans 6.57 square cm
1. You have a rectangle with $A = 3 \cdot 4 = 12$

2. "Cut off" the 4 areas A1 to A4. What's left is the area in question:

$A1 = 3^2-\frac14 \cdot \pi \cdot 3^2$

$A2 = 1^2-\frac14 \cdot \pi \cdot 1^2$

$A3 = 4^2-\frac12 \cdot \pi \cdot 1^2$

$A4 = 2^2-\frac14 \cdot \pi \cdot 2^2$

3. Subtract these areas from A.

EDIT: ... too late - as usual

4. Nah, that's okay, your post has a pretty drawing, so... it's kind of better than mine. If I had to draw that, I'd have taken more time

5. ## Thanks

Thanks very much for the solution but I really wonder whether there is any shortcut by looking at the symmetry of the diagtram. Also I wonder how you can replicate the diagram so nice!.
thanks

6. Hi earboth,

A3 = 2^2 - 1/2*π*1^2

7. Hi earboth,

A3 = 2^2 - 1/2*π*1^2

8. Originally Posted by sa-ri-ga-ma
Hi earboth,

A3 = 2^2 - 1/2*π*1^2
... of course you are right - but as you can see my brain is calculating faster than I can type

9. Ok, since you're mentioning this, I think we can use, not symmetry, but similarity.

I've made a drawing this time ( though not as good as yours earboth)

The red and green parts together represent one segment, the pale blue another segment, and the green, a third segment.

The area of the largest segment is given by:

$Area = \frac14 \pi r^2 - \frac12 (base)(height)= \frac14 \pi (3)^2 - \frac{(3\times3)}{2}= \frac94 \pi - \frac92 = 9(\frac{\pi - 2}{4})$

Through similarity, we know then that segments, from largest to smallest are in the ratio:

9 : 4 : 1

This means, the area of the smaller segments are $\pi - 2$ and $\frac{\pi-2}{4}$ respectively.

Now, we have all three segments. Take the largest area, add the second largest area and remove the smallest area. We only need to add the area of the smallest circle (yellow + green), which we get by:

$Area = \pi (1)^2 = \pi$

The total shaded area then is:

$Total\ Area = (\frac{\pi - 2}{4})(9+4-1) + \pi= 4\pi - 6$

10. Thanks very very much.,
You have really provided a very neat and creative solution
Big thanks
Kingman

11. dear Sir,
I would be glad if you showus what software you used to draw such nice diagram.
thanks

12. Originally Posted by kingman
dear Sir,
I would be glad if you showus what software you used to draw such nice diagram.
thanks
Uh... is it directed to me?

If so, I first used Open Office Drawing to draw the geometric lines, and used GIMP image editor to add colors...

13. Originally Posted by kingman
dear Sir,
I would be glad if you showus what software you used to draw such nice diagram.
thanks
...only in case I was meant, have a look here: euklid dynageo homepage

I don't know if there exists an English version of the program.