1. ## area of figure

Dear sir,
i need help in the below question
Thanks

The attached figure is made of a circle , a triangle and 2 identical semi-circles. The diameter of the circle is 8cm and the diameter of the semi-circle is 4-cm. find the area of the shaded regions.(Take pi=3.14)

2. If you were to take the area of the big circle, subtract the area of the triangle, then add the area of the two semi-circles, you have your answer.

Do you see why this is the case and is this enough of a hint to finish the problem yourself?

3. ## thanks

Thanks but can you explain how you reason it out.
thanks

4. ## area of figure

Originally Posted by kingman
Dear sir,
i need help in the below question
Thanks

The attached figure is made of a circle , a triangle and 2 identical semi-circles. The diameter of the circle is 8cm and the diameter of the semi-circle is 4-cm. find the area of the shaded regions.(Take pi=3.14)
Hi kingman,
Pims solution may be a good approximation but Ithink it does not account for the small segments of the 4cm circle. Complete solution requires trig to calculate central anglesof three chords.
longest chord angle 126.9
medium chord " 53.1 and126.9
shortest chord " 53.1
You will need to calculate the altitudes of each sector triangle using trig. Happy calculating !

bjh

5. I really like this graph, it was a nice find. Thanks for posting!

Actually I took for granted that Pims suggestion was correct. But it is only correct if the triangle intersect at the very precise same coordinate - as the circles intersect. (…and as fas as I can tell after calculating on it - they do?)

Radius of circles and their centerpoints
$R_1 = 4, h_1, k_1 = [4,0]$
$R_2 = 2, h_2, k_2 = [8,2]$

triangle segment as function:
$y = x/2$

Circle formula and substitution with
$R_1 = (h_1-x)^2 + (k_1-y)^2$
$16=(4-x)^2+(x/2)^2$

$=> x=0 , x=32/5$

$R_2 = (h_1-x)^2 + (k_1-y)^2$
$4=(8-x)^2 + (2-x/2)^2$

$=> x=32/5 , x =8$

I added a diagram to with a symbolic interpretation rather than a color difference.

And, if an area had a resulting ++ or --, i.e 'double layers'
You'd had to calculate that area.

6. Originally Posted by ellensius
I really like this graph, it was a nice find. Thanks for posting!

Actually I took for granted that Pims suggestion was correct. But it is only correct if the triangle intersect at the very precise same coordinate - as the circles intersect. (…and as fas as I can tell after calculating on it - they do?)

Radius of circles and their centerpoints
$R_1 = 4, h_1, k_1 = [4,0]$
$R_2 = 2, h_2, k_2 = [8,2]$

triangle segment as function:
$y = x/2$

Circle formula and substitution with
$R_1 = (h_1-x)^2 + (k_1-y)^2$
$16=(4-x)^2+(x/2)^2$

$=> x=0 , x=32/5$

$R_2 = (h_1-x)^2 + (k_1-y)^2$
$4=(8-x)^2 + (2-x/2)^2$

$=> x=32/5 , x =8$

I added a diagram to with a symbolic interpretation rather than a color difference.

And, if an area had a resulting ++ or --, i.e 'double layers'
You'd had to calculate that area.
After seeing yourwork Icalculatedthe shaded area in a normal way andthen using Pim's method. Result was the same (30.8 cm^2)
I commend Pim for determining that there was an easy way to the answer.Looking at a half figure carefully I now see this as well

bjh

7. if i label the area counter clockwise starting from the largest, as a, b, c, d and e; then
a + b + c = pi (r^2) = 16pi - eqn 1
b + c + d = (1/2)(8)(4) = 16 - eqn 2
c + d + e = pi (r^2) = pi (2^2) = 4pi

What we need to solve is (a + c + e), anyone knows how to solve this 3 equations in 5 unknowns?

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