stanisbaratheon:


Cc: Mrshape

Having seen your solution, I want to bring your attention to this question I asked earlier about the line you said was a chord to the shaded circle.

The question is, if it's a chord to the shaded circle, why did you take your reference point at the center of the shaded circle?? You even went as far as designating it "R+7". That means you recognised that line going through both shapes as radii.

I did that because there is a geometric theory that state that if two circles touch externally the distance between the two centers is sum of there radi

You made an assumption about the diameter of the shaded circle that's almost correct but not exact.

stanisbaratheon:


Diameter AB= 56cm
Radius of AB=56/2=28cm
Diameter of AE=EF=FG=GB=56/4=14cm
Radius of AE=EF=FG=GB=14/2=7cm

Diameter of shaded portion=28-7=21cm

Area of shaded portion=(πD^2)/4=(π×21^2)/4

Area=(3.14159×21^2)/4=1385.44/4 =346.36sq.cm

Mrshape:

I did that because there is a geometric theory that state that if two circles touch externally the distance between the two centers is sum of there radi

hmmmm....okay

I guess what you should have said was that line is a chord to the big semicircle.

Well whichever way it is, I'm good with your solution.

oyatz:
You made an assumption about the diameter of the shaded circle that's almost correct but not exact.

Okay

Martinez39s:
In Maths, you don't just make assumptions without proof and present them as solutions. Even by mere intuitive geometric consideration, |OF| cannot equal |ON|. If |OF| = |ON|, then the circle centred at O would touch F. |ON| is the radius of the circle centred at O, how then is |OF|, which is clearly bigger that the radius of the circle centred at O, equal to |ON|?


MrShape.

You are absolutely correct on the point u made dat the shaded circle is not really inscribed in the semi-circle and so the shaded circle cannot have its center where OF and ON become equal and that automatically OF and ON cannot be 14cm. It escaped my notice. Thumbs up! So d search for d solution continues, Mr Shape!

Rezos017:
You are absolutely correct on the point u made dat the shaded circle is not really inscribed in the semi-circle and so the shaded circle cannot have its center where OF and ON become equal and that automatically OF and ON cannot be 14cm. It escaped my notice. Thumbs up! So d search for d solution continues, Mr Shape!

I have answered it already check the first comment of this thread

Rezos017:
You are absolutely correct on the point u made dat the shaded circle is not really inscribed in the semi-circle and so the shaded circle cannot have its center where OF and ON become equal and that automatically OF and ON cannot be 14cm. It escaped my notice. Thumbs up! So d search for d solution continues, Mr Shape!

MrShape has already given the solution in the first page of this thread. The isosceles triangle MrShape drew and its dimensions ain't based on assumptions. Read the question to understand why the triangle is valid (basic geometry on circles might also help).

Mr Shape,i have a question for you sir.




Can we do it this way?

1) Calculate the area of the biggest semi circle.

2) Calculate the areas of each small semi-circles.

3) Add the areas of the four semi circles.

4) Subtract the summation from the
area of the biggestt semi-circle

Mrshape:
Take necessary precaution, stay safe and relax your mind with this shape

How I got my answer @ Mrshape

Kylekent59:
How I got my answer @ Mrshape

Fantastic

oyatz:
Mr Shape,i have a question for you sir.




Can we do it this way?

1) Calculate the area of the biggest semi circle.

2) Calculate the areas of each small semi-circles.

3) Add the areas of the four semi circles.

4) Subtract the summation from the
area of the biggestt semi-circle

You still need to subtract the area of the arcs inside the semi big semicircular

Kylekent59:
How I got my answer @ Mrshape

Nice

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Rezos017:
Overlook the poor sketch...not drawn to scale anyway.

You did a nice sketch to solve the problem, but missed the workings at a point. If you look well again, it would have been better to do this, "from the center of the big semi circle to the tip of the shaded circle is radius of the big semi circle which is 28 cm, and the radius of the small semi circle is 7cm, hence ur FC is also 7cm, therefore the diameter of the shaded circle is
= 28-7
=21cm
Hence, the radius of the shaded circle is 10.5 cm".

NB: I choose FC to represent the distance from F to the circumference of the shaded circle (the line joining F to the tip of the shaded circle)

BrutusOj:
You did a nice sketch to solve the problem, but missed the workings at a point. If you look well again, it would have been better to do this, "from the center the of the big semi circle to the tip of the shaded circle is radius of the big semi circle which is 28 cm, and the radius of the small semi circle is 7cm, hence ur FC is also 7cm, there diameter of the shaded circle is = 28-7
=21cm
Hence, the radius of the shaded circle is 10.5 cm".

NB: I choose FC to represent the distance from F to the circumference of the shaded circle (the line joining F to the tip of the shaded circle)

Not correct

Mrshape:
Take necessary precaution, stay safe and relax your mind with this shape

The area is 394.24 square cm

Rexmy:

The area is 394.24 square cm

Good

Rexmy:

The area is 394.24 square cm

TRG my other shape

Mrshape:
TRG my other shape

Biko define your TRG

Rexmy:
Biko define your TRG

Lol typo
Try my other shape is what I mean

Mrshape:

Not correct

yea..I was trying to reason from his angle. I got your working clearly.