Martinez39:
Mrshape is back!

Yes
Thank u sir.
Lets fire down
EVREY SHAPE MUST BE SOLVED

Waiting

Kkings11:
Waiting

I am waiting for your solution sir

A = πr²/(10√2 - 8 )

Cc. Mrshape

Area of green shaded portion = (πr^2)/6

Martinez39:
A = πr²/(10√2 - 8 )
Cc. Mrshape

you guy should solve it step by step

Kkings11:

you guy should solve it step by step

I used coordinate geometry. I can't draw the picture.

Darivie04:
Area of green shaded portion = (πr^2)/6

Correct,
√√√√√√√√√√
show workings

Martinez39:

A = πr²/(10√2 - 8 )

Cc. Mrshape

Check ur work again,,
It is different from the answer from thousandth downward.

Mrshape:

Check ur work again,,
It is different from the answer from thousandth downward.

I made a mistake in my working, on cross-checking, I found out that Darivie04 is correct
A = πr²/6

Martinez39:

I made a mistake in my working, on cross-checking, I found out that Darivie04 is correct
A = πr²/6

Yes yes
√√√√√√√√
Correct

U guys should show workings,,
Or should I bring mine?

Kkings11:

you guy should solve it step by step

Here is the solution guys.
ask questions were u don't get.

NB: refer to the image below. #Mrshape

The line through point (α, β) and (r₁, r₁) perpendicular to y = x hence

(r₁ - β)/(r₁ - α) = -1

Resolving will lead to

α + β = 2r₁ ——— (1)

Since point (α, β) lies on the circlepoint x² + y² = r² therefore

α² + β² = r² ——— (2)

Also observe that

β = ((√2 - 1)/√2)r₁ ——— (3)
therefore β² = ((3 - 2√2 )/2)r₁²

Substituting (3) in (1), you have,

α = ((√2 - 1)/√2)
therefore α² = ((3 + 2√2 )/2)r₁²

Substituting both values for α and β into (1), you have

α² + β² = r² = 3r₁²
therefore r₁² = ⅓r²

The shaded area(a quarter of a circle or radius r₁²) is half of the area required. The area, A , required is

A = (πr₁²)/2 = πr²/6

Therefore,
A = πr²/6

Martinez39:
NB: refer to the image below. #Mrshape

The line through point (α, β) and (r₁, r₁) perpendicular to y = x hence

Resolving will lead to


Since point (α, β) lies on the circlepoint x² + y² = r² therefore

Also observe that


Substituting (3) in (1), you have,

Substituting both values for α and β into (1), you have


The shaded area(a quarter of a circle or radius r₁²) is half of the area required. The area, A , required is

Nice perspective,
The coordinate geometry boss,

Mrshape:

Nice perspective,
The coordinate geometry boss,

You are welcome sir. Expecting more shapes. Your questions shows that it is one thing to understand maths and it is another thing to apply and think creatively.

Martinez39:
You are welcome sir. Expecting more shapes. Your questions shows that it is one thing to understand maths and it is another thing to apply and think creatively.

Sure tomorrow I will post another.

Ho2 I wish I'm good at Maths

@Darivie04
you can also post your solution their is beauty in variety

Mrshape:
@Darivie04
you can also post your solution their is beauty in variety

I will. I'm occupied with something now.

Mrshape:
Here is the solution guys. ask questions were u don't get.

What method is this

Kkings11:

What method is this

No specific name,
Just an application of Pythagoras rule
And area of circle formula,
And more of mathematical thinking ability

Darivie04:

I will. I'm occupied with something now.

Ok

Mrshape:
Here is the solution guys. ask questions were u don't get.

Why is the Line EF = xroot2

Kkings11:

Why is the Line EF = xroot2

He drew a square with dimensions labeled "x". Line EF which is the diagonal of the square must therefore be xroot2.
You can get it by diving the square across the diagonal to get two triangles. Each triangle would have 2 equal sides labeled "x". Pythagorean theorem will then get you the hypothenuse of the triangle which is the diagonal of the square, which is Line EF.
I hope there's no need to explain Pythagoras as well.

Una weldone o