Mathematics (Mensuration)

[usman]

:shock:
Given that GHIJ is a square with GH = HI = IJ = GJ = 1cm
Also,
HJI, GIJ, GIH & GHJ are all identical quadrants centred at I, J, H & G respectively. They are all inscribed
in the square GHIJ as shown in the diagram.
Now, the question is to find the area of the region ABCD (which is in the centre of the square).
Can anybuddy help :mrgreen:

 

[Fj{Death-Hunter}Fj]

Where did u find this Question? It doesn't come in O'Levels, its a SAT question (something to worry next year )

 

[usman]

any ways .... (let's try it buddy) ...

 

[Angraiz]

:shock:
Given that GHIJ is a square with GH = HI = IJ = GJ = 1cm
Also,
HJI, GIJ, GIH & GHJ are all identical quadrants centred at I, J, H & G respectively. They are all inscribed
in the square GHIJ as shown in the diagram.
Now, the question is to find the area of the region ABCD (which is in the centre of the square).
Can anybuddy help :mrgreen:

Area of quadrant= Pi x R square / 4
Remaining area = 1 - 0.78 = 0.21
Area of ABCD = 0.78 - 0.21 = 0.57 Ans

 

[usman]

nyc try @ Angraiz:
bt area of ABCD in this given question refers only to the centre of the circle
i.e. the area you calculated minus the ( area DAH + area BCJ )

 

[usman]

The area of region ABCD is 1+ pi/3 - sqrt(3)

This is how the solution goes.

Denote region AGB, region GAH, and region ABCD by x, y and z respectively.

We can equate the sum of of all the regions to the total area like this:

4x + 4y + z =1

Also, you can equate the sum of all regions inside one quadrant to the area of the quadrant:

3x + 2y + z = pi/4

Now, triangle AJI is equilateral, so angle AJI is pi/3 radian. This means angle GJA is pi/6 radian.

So,
y = 1 - 1/2 * 1 * 1 * sin pi/3 - 2 * 1/2 * 1 * 1 * pi/6

Evaluate y and get the value of z from the simultaneous equations.