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If ABCD is a square of area 1. E, F are mid points of AB and BC respectively. What is the area of blue region?

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If ABCD is a square of area 1. E, F are mid points of AB and BC respectively. What is the area of blue region?

Area of blue region

posted Jul 1, 2017 by anonymous

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1 Answer

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Size of highlighted area equal = 0.116 squr meter

answer Oct 9, 2017 by Emidway
What is the logic (please share the working)
Is it correct?!
Trying to send photo of paper can’t ... and it’s horrible lol
Ok here is my logic
But if your confused ( I know !!! Loool )

The logic in solving the question
The information that I have is that the square ABCD = 1 m Sqr
Which makes each line in the squr = 1
And that the points F and E are the exact middle of lines CB , AB respectively
Please note :
(^ )means triangle and (< ) means angle
First I took
^ FBE which I have < B = 90 and line AB = 1 abd line FB=0.5

To find FE^2 = FB^2 + AB^2
                       = 1+0.25 = 1.25
                FE= 1.118

Tan^-1 (<A)= FB/AB 0.5 /1 = 0.5
Tan^-1(<A) = 26.565

<F = 63.435

Area of ^FBE = 0.25
(I know that this is long I could’ve divided the square from point E to its opposite side on which would give me an area of 0.5 and blah blah blah divide by two =0.25 but I needed the angles of f and A )

Ok then I took ^ BCD similar calculation
Given BC= 1 CD=1 < C = 90
< B = 45  < D = 45
Line DB = 1.414

 Please note that point G is the cross section between the lines AF and DB
 And point H is the cross section of Af and De

3rd ^ BGF
Given BF=0.5 <F 63.435 ( from first calculation) <B 45
Then < G = 71.565 logic (180-<F+<B in ^ BGF)
Calculation line GF= 0.373 line GB = 0.373
Area = 0.069
Note I drew a line from point E to <C going through point G
4th ^ BEG  given ( line BG=0.373 line BE=0.5 <B = 45 from ^ BCD <B= 45 {90-45=45}

Result line EG=0.373 <E= 63.376 < G =71.624
Area of ^BEG=0.083
 5th triangle ^ AEB given line Eb= 0.5 line BD=1.414 from 2nd calculation angle B=45 ( 90-45 )
<D=18.433 <E=116.562 line DE=1.118

Final ^ EHG given (from 3rd and 4th triangles <G = 71.565 , <G 71.624 - 180 line AF =36.811 ) so < G in triangle EHG= 36.811 )
From fifth triangle < E = 116.562 subtract <E from ^ EGB WHICH IS 63.376 result =53.186 and line EG= 0.373 ) and finally <G found it from subtracting 180(line FA ) - < G from ^ BGf 3rd triangle = 71. 565 and <G from 4th triangle = 71.624
{180-71.565+71.624= 36.811 }
So < G in triangle EHG = 36.811
We calculate final triangle ASA
Area of ^ EHG =0.033

Result = Area of ^BEG=0.083 + Area of ^EHG =0.033

Area BHGB = 0.116
cool superb...

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