Concept used:
1) Diagonal divides the square into 2 triangles of equal area
2) Median divides the triangle into 2 triangles of equal areaAttachment:
WhatsApp Image 2018-05-17 at 19.35.09.jpeg [ 20.72 KiB | Viewed 1732 times ]
let the Area of Square ABCD be 100
Area of triangle ABD = \(\frac{100}{2}\) =50
in triangle ABD, PD is median, So area of triangle BDP = 50/2 =25
Now similarly, Area of triangle BDS = 25
Area of Shaded region is less than Sum of Area of triangle BDS & Atea od triangle BDP
= Area of shaded region < 25+25
= Area of shaded region < 50... (eq 1)
Similarly, it is clear from FIgure,
Area of triangle BDM>area of triangle BMP
hence, Area of triangle BDM> (area of triangle BDP)/2 = 25/2=12.5
Similarly, Area of triangle BND > 12.5
Area of shaded region >12.5+12.5 =25
Area of shaded region >25 ... (eq2)
From eq 1 & 2. 25<Area of shaded region<50
or, \(\frac{25}{100}\) <\(\frac{Area of shaded region}{Area of Square}\)<\(\frac{50}{100}\)
Hence required fraction is between 1/4 and 1/2.
So we get from the options , Fraction =\(\frac{1}{3}\)
Princ wrote:
Attachment:
The attachment question.PNG is no longer available
[A]\(\frac{1}{5}\)
[B]\(\frac{1}{4}\)
[C]\(\frac{1}{3}\)
[D]\(\frac{1}{2}\)
[E]\(\frac{4}{5}\)
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