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by eagleeye » Fri Jun 01, 2012 7:30 pm
Hi phoenix9801:

In the triangle HIJ wwhere angle HIJ = 90 (angle of a rectangle), we use the Pythagoras theorem to find JH:

JH== sqrt(HI^2+IJ^2) = sqrt(12^2+5^2) = 13.
[spoiler]
Since JH = 13 and JK = 3, Area of HGKJ = 13*3 = 39[/spoiler]

Let me know if this helps :)
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by Anurag@Gurome » Fri Jun 01, 2012 11:28 pm
In rectangle HGKJ,
  • HG = JK = 3 and
    GK = HJ = √(HI² + IJ²) = √(5² + 12²) = √(25 + 144) = √169 = 13
Hence, area of the rectangle HGKJ = 3*13 = 39
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