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05 May 2006, 23:12
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A square has side length 8. Above it, there is a right isosceles triangle (The hypotenuse is 8). What is the total area?
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17 May 2006, 00:50
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A bit easier approach
The height of the triangle is 4, since it is equal to the half of the length of the square( assume BD is the height of the triangle ABC , so BDC is an isosceles triangle ).
So the total area is 8^2 + 16 = 80
The height of the triangle is 4, since it is equal to the half of the length of the square( assume BD is the height of the triangle ABC , so BDC is an isosceles triangle ).
So the total area is 8^2 + 16 = 80
