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Old 07-31-2007   #1
an Indian soul
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Default Pl solve this....(Geometry)?

Assume EADF is a rectangle and ABC is triangle whose vertices lie on the sides of EADF. AE =22, BE= 6, CF=16 and BF= 2, then find the area of the triangle ABC .( A is at right side top, B is at left side between E and F, & C is at base between F and D). The answer options are a) 60 sq.units b) 70 sq.units c)80 sq.units d)-----.
Old 07-31-2007   #2
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Okay, its ((6+2)*(22))-((22*6)/2) -((16*2)/2)-((8*(22-16))/2) = 70.SO, its b), if I didnt make any mistakes!
Old 07-31-2007   #3
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if you draw the rectangle with the triangle inside of it and then fill in all the lengths of segments that you have given, you can see that to find the area of the triangle ABC you need to determine the total area of the rectangle and subtract out the area of the 3 right triangles that have been formed by making triangle ABC.area of triangle ABC = area of rectangle EADF - area of right triangle EAB - area of right triangle BCF - area of right triangle CAD=(22*8) - (.5*6*22) - (.5*2*16) - (.5*6*8)=70the area of the triangle ABC is b) 70 square units

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