Question

IF D[-1/2 , 5/2 ] , E [ 7,3 ] , F [7/2 ,7/2 ] are the mid points of triangle ABC , then find the area of triangle ABC .ITS URGENT !!

Answer 1

     D (-1/2,  5/2),   E (7, 3)    and   F (7/2,  7/2)    are the midpoints of sides AB, BC and CA respectively. ABC is a triangle.

Area of triangle DEF :
$=\frac{1}{2} | (x_D y_E + x_E y_F + x_F y_D) - ( y_D x_E + y_E x_F + y_F x_D ) |\\\\=\frac{1}{2} |\ \frac{-1}{2}*3+7*\frac{7}{2} +\frac{7}{2}*\frac{5}{2}-\ ( \frac{5}{2}*7+3*\frac{7}{2}+\frac{7}{2}*\frac{-1}{2} )\ |\\\\=\frac{57}{8}$

The triangles ADE, BDF, DEF, EFC are all of equal size.  The reason is DE = 1/2 BC and Δ ADE is similar to Δ ABC.  Hence, area of Δ ADE = 1/4 * area of Δ ABC.

So area of Δ ABC = 57/2

Answer 2

Area of triangle ABC = 1/2[ x1(y2 - y3) + x2 (y3 - y1) + x3 (y1 - y2) ] Area of triangle ABC = 4 area of triangle DEF Area of triangle DEF = 1/2[-1/2(3 - 7/2) + 7(7/2 - 5/2) + 7/2(5/2 - 3)] Area of triangle DEF = 1/2[ 1/4 +7 -7/4] Area of triangle DEF = 22/8 Area of triangle ABC = 4 area of triangle DEF Area of triangle ABC = 4*22/8 Area of triangle ABC = 11