# .If (3,-1) ,(2,6),(-5,7) are the midpoints of the sides of the triangle ABC, find the area of Triangle ABC.

by anudeepd2

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by anudeepd2

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let,

midpoints of triangle ABC be X(3,-1) , Y(2,6) ,Z(-5,7)

area of the triangle =I/2| x1(y2-y3)+x2(y3-y1)+x3(y1-y2)|

area of triangle abc=I/2|3(6-7)+ 2{7-(-1)}+ {(-5)(-1-6)}|

=1/2|-3+16+35|

=1/2|48|

=48/2

=24

we have a theorem that area of triangle formed by midpoints of triangle is four times the area of the whole triangle

so,area of triangle formed by midpoints of ABC is 24 sq.units

then area of triangle ABC is** **24×4=**96 sq.units**

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D(3, -1) E (2, 6) F(-5, 7)

DE = √(7² + -1²) = 5√2 DE = (y-6) = -7(x-2) => y - 20 + 7x = 0

Length of perpendicular from F to DE, = | 7 - 20 -35 | / √(7²+1²) = 48/5√2

Area of DEF = 1/2 * 5√2 * 48/5√2 = 24 units

Total area of ABC is 4 times area of DEF => 96 units.

DE = √(7² + -1²) = 5√2 DE = (y-6) = -7(x-2) => y - 20 + 7x = 0

Length of perpendicular from F to DE, = | 7 - 20 -35 | / √(7²+1²) = 48/5√2

Area of DEF = 1/2 * 5√2 * 48/5√2 = 24 units

Total area of ABC is 4 times area of DEF => 96 units.