Answers

2014-10-23T11:51:38+05:30


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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2014-10-25T01:06:10+05:30

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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.

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