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Question

A (4,6 ), B (3,2 ) and C (5,2) are the vertices of a ΔABC and AD is its median. Prove that the median AD divides Δ ABC into two triangles of equal areas.

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Solution

Given: AD be the median of ΔABC

BD = DC

Sice D is the mid-point of BC

Using the mid point theorem to find the coordinates of D,

D=[x2+x32,y2+y32]

D=[3+52,2+22]=[4,0]

Area of ΔABD=12|[(x1(y2y4)+x2(y4y1)+x4(y1y2))]|

=12|[(4(20)+3(0(6))+4(6(2)))]|

=12|[8+1826]|

=12|[6]|=3 Sq.units

Area of ΔADC=12|[(x1(y4y3)+x4(y3y1)+x3(y1y4))]|

=12|[(4(02)+4(2(6))+5(60))]|

=12|[8+3230]|

=12|[6]|=3 Sq.units

Area od ΔABD=ΔADC

Hence proved that the median of triangle AD divides the triangles into equal areas.


553948_494206_ans_fe483ffdb93a4031862d0b0cede07928.png

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