In the figure, the vertices of ◇ ABC are A (5, 6), B (1, 4) and C (7, 2). Points D and E are mid points of sides AB and AC respectively. Find the area of ◇ ADE.
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the area of toangle forming by the mid points of two sides is 1/4th of the triangle
area of tiangle abc= 1/2[x1(y2-y3)+ x2(y3-y1)+ x3(y1-y2)
....by area of triangle formula from coordinate geometry
=1/2[5(4-2)+1(2-6)+7(6-4)]
=1/2[5(2)+1(-4)+7(2)]
=1/2(10-4+14)
=1/2(20)
=10unit sq.
coordinates of point D
by mid point formula
x=5/2+1/2 ; y=6/2+4/2
x=3 ; y=5
coordinates of pint D(3,5)
similarly,
coordinates of point E(6,4)
area of triangle ADE
=1/2[5(5-4)+3(4-6)+6(6-5)]
=1/2(5-6+6)
=1/2(5)
=5/2unit sq.
or,
2.5 unit sq.
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