Question

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.​

Answer 1

Answer:

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.