  Question

# Question 6 The vertices of triangle ABC are A (4, 6), B (1, 5) and C (7, 2). A line is drawn to intersect sides AB and AC at D and E respectively, such that ADAB=AEAC=14. Calculate the area of triangle ADE and compare it with area of triangle ABC.

Solution

## Point D divides AB in ratio 1 : 3 Point E divides AC in the same ratio. Hence, m1 = 1 and m2 = 3 Coordinates of D can be calculated as follows: x=m1x2+m2x1m1+m2 =3×4+1×14=134 y=m1y2+m2y1m1+m2 =3×6+1×54=234 Coordinates of E can be calculated as follows: x=1×7+3×44=194 y=1×2+3×64=5 Area of triangle ABC can be calculated as follows: =12[x1(y2−y3)+x2(y3−y1)+x3(y1−y2)] =12[4(5−2)+1(2−6)+7(6−5)] =12(12−4+7)=152sq unit Area of triangle ADE can be calculated as follows: =12[x1(y2−y3)+x2(y3−y1)+x3(y1−y2)]  =12[4(234−5)+134(5−6)+194(6−234)] =12[4(23−204)−134+194(24−234)] =12(3−134+1916) =12[48−52+1916] =12×1516=1532sq unit Hence, ratio of area of triangle ADE to area of triangle ABC = 1532152 = 116  = 1 : 16  Suggest corrections  Similar questions
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