- Area of a Triangle
Find the values of k for which the points A(k + 1, 2k), B(3k, 2k + 3) and (5k – 1, 5k) are collinear.
Find the values of k so that the area of the triangle with vertices (k + 1, 1), (4, -3) and (7, -k) is 6 sq. units.
Find the values of k so that the area of the triangle with vertices (1, -1), (-4, 2k) and (-k, -5) is 24 sq. units.
A(4, - 6), B(3,- 2) and C(5, 2) are the vertices of a 8 ABC and AD is its median. Prove that the median AD divides Δ ABC into two triangles of equal areas.
In Fig. 8, the vertices of ΔABC are A(4, 6), B(1, 5) and C(7, 2). A line-segment DE is drawn to intersect the sides AB and AC at D and E respectively such that `(AD)/(AB)=(AE)/(AC)=1/3 `Calculate th area of ADE and compare it with area of ΔABCe.
In Fig. 6, ABC is a triangle coordinates of whose vertex A are (0, −1). D and E respectively are the mid-points of the sides AB and AC and their coordinates are (1, 0) and (0, 1) respectively. If F is the mid-point of BC, find the areas of ∆ABC and ∆DEF.