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Home » You have studied in Class IX that a median of a triangle divides it into two triangles of equal areas. Verify this result for ΔABC whose vertices are A (4, – 6), B (3, – 2) and C (5, 2).
You have studied in Class IX that a median of a triangle divides it into two triangles of equal areas. Verify this result for ΔABC whose vertices are A (4, – 6), B (3, – 2) and C (5, 2).
CBSE | Class 10 | Coordinate Geometry | Maths | NCERT
You have studied in Class IX that a median of a triangle divides it into two triangles of equal areas. Verify this result for ΔABC whose vertices are A (4, – 6), B (3, – 2) and C (5, 2).

Solution:

Let A (4, -6), B (3, -2), and C (5, 2) be the vertices of the triangle.

Let the side BC of triangle ABC has a midpoint as D. As a result, in ΔABC, AD is the median.

Midpoint of BC = Coordinates of point D = ((3+5)/2, (-2+2)/2) = (4, 0)

The area of triangle  = 1/2 × [x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]

Area of ΔABD = 1/2 [(4) {(-2) – (0)} + 3{(0) – (-6)} + (4) {(-6) – (-2)}]

= 1/2 (-8 + 18 – 16)

= -3 square units

Area, on the other hand, cannot be negative. Hence, the area of ΔABD is 3 square units.

Area of ΔACD = 1/2 [(4) {0 – (2)} + 4{(2) – (-6)} + (5) {(-6) – (0)}]

= 1/2 (-8 + 32 – 30) = -3 square units

Area, on the other hand, cannot be negative. As a result, the ΔACD has a area of 3 square units.

Both sides have the same area. As a result, median AD has divided triangle ABC into two equal-sized triangles.

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