You have studied in Class IX, (Chapter 9, Example 3), that a median of a triangle dividesit into two triangles of equal areas. Verify this result for Δ ABC whose vertices areA(4, – 6), B(3, –2) and C (5, 2)

Question

You have studied in Class IX, (Chapter 9, Example 3), that a median of a triangle dividesit into two triangles of equal areas. Verify this result for &#x394; ABC whose vertices areA(4, &#x2013; 6), B(3, &#x2013;2) and C (5, 2)

Philoid · Accepted Answer

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

Let D be the mid-point of side BC of ΔABC Therefore, AD is the median in ΔABC

Coordinates of point D = () = (4, 0)

Area of the triangle (ABC) = = [4 (-2 - 0) + 3 (0+ 6) + 4 (-6+ 2)]

= (-8 + 18 – 16)

= -3 Square units

However, area cannot be negative. Therefore, area of ΔABD is 3 square units

Area of the triangle (ADC) = = [4 (0 - 2) + 4 (2 + 6) + 5 (-6 - 0)]

= (-8 + 32 – 30)

= - 3 Square units

However, area cannot be negative. Therefore, area of ΔADC is 3 square units

Clearly, median AD has divided ΔABC in two triangles of equal areas