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).

Asked by Pragya Singh | 1 year ago |  84

1 Answer

Solution :-

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

NCERT Solutions for Class 10 Chapter 7-28

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

Coordinates of point D = Midpoint of BC 

=\( \dfrac{1(3+5)}{2}\),\(\dfrac{(-2+2)}{2}\) = (4, 0)

Formula, to find Area of a triangle = \( \dfrac{1}{2}\) × [x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]

Now, Area of ΔABD = \( \dfrac{1}{2}\) [(4) {(-2) – (0)} + 3{(0) – (-6)} + (4) {(-6) – (-2)}]

\( \dfrac{1}{2}\) (-8 + 18 – 16)

= -3 square units

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

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

\( \dfrac{1}{2}\) (-8 + 32 – 30) = -3 square units

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

The area of both sides is the same. Thus, median AD has divided ΔABC in two triangles of equal areas.

Answered by Abhisek | 1 year ago

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