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

##### Solution :-

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 = 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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