Question

- You have studied in Class IX, (Chapter 9, Example 3), that a median of a triangle divides
it into two triangles of equal areas. Verify this result for A ABC whose vertices are
A(4, -6), B(3,-2) and C(5,2).​

Answer 1

Step-by-step explanation:

Let AD is the median of △ABC. Then D is the midpoint of BC.

We know that the coordinates of the midpoint of the line segment joining (x

1

,y

1

) and (x

2

,y

2

) are:

P(x,y)=(

2

x

1

+x

2

,

2

y

1

+y

2

)

∴ Coordinates of D=(

2

3+5

,

2

−2+2

)=(4,0)

Median AD divides the △ABC into two triangles.

∴ Area of △ABD=

∣

∣

∣

∣

∣

2

1

[x

1

(y

2

−y

3

)+x

2

(y

3

−y

1

)+x

3

(y

1

−y

2

)]

∣

∣

∣

∣

∣

Here, (x

1

,y

1

)=(4,−6), (x

2

,y

2

)=(3,−2) and (x

3

,y

3

)=(4,0)

Hence, area of △ABD=

2

1

[4(−2−0)+3(0+6)+4(−6+2)]

=

2

1

[−8+18−16]=

∣

∣

∣

∣

∣

2

−6

∣

∣

∣

∣

∣

=3

Also, area of △ADC=

∣

∣

∣

∣

∣

2

1

[x

1

(y

2

−y

3

)+x

2

(y

3

−y

1

)+x

3

(y

1

−y

2

)]

∣

∣

∣

∣

∣

Here, (x

1

,y

1

)=(4,−6), (x

2

,y

2

)=(4,0) and (x

3

,y

3

)=(5,−2)

Hence, area of △ADC=

2

1

[4(0+2)+4(−2+6)+5(−6−0)]

=

2

1

[8+16−30]=

∣

∣

∣

∣

∣

2

−6

∣

∣

∣

∣

∣

=3

∴ Area of △ABD= Area of △ADC=3

Hence, it is proved that a median of a triangle divides it into two triangles of equal areas.