English, asked by aprakasha81, 5 months ago

In a AXYZ, KA is the median, the vertices of the Axyz are
(5,2) (4,7) and (6,3) show that medias XA divides the given
triangle intog 2 equal areas​

Answers

Answered by Itzkrushika156
0

Explanation:

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

∴Co-ordinates of D=(

2

x

1

+x

2

,

2

y

1

+y

2

)=(

2

3+5

,

2

−2+2

)=(4,0)

Median AD divide 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)

(x

3

,y

3

)=(4,0)

Area of △ABD=

2

1

[4(−2−0)+3(0+6)+4(−6+2)]

=

2

1

[−8+18−16]=

2

−6

=3

∴ 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)

(x

3

,y

3

)=(5,−2)

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

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

Attachments:
Similar questions