you have studied in class 9 (chapter 9,example 3)that a median of a triangle divides it into two triangle of equal areas. verify this result for adc whose vertices are A (4,_5),B (3,_2),C (5,2)
Answers
you have studied in class 9 (chapter 9,example 3)that a median of a triangle divides it into two triangle of equal areas. verify this result for adc whose vertices are A (4,_5),B (3,_2),C (5,2)
Answer:
Step-by-step explanation:
Let AD be the median that divides Bc into two equal parts : CD and BD
Therefore the coordinates of D will be
(3+5/2, -2+2/2 )
=(8/2 , 0/2)
=(4,0)
Answer:
AD median is drawn to side BC which is the vertex of △ABC.
Median AD divides △ABC into two triangles △ABD and △ADC which are equal in area.
Now Median AD bisects BC.
∴BD=DC.
(i) As per Mid-Points formula,
Coordinates of D are
=
(4)
2
,(0)
2
=
16,0
=(4,0)
∴ Coordinate of D are (4,0)
(ii) Now, Area of △ABD:
=
2
1
[x
1
(y
2
−y
3
)+x
2
(y
3
−y
1
)+x
3
(y
1
−y
2
)]
=
2
1
[4(−2−0)+3(0+6)+4(−6+2)]
=
2
1
[4(−2)+3(6)+4(−4)]
=
2
1
[−8+18−16]
Area of △ABD=
2
1
(−6)
=−3 sq. units
(iii) Now, Area of ADC :
=
2
1
[x
1
(y
2
−y
3
)+x
2
(y
3
−y
1
)+x
3
(y
1
−y
2
)]
=
2
1
[4(0−2)+4(2+6)+5(−6+0)]
=
2
1
[4(−2)+4(8)+5(−6)]
=
2
1
[−8+32−30]
=
2
1
×(−6)
=−3 sq. units
∴ Area of △ADC=−3 sq. units.
∴ Area of △ABC:
Area of △ABD+Areaof△ADC
=(−3)+(−3)
=−6 sq. units.
Since area can not be negetive
=6 sq. unit
(iv) Now, Area of △ABC : (Direct Method)
=
2
1
[x
1
(y
2
−y
3
)+x
2
(y
3
−y
1
)+x
3
(y
1
−y
2
)]
=
2
1
[4(−2−2)+3(2+6)+5(−6+2)]
=
2
1
[4(−4)+3(8)+5(−4)]
=
2
1
[−16+24−20]
=
2
1
[−36+24]
=
2
1
[−12]
=−6 sq. units.
Since area can not be negetive
=6 sq. unit
Step-by-step explanation:
hope it's help you