Find the area. of triangle whose verteses are (4,-6),(3,-2),(5,2)
Answers
Step-by-step explanation:
Given: AD be the median of ΔABC
BD = DC
Sice D is the mid-point of BC
Using the mid point theorem to find the coordinates of D,
D=[
2
x
2
+x
3
,
2
y
2
+y
3
]
D=[
2
3+5
,
2
−2+2
]=[4,0]
Area of ΔABD=
2
1
∣[(x
1
(y
2
−y
4
)+x
2
(y
4
−y
1
)+x
4
(y
1
−y
2
))]∣
=
2
1
∣[(4(−2−0)+3(0−(−6))+4(−6−(−2)))]∣
=
2
1
∣[−8+18−26]∣
=
2
1
∣[−6]∣=3 Sq.units
Area of ΔADC=
2
1
∣[(x
1
(y
4
−y
3
)+x
4
(y
3
−y
1
)+x
3
(y
1
−y
4
))]∣
=
2
1
∣[(4(0−2)+4(2−(−6))+5(−6−0))]∣
=
2
1
∣[−8+32−30]∣
=
2
1
∣[−6]∣=3 Sq.units
Area od ΔABD=ΔADC
Hence proved that the median of triangle AD divides the triangles into equal areas.
Answer:
Answer
Open in answr app
Open_in_app
Given: AD be the median of ΔABC
BD = DC
Sice D is the mid-point of BC
Using the mid point theorem to find the coordinates of D,
D=[
2
x
2
+x
3
,
2
y
2
+y
3
]
D=[
2
3+5
,
2
−2+2
]=[4,0]
Area of ΔABD=
2
1
∣[(x
1
(y
2
−y
4
)+x
2
(y
4
−y
1
)+x
4
(y
1
−y
2
))]∣
=
2
1
∣[(4(−2−0)+3(0−(−6))+4(−6−(−2)))]∣
=
2
1
∣[−8+18−26]∣
=
2
1
∣[−6]∣=3 Sq.units
Area of ΔADC=
2
1
∣[(x
1
(y
4
−y
3
)+x
4
(y
3
−y
1
)+x
3
(y
1
−y
4
))]∣
=
2
1
∣[(4(0−2)+4(2−(−6))+5(−6−0))]∣
=
2
1
∣[−8+32−30]∣
=
2
1
∣[−6]∣=3 Sq.units
Area od ΔABD=ΔADC
Hence proved that the median of triangle AD divides the triangles into equal areas.
solution