In AABC, if AB = BC = CA = 2 units, then the length of the median of AABC is
Answers
Answered by
1
Answer:
Given, the vertices of △ABCare A(0,−1),B(2,1) and C(0,3)
Taking D, E, F as the midpoints of the sides AB, BC, CA respectively. Then DC, EA, FB will be the medians of the sides AB, BC, CA respectively.
∴Coordinates of D=(
2
0+2
,
2
−1+1
)=(1,0),
Coordinates of E=(
2
2+0
,
2
1+3
)=(1,2)
Coordinates of F=(
2
0+0
,
2
3−1
)=(0,1)
∴Length of DC=
(1−0)
2
+(0−3)
2
=
1+9
=
10
units
Length of EA =
(1−0)
2
+(2+1)
2
=
10
units
Length of FB =
(0−2)
2
+(1−1)
2
=
(−2)
2
=2 units.
Answered by
6
Answer:
In △ABC,
AQ is the median.
∴ by Apollonius theorem,
AB
2
+AC
2
=2AQ
2
+2QC
2
122=2AQ
2
+2(
2
1
BC)
2
122=2AQ
2
+2(
2
1
×10)
2
122=2AQ
2
+2(5)
2
122−50=2AQ
2
2AQ
2
=72
AQ
2
=36
∴AQ=6
Step-by-step explanation:
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