If A(4,2),B(6,2),C(6,8) are the vertices of a rectangle ABCD,then find the are of quadrilateral formed by joining the mid points of AB,BC,CDandDA
Answers
Answer:
Let P,Q,R and S are the midpoint of AB,BC,CD andDA.
∴ Co-ordinates of P=[
2
x
1
+x
2
,
2
y
1
+y
2
]=[
2
−1−1
,
2
−1+4
]=[−1,
2
3
]
∴ Co-ordinates of Q=[
2
x
1
+x
2
,
2
y
1
+y
2
]=[
2
−1+5
,
2
4+4
]=[2,4]
∴ Co-ordinates of R=[
2
x
1
+x
2
,
2
y
1
+y
2
]=[
2
5+5
,
2
4−1
]=[5,
2
3
]
∴ Co-ordinates of S=[
2
x
1
+x
2
,
2
y
1
+y
2
]=[
2
5−1
,
2
−1−1
]=[4,−1]
Now, length of PQ=
(−1−2)
2
+(
2
3
−4)
2
=
(−3)
2
+(−
2
5
)
2
=
9+
4
25
=
4
61
Length of QR=
(2−5)
2
+(4−
2
3
)
2
=
(−3)
2
+(
2
5
)
2
=
9+
4
25
=
4
61
Length of RS=
(5−2)
2
+(
2
3
+1)
2
=
(3)
2
+(
2
5
)
2
=
9+
4
25
=
4
61
Length of S=
(2+1)
2
+(−1−
2
3
)
2
=
(3)
2
+(−
2
5
)
2
=
9+
4
25
=
4
61
Length of diagonal PR=
(−1−5)
2
+(
2
3
−
2
3
)
2
=
(6
2
)
=
36
=6
Length of diagonal QS=
(2−2)
2
+(4+1)
2
=
(5)
2
=
25
=5
Hence the all the sides of the quadrilateral PQRS are equal but the diagonals are not equal then PQRS is a rhombus.
Step-by-step explanation: