Math, asked by bheemanianudeep, 6 months ago

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

Answered by Ameya09
0

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:

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