four vertices of a quadrilateral are (-1,3),(5,3),(4,8) and (-3,8) .prove that the quadrilateral is a trapezium.also prove that the sum of the lengths of the parallel sides is equal to twice the length of the line segment joining the midpoints of the oblique sides
Answers
Answer:
The given quadrilateral is a trapezium.
The sum of lengths of parallel sides of the given trapezium is equal to twice the length of the line segment joining the midpoints of the oblique sides.
Step-by-step explanation:
NOTE: Refer to the attachment for the diagram.
Let the given points of the quadrilateral be A, B, C & D respectively.
In figure, □ABCD is given.
A ≡ ( - 1, 3 ) ≡ ( x₁, y₁ )
B ≡ ( 5, 3 ) ≡ ( x₂, y₂ )
C ≡ ( 4, 8 ) ≡ ( x₃, y₃ )
D ≡ ( - 3, 8 ) ≡ ( x₄, y₄ )
Now, we know that,
Slope of line = ( y₂ - y₁ ) / ( x₂ - x₁ )
∴ Slope of line AB = ( y₂ - y₁ ) / ( x₂ - x₁ )
⇒ Slope of line AB = ( 3 - 3 ) / [ 5 - ( - 1 )
⇒ Slope of line AB = 0 / ( 5 + 1 )
⇒ Slope of line AB = 0 / 6
⇒ Slope of line AB = 0 - - ( 1 )
Now,
Slope of line BC = ( y₃ - y₂ ) / ( x₃ - x₂ )
⇒ Slope of line BC = ( 8 - 3 ) / ( 4 - 5 )
⇒ Slope of line BC = 5 / - 1
⇒ Slope of line BC = - 5 - - ( 2 )
Now,
Slope of line CD = ( y₄ - y₃ ) / ( x₄ - x₃ )
⇒ Slope of line CD = ( 8 - 8 ) / ( - 3 - 4 )
⇒ Slope of line CD = 0 / - 7
⇒ Slope of line CD = 0 - - ( 3 )
Now,
Slope of line AD = ( y₄ - y₁ ) / ( x₄ - x₁ )
⇒ Slope of line AD = ( 8 - 3 ) / [ - 3 - ( - 1 ) ]
⇒ Slope of line AD = 5 / ( - 3 + 1 )
⇒ Slope of line AD = 5 / - 2
⇒ Slope of line AD = - 5 / 2 - - ( 4 )
From ( 1 ) & ( 3 ),
Slope of line AB = Slope of line CD
∴ Line AB ∥ line CD
From ( 2 ) & ( 4 ),
Slope of line BC ≠ Slope of line AD
∴ Line BC ∦ line AD
∴ In □ABCD, there is only one pair of parallel sides.
∴ □ABCD is a trapezium. - - [ By definition ]
The given quadrilateral is a trapezium.
Hence proved!
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Now, we know that, by distance formula,
d ( A, B ) = √[ ( x₁ - x₂ )² + ( y₁ - y₂ )² ]
⇒ d ( A, B ) = √[ ( - 1 - 5 )² + ( 3 - 3 )² ]
⇒ d ( A, B ) = √[ ( - 6 )² + ( 0 )² ]
⇒ d ( A, B ) = √( 36 + 0 )
⇒ d ( A, B ) = √36
⇒ d ( A, B ) = 6 - - ( 5 )
Now,
d ( C, D ) = √[ ( x₃ - x₄ )² + ( y₃ - y₄ )² ]
⇒ d ( C, D ) = √{ [ 4 - ( - 3 ) ]² + ( 8 - 8 )² }
⇒ d ( C, D ) = √{ [ 4 + 3 ]² + ( 0 )² }
⇒ d ( C, D ) = √[ ( 7 )² + 0 ]
⇒ d ( C, D ) = √( 49 + 0 )
⇒ d ( C, D ) = √49
⇒ d ( C, D ) = 7 - - ( 6 )
Now,
AB + CD = 6 + 7
⇒ AB + CD = 13 - - ( 7 )
Now,
In figure, P & Q are midpoints of AD & BC respectively.
P ≡ ( x, y )
Q ≡ ( x', y' )
Now, by midpoint formula,
x = ( x₁ + x₄ ) / 2 , y = ( y₁ + y₄ ) / 2
⇒ x = [ - 1 + ( - 3 ) ] / 2 , y = ( 3 + 8 ) / 2
⇒ x = ( - 1 - 3 ) / 2 , y = 11 / 2
⇒ x = - 4 / 2 , y = 11 / 2
⇒ x = - 2 , y = 11 / 2
∴ P ≡ ( - 2, 11 / 2 )
Now,
x' = ( x₂ + x₃ ) / 2 , y' = ( y₂ + y₃ ) / 2
⇒ x' = ( 5 + 4 ) / 2 , y' = ( 3 + 8 ) / 2
⇒ x' = 9 / 2 , y' = 11 / 2
∴ Q ≡ ( 9 / 2, 11 / 2 )
Now, by distance formula,
d ( P, Q ) = √[ ( x - x' )² + ( y - y' )² ]
⇒ d ( P, Q ) = √[ ( - 2 - 9 / 2 )² + ( 11 / 2 - 11 / 2 )² ]
⇒ d ( P, Q ) = √{ [ ( - 4 - 9 ) / 2 ]² + ( 0 )² }
⇒ d ( P, Q ) = √[ ( - 13 / 2 )² + 0 ]
⇒ d ( P, Q ) = √( - 13 / 2 )²
⇒ d ( P, Q ) = 13 / 2
Now,
2 × PQ = 2 × ( 13 / 2 )
⇒ 2 × PQ = 13 / 2 × 2
⇒ 2 × PQ = 13 × 1
⇒ 2 PQ = 13 - - ( 8 )
From ( 7 ) & ( 8 )
AB + CD = 2 PQ
∴ The sum of lengths of parallel sides is equal to twice the length of the line segment joining the midpoints of the oblique sides.
Hence proved!
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