Prove that the quadrilateral ABCD whose vertices are (8, 3), (7, 8), (2, 7) and (3, 2) taken in order is a square.
Also, find its area.
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Prove that the quadrilateral ABCD whose vertices are (8, 3), (7, 8), (2, 7) and (3, 2) taken in order is a square.
Also, find its area.
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Question :- Prove that the quadrilateral ABCD whose vertices are (8, 3), (7, 8), (2, 7) and (3, 2) taken in order is a square. Also, find its area. ?
Solution :-
Let the given points be A(8,3), B(7,8) , C(2,7) and D(3,2) .
Length of Line segment A(8,3), B(7,8) :-
→ AB = √{(x2 - x1)² + (y2 - y1)²}
→ AB = √{(7 - 8)² + (8 - 3)²}
→ AB = √{(-1)² + (5)²}
→ AB = √(1 + 25)
→ AB = √26
Similarly,
→ Length of Line segment B(7,8) , C(2,7) :-
→ BC = √{(x2 - x1)² + (y2 - y1)²}
→ BC = √{(2 - 7)² + (7 - 8)²}
→ BC = √{(-5)² + (-1)²}
→ BC = √(25 + 1)
→ BC = √26
Similarly,
→ Length of Line segment C(2,7) and D(3,2) :-
→ CD = √{(x2 - x1)² + (y2 - y1)²}
→ CD = √{(3 - 2)² + (2 - 7)²}
→ CD = √{(1)² + (-5)²}
→ CD = √(1 + 25)
→ CD = √26
Similarly,
→ Length of Line segment D(3,2), A(8,3) :-
→ DA = √{(x2 - x1)² + (y2 - y1)²}
→ DA = √{(8 - 3)² + (3 - 2)²}
→ DA = √{(5)² + (1)²}
→ DA = √(25 + 1)
→ DA = √26
Now, Checking Both Diagonals AC and BD,
→ Length of Line segment A(8,3), C(2,7) :-
→ AC = √{(x2 - x1)² + (y2 - y1)²}
→ AC = √{(2 - 8)² + (7 - 3)²}
→ AC = √{(-6)² + (4)²}
→ AC = √(36 + 16)
→ AC = √52
→ AC = 2√13.
Similarly,
→ Length of Line segment B(7,8) , D(3,2) :-
→ BD = √{(x2 - x1)² + (y2 - y1)²}
→ BD = √{(3 - 7)² + (2 - 8)²}
→ BD = √{(-4)² + (-6)²}
→ BD = √(16 + 36)
→ BD = √52
→ BD = 2√13.
Conclusion :-
we know that ,
Therefore, we can conclude that, Given Quadrilateral ABCD is a Square.
Now,
→ Each Side of Square = √26
→ Area of Square = (side)²
→ Area of Square = (√26)²
{(√a)² = a} .
→ Area of Square = 26 units. (Ans..)