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.

no span please. and brainly teacher can also answe​

Answer 1

$\bf { \underline{ \underline{ \red{Concept \: \: used:-}}}}$

${ \underline{ \purple{Distance \: \: Formula:-}}} \\ \\ </p><p></p><p>Distance \: \: between \: \: the \: \: points \: \: A(x_1,y_1) \: \: and \: \: B (x_2,y_2) is \\ \\ AB = \sqrt{ {(x_2 - x_1)}^{2} + {(y_2 -y_1) }^{2} }$

Solution:

Let the given points A(8,3),B(7,8),C(2,7),D(3,2)

Now,

By using distance formula:

Distance between A and B is :

$A(8,3) \: \: B(7,8) \\ \\ D= \sqrt{ {(7 - 8)}^{2} + {(8 - 3)}^{2} } \\ \\ D = \sqrt{ {( - 1)}^{2} + {(5)}^{2} } \\ \\ D = \sqrt{1 + 25} \\ \\ D = \sqrt{26}$

Distance between B and C is:

$B(7,8) \: \: C(2,7) \\ \\ D = \sqrt{ {(2 - 7)}^{2} + {(7 - 8)}^{2} } \\ \\ D = \sqrt{ {( - 5)}^{2} + {( - 1)}^{2} } \\ \\ D= \sqrt{1 + 25} \\ \\ D= \sqrt{26}$

Distance between C and D is:

$C(2,7) \: \: D(3,2) \\ \\ D= \sqrt{ {(3 - 2)}^{2} + {(2 - 7)}^{2} } \\ \\ D= \sqrt{ {(1)}^{2} + {( - 5)}^{2} } \\ \\ D= \sqrt{1 + 25} \\ \\ D= \sqrt{26}$

Distance between D and A :

$A(8,3) \: \: D (3,2) \\ \\ D = \sqrt{ {(8 - 3)}^{2} + {(3 - 2)}^{2} } \\ \\ D = \sqrt{ {(5)}^{2} + {( - 1)}^{2} } \\ \\ D= \sqrt{25 + 1} \\ \\ D= \sqrt{26}$

Therefore, all the side of quadrilateral is equal , so it is a square.

Area of square = (side)²

Area of square= (√26)²

Area of square = 26 sq.units....

Answer 2

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 :-

• All sides of Quadrilateral are Equal :- AB = BC = CD = DA = √26.
• Both Diagonals are Equal :- AC = BD = 2√13.
• Length of Diagonals > Length of side of Quadrilateral :- 2√13 > √26 .

we know that ,

• All the sides of a Square are Equal in Length.
• Both diagonals of a Square are Equal in Length.
• The length of diagonals is greater than the sides of the square .

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.)