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. Plz friends answer fast I WILL MARK AS BRAINLIEST
Answers
Given,
The coordinates of the four vertices of the quadrilateral are (8,3) , (7,8) , (2,7) and (3,2).
To find,
The quadrilateral is a square and the area of the square.
Solution,
We can prove the quadrilateral is a square by showing that the four sides are equal and one angle is 90°.
Length of AB = ✓(8-7)²+(3-8)² = ✓1+25 = ✓26 units
Length of BC = ✓(7-2)²+(8-7)² = ✓25+1 = ✓26 units
Length of CD = ✓(2-3)²+(7-2)² = ✓1+25 = ✓26 units
Length of AD = ✓(3-8)²+(2-3)² = ✓25+1 = ✓26 units
Now, if the angle ABC is 90°, then it's components will satisfy the Pythagoras theorem.
Length of AC diagonal = ✓(8-2)²+(3-7)² = ✓36+16 = ✓52 units
(AC)² = (✓52)² = 52 units
Now,
(AB)²+(BC)² = (✓26)²+(✓26)² = 52 units
So,
(AC)² = (AB)²+(BC)²
We can say that, the angle ABC is 90°
And, ABCD is a square.
Area of ABCD = (✓26)² = 26 unit²
Hence, it is proved that ABCD is a square and it's area is 26 unit².