Find the remaining coordinates of the vertices of a square whose diagonal intersect at origin and one of the vertices is (-3,0) and plot these points. Also find the perimeter of the square.
Answers
Answer:
( 3, 0) , ( 0 , 3) & (0 , -3)
12√2
Step-by-step explanation:
Find the remaining coordinates of the vertices of a square whose diagonal intersect at origin and one of the vertices is (-3,0) and plot these points. Also find the perimeter of the square.
Diagonal from vertices is (-3,0) and passing through origion
Slope of the diagonal
(0 - 0)/(0-(-3)) = 0
y = c
=> Diagonal is over x axis
as Diagonal bisect
So coordinate of other vertices of diagonal ( 3, 0)
Diagonal are perpendicular to each other so
other diagonal would be at y axis
so other vertices will be ( 0 , 3) & (0 , -3)
Side of Square √( 3- 0)² + (0-3)² = 3√2
Perimeter = 4 * 3√2 = 12√2
Answer:
Perimeter of square = 4√18
Step-by-step explanation:
In the graph below you can see easily the diagonals are intersecting the are original and
Vertics are
A(-3,0)
B(0,-3)
C(3,0)
D(0,3)
Length of side can be find as
AB = (0,-3) - (-3,0) = (3 , -3)
And length of AB using pathogorus theorem is given as
AB = √( (3)² + (-3)² ) = √(9 + 9) = √18
So
Perimeter of square = 4 × side of length
Putting the values we get
Perimeter of square = 4 × √18 = 4√18
So
Perimeter of square = 4√18
