three vertices of a rectangle abcd are A(3,1) B(-3,1) and C(-3,3). find the fourth vertex and the area of the rectangle
2 answers:
<u>Answer: </u>
The Fourth vertex of the rectangle ABCD is (3, 3) and the area of the rectangle is 12 unit.
<u>Solution: </u>
Given three vertices of a rectangle are
A= (3, 1)
B= (-3, 1)
C= (-3, 3)
Step 1:
The graph for this sum is attached below.
By looking at the graph we see that coordinates of D are (3, 3)
We found the fourth vertex and now we can find the Area of rectangle.
Step 2:
The distance between B and C
We know the co-ordinates of B and C are B = (-3, 1) and C= (-3, 3)
BC= (-3+3) + (3-1) = 2
Step 3:
The distance between D and C
We know the co-ordinates of D and C are (3, 3) and C= (-3, 3)
DC= (-3-3) + (3-3) = 6
Step 4:
Area of ABCD rectangle = length breadth = DC
BC
= 6 × 2 = 12 units
Therefore area of Rectangle is 12 cm.
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3:5 ratio mean there are 8 "parts" (3 + 5 = 8)
The distance between the x-coordinates is |-3 - 5| = 8.
So each "part" is 8/8 = 1 unit long in the x-direction.
You want I to be 3(1) = 3 units from D, so the x-coordinate of I is -3 + 3 = 0.
Same deal for y.
|2 - 5| = 3 is the distance between D and E
Each part is 3/8.
3(3/8) = 9/8
2 + 9/8 = 25/8
So the point I is (0, 25/8)
The distance between the x-coordinates is |-3 - 5| = 8.
So each "part" is 8/8 = 1 unit long in the x-direction.
You want I to be 3(1) = 3 units from D, so the x-coordinate of I is -3 + 3 = 0.
Same deal for y.
|2 - 5| = 3 is the distance between D and E
Each part is 3/8.
3(3/8) = 9/8
2 + 9/8 = 25/8
So the point I is (0, 25/8)