A(-4,0) and (-1,4) are two given points.C and d are points which are symmetric to the given points A and B respectively with respect to Y-axis.Calculate the perimeter of Trapezium ABDC.
Did you try drawing the figure?
clearly, the other two points are
(1,4) and (4,0)
So, the trapezium has bases 2 and 8, and height 4.
each sloping side has length z, where
z^2 = 4^2 + 3^2
z = 5
To find the perimeter of trapezium ABDC, we need to calculate the lengths of its four sides.
Given points A(-4,0) and B(-1,4), we can find points C and D, which are symmetric to A and B, respectively, with respect to the y-axis.
To find C, we take the x-coordinate of A and change its sign. Since A(-4, 0), C would have the coordinates (4, 0).
To find D, we take the x-coordinate of B and change its sign. Since B(-1, 4), D would have the coordinates (1, -4).
Now, we can calculate the lengths of the sides of the trapezium.
Side AB = distance between points A and B
= √((x2 - x1)^2 + (y2 - y1)^2)
= √((-1 - (-4))^2 + (4 - 0)^2)
= √((3)^2 + (4)^2)
= √(9 + 16)
= √25
= 5
Side BC = distance between points B and C
= √((x2 - x1)^2 + (y2 - y1)^2)
= √((4 - (-1))^2 + (0 - 4)^2)
= √((5)^2 + (-4)^2)
= √(25 + 16)
= √41
Side CD = distance between points C and D
= √((x2 - x1)^2 + (y2 - y1)^2)
= √((1 - 4)^2 + (-4 - 0)^2)
= √((-3)^2 + (-4)^2)
= √(9 + 16)
= √25
= 5
Side DA = distance between points D and A
= √((x2 - x1)^2 + (y2 - y1)^2)
= √((4 - (-4))^2 + (0 - (-4))^2)
= √((8)^2 + (4)^2)
= √(64 + 16)
= √80
= 4√5
Finally, we can calculate the perimeter of the trapezium ABDC.
Perimeter = AB + BC + CD + DA
= 5 + √41 + 5 + 4√5
= 10 + √41 + 4√5
Therefore, the perimeter of trapezium ABDC is 10 + √41 + 4√5.
To find the coordinates of C and D, we need to reflect the given points A and B with respect to the y-axis. This can be done by changing the sign of the x-coordinate while keeping the y-coordinate the same.
Given points A(-4,0) and B(-1,4), the coordinates of C will be (4,0) and the coordinates of D will be (1,-4).
Now, we can calculate the lengths of the sides of the trapezium ABDC using the distance formula:
Length of AB = √[(x2 - x1)^2 + (y2 - y1)^2]
Length of BC = √[(x2 - x1)^2 + (y2 - y1)^2]
Length of CD = √[(x2 - x1)^2 + (y2 - y1)^2]
Length of DA = √[(x2 - x1)^2 + (y2 - y1)^2]
Length of AB = √[(-1 - (-4))^2 + (4 - 0)^2]
= √[3^2 + 4^2]
= √[9 + 16]
= √25
= 5
Length of BC = √[(4 - (-1))^2 + (0 - 4)^2]
= √[5^2 + (-4)^2]
= √[25 + 16]
= √41
Length of CD = √[(1 - 4)^2 + (-4 - 0)^2]
= √[(-3)^2 + (-4)^2]
= √[9 + 16]
= √25
= 5
Length of DA = √[(-4 - 1)^2 + (0 - 4)^2]
= √[(-5)^2 + (-4)^2]
= √[25 + 16]
= √41
The perimeter of the trapezium ABDC is the sum of the lengths of its sides:
Perimeter = AB + BC + CD + DA
= 5 + √41 + 5 + √41
= 10 + 2√41
Therefore, the perimeter of Trapezium ABDC is 10 + 2√41.