Question

A, B, C, and D form the coordinates of a trapezium. What would be the perimeter of the trapezium?

• A. 13 units
• B. 11 units
• C. 11 + √13 units
• D. 13 + √11 units

Answer 1

The correct option is C

11 + √13 units

The perimeter of the trapezium is the sum of the sides.
The distance between AB = $\sqrt{(x_b-x_a{)}^2+(y_b-y_a{)}^2}$
= $\sqrt{(5-2{)}^2+(3-5{)}^2}$
= $\sqrt{1}3$ units
The distance between BC = $\sqrt{(x_c-x_b{)}^2+(y_c-y_b{)}^2}$
= $\sqrt{(5-5{)}^2+(0-3{)}^2}$
= 3 units
The distance between CD = $\sqrt{(x_d-x_c{)}^2+(y_d-y_c{)}^2}$
= $\sqrt{(2-5{)}^2+(0{)}^2}$
= 3 units
The distance between AD = $\sqrt{(x_a-x_d{)}^2+(y_a-y_d{)}^2}$
= $\sqrt{(2-2{)}^2+(5-0{)}^2}$
= 5 units
The perimeter of the trapezium = sum of the distances (AB, BC, CD, DA)
= $\sqrt{1}3$ + 3 + 3 + 5
= 11 + $\sqrt{1}3$ units

If the distance to be measured is along a line parallel to X-axis or Y-axis, then the distance is equal to the positive difference in the coordinates along the line.
Alternate approach: The distance parallel to X-axis or Y-axis can be found simply by finding the positive difference between the x-coordinates and the y - coordinates, resp.
Distance AD = $|$ $y_2$ - $y_1$ $|$ = 5 - 0 = 5 units (parallel to y axis )
Distance CD = $|$ $x_2$ - $x_1$ $|$ = 5 - 2= 3 units (along x axis )
Distance BC = $|$ $y_2$ - $y_1$ $|$ = 3 - 0 = 3 units (parallel to y axis )
Distance AB = $\sqrt{1}3$ units (From the distance formula)