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.