Question

Trapezoid ABCD has vertices A(1,6) B(-2,6) C(-10,-10) and D(20,-10). Find the measure of ABCD’s midsegment to the nearest tenth

Answer 1

Answer:

16.5 units

Step-by-step explanation:

The midsegment is the distance between the midpoints of the nonparallel sides of the trapezoid.

The trapezoid ABCD has vertices A(1,6) B(-2,6) C(-10,-10) and D(20,-10).

We want to find the midsegment of ABCD to the nearest tenth.

The midpoint of BC is;

$( \frac{ - 2 + - 10}{2} , \frac{6 + - 10}{2} ) = ( - 6, - 2)$

The midpoint of AD is :

$( \frac{1 + 20}{2} , \frac{6 + - 10}{2} ) = ( 10.5, - 2)$

The length of the midsegment is the distance from (-6,2) to (10.5,2)

$= |10 .5 - - 6| = 16.5$