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Lelu [443]
3 years ago
11

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

Mathematics
1 answer:
Elina [12.6K]3 years ago
4 0

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

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