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3 years ago

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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3 years ago

D)) The ratio of the monthly income and expenditure of Sunayana is 5 : 3. If she

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Answer:

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Step-by-step explanation:

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Substitute 15000 for e. Evaluate:

$i=\frac{5}{3}(15000)=\text{Rs }25000$

So, Sunayana's income is ₹25000 and expenditures are ₹15000.

And we're done!

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