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1. What are the endpoint coordinates for the midsegment of △JKL that is parallel to JL⎯⎯⎯⎯?

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( ) ( ), and ( ) ( )
2. In parallelogram ABCD , diagonals AC⎯⎯⎯⎯⎯ and BD⎯⎯⎯⎯⎯ intersect at point E, BE=2x2−3x , and DE=x2+10 .

What is BD ?
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units
3.JKLM is a parallelogram.

What is the measure of ∠KLJ?

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1. What are the endpoint coordinates for the midsegment of △JKL that is parallel to-example-1
1. What are the endpoint coordinates for the midsegment of △JKL that is parallel to-example-1
1. What are the endpoint coordinates for the midsegment of △JKL that is parallel to-example-2

1 Answer

3 votes

Answer:

Part 1)
(2.5,2.5) and
(2,0)

Part 2)
BD=70\ units

Part 3) m∠KLJ=
25\°

Explanation:

Part 1)

we have


J(1,4),K(4,1),L(0,-1)

Find the coordinates of the midpoint JK

the x-coordinate of the midpoint JK is equal to


x=(1+4)/(2)\\\\x=2.5

the y-coordinate of the midpoint JK is equal to


y=(4+1)/(2)\\\\y=2.5

The midpoint JK is the point
(2.5,2.5)

Find the coordinates of the midpoint LK

the x-coordinate of the midpoint LK is equal to


x=(0+4)/(2)\\\\x=2

the y-coordinate of the midpoint LK is equal to


y=(-1+1)/(2)\\\\y=0

The midpoint LK is the point
(2,0)

The answer part 1) is

the endpoint coordinates for the midsegment of △JKL that is parallel to JL are the points
(2.5,2.5) and
(2,0)

Part 2)

we know that

The diagonals bisect the parallelogram into two congruent triangles

In the parallelogram ABCD


BE=DE

substitute the values


2x^(2)-3x=x^(2)+10\\x^(2)-3x-10=0

Solve the quadratic equation

The formula to solve a quadratic equation of the form
ax^(2) +bx+c=0 is equal to


x=\frac{-b(+/-)\sqrt{b^(2)-4ac}} {2a}

in this problem we have


x^(2)-3x-10=0

so


a=1\\b=-3\\c=-10

substitute in the formula


x=\frac{3(+/-)\sqrt{(-3)^(2)-4(1)(-10)}} {2(1)}


x=\frac{3(+/-)√(49)} {2}


x=\frac{3(+/-)7} {2}


x=\frac{3+7} {2}=5


x=\frac{3-7} {2}=-2

Find the value of BD


BD=2x^(2)-3x+x^(2)+10

Substitute the value of
x=5


BD=2(5)^(2)-3(5)+(5)^(2)+10=70\ units

Part 3)

we know that

The diagonals bisect the parallelogram into two congruent triangles

In the parallelogram JKLM

m∠KLJ=m∠MLJ

we have that

m∠MLJ=
25\°

therefore

m∠KLJ=
25\°

User Shorty
by
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