Abcd  is an isosceles trapezoid. if ad||bc. m< b =(x+10)  and m< c =(2x-30)  find the measure of < d.

The measure of <D is 130°.

Solution:

m<B=x+10

m<C=2x-30

If ABCD is an isosceles trapezoid with AD||BC, then:

(1) m<B=m<C

(2) m<A=m<D

Using equation (1) and replacing m<B by x+10 and m<C by 2x-30

(1) x+10=2x-30

Solving for x: Subtracting x and adding 30 both sides of the equation:

x+10-x+30=2x-30-x+30

Adding similar terms:

40=x

x=40

Replacing x in  m<B=x+10:

m<B=40+10

m<B=50°

Replacing x in m<C:

m<C=2x-30

m<C=2(40)-30

m<C=80-30

m<C=50°

Now, we need to find m<D. Using that the sum of the interior angles of any quadrillateral is equal to 360°:

(3) m<A+m<B+m<C+m<D=360°

Replacing: m<A=m<D; m<B=50°; and m<C=50° in equation (3):

(3) m<D+50°+50°+m<D=360°

Solving for m<D: Adding smilar terms:

2 m<D+100°=360°

Subtracting 100° both sides of the equation:

2 m<D+100°-100°=360°-100°

2 m<D=260°

Dividing both sides of the equation by 2:

2 m<D / 2 =260°/2

m<D=130°

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