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3 years ago

what is the area of a trapezoid with bases of 8.2 CM + 5.9 cm and a height of 10 cm will mark brainiest

Mathematics

1 answer:

baherus [9]3 years ago

5 0

Trapezoid area = ((sum of the bases) ÷ 2) • height
Trapezoid area = [(8.2 + 5.9) /2] * 10
Trapezoid area = (14.1 / 2) * 10
Trapezoid area =7.05 * 10
Trapezoid area = 70.50

Source:
http://www.1728.org/quadtrap.htm

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If the measure of angle BCD=51 degrees, solve for x.

BabaBlast [244]

Answer:

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Step-by-step explanation:

step 1

Find the measure of angle EFD

In this problem I will assume that ABCD is a parallelogram

In a parallelogram opposite angles are congruent and consecutive angles are supplementary

$m\ angle BCD=m\angle BED=51^o$

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Remember that the sum of the interior angles in any triangle must be equal to 180 degrees

In the triangle EFD

$m\ angle FED+m\ angle FDE+m\ angle EFD=180^o$

substitute the given values

$25.5^o+55^o+m\ angle EFD=180^o$

$80.5^o+m\ angle EFD=180^o$

$m\ angle EFD=180^o-80.5^o$

$m\ angle EFD=99.5^o$

step 2

Find the measure of angle EFB

we know that

$m\angle EFB+m\angle EFD=180^o$ ---> by supplementary angles

we have

$m\ angle EFD=99.5^o$

substitute

$m\angle EFB+99.5^o=180^o$

$m\angle EFB=180^o-99.5^o$

$m\angle EFB=80.5^o$

step 3

Find the value of x

Remember that the sum of the interior angles in any triangle must be equal to 180 degrees

In the triangle EBF

$m\ angle BEF+m\ angle EFB+m\ angle EBF=180^o$

we have

$m\angle BEF=m\ angle FED=25.5^o$

$m\angle EFB=80.5^o$

$m\angle EBF=(14x+4)^o$

substitute

$25.5^o+80.5^o+(14x+4)^o=180^o$

solve for x

Combine like terms

$(14x+110)^o=180^o$

$14x=180-110$

$14x=70$

$x=5$

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