Let ABCD be a parallelogram such that AB = 10,BC = 14, and \angle A = 45. Find the area of the parallelogram. please write your answer as a root. NOT a decimal.

Respuesta :

Answer:

[tex]70\sqrt{2}[/tex]

Step-by-step explanation:

Since angle A is 45, the height of the base BC is 10/[tex]\sqrt{2}[/tex] by 1-1-sqrt(2) triangles.

Then we can apply parallelogram area formula.

Answer: 70*sqrt(2)

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Explanation:

One formula you can use is

A = b*c*sin(theta)

where angle theta is between sides b and c

Note how angle A is between sides AB = 10 and AD = BC = 14.

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Using that formula gives

A = b*c*sin(theta)

A = 10*14*sin(45)

A = 140*sin(45)

A = 140*sqrt(2)/2 ... use unit circle

A = (140/2)*sqrt(2)

A = 70*sqrt(2)