Question

Adjacent sides of a parallelogram have lengths 12.5 cm and 8 cm. The measure of the included angle is 40º. Find the area of the parallelogram to the nearest tenth of a cm?

Answer 1

Please read the explanation.

Explanation:

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Adjacent sides of a Parallelogram and the included angle are given.

Draw a sketch:

`ABCD` is the parallelogram with

`bar(AB)="12.5 cm" and bar(AD)="8 cm"`

`/_BAD = 40^@`

Altitude `DE`, a perpendicular line segment is drawn through point `D` to the base `AB`.

Let `bar(DE)= "h cm"`.

Observe that `/_AED = 90^@` with `/_A=40^@`

`/_AED` is a right-triangle.

Area of the Parallelogram `color(blue)(ABCD = AB*DE=base * h`

Observe that

`h/(AD) = sin(theta)`

`h/(AD) = sin(40^@)`

`h=AD*Sin(40^@)`

`h=8*sin(40^@)`

`h~~8*(0.64278760968)`

`h~~5.142301`

Hence, Height of the Parallelogram `DE ~~ "5.142301 cm"`

Hence,

Area of the Parallelogram `=Base * Height`

`rArr (12.5)*(5.142301)`

Hence,

Area of the Parallelogram `ABCD ~~ "64.3 cm"^2`.