A parallelogram has two adjacent sides of length 4 cm and 6 cm respectively. If the included angle measures 52 degrees, find the area of the parallelogram.
please show all working out.
if $\displaystyle a$ and $\displaystyle b$ are the lengths of two adjacent sides of a parallelogram and $\displaystyle \theta$ is the included angle between those two sides, then the area of the parallelogram is given by
$\displaystyle
A = a \cdot b \cdot \sin{\theta}
$
../------------|--/
/Theta------- |/
6cm
Area of a parallelogram is given by Base into corresponding altitude..k?
Now, draw perpendicular to one side.. the other side is found out by:
sin theta=opposite leg
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hypotenuse
multiplying sin theta by hypotenuse gives us opposite leg.. which in the case of the triangle with one vertex, theta and altitude.. is the altitude..
so area=a.b.sin theta
=4*6*sin 52
The area of a triangle is:
$\displaystyle A = \frac{1}{2}a * b * \sin{C}$
From your parallelogram, there are 2 identical triangles if you divide your parallelogram in half.
From your information, a = 4, b = 6, C = $\displaystyle 54^{\circ}$.
There you can work out the area of a triangle, then double it to find the area of your parallelogram.