# Areas of Triangles

• May 26th 2010, 02:46 AM
Tessarina
Areas of Triangles
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.

• May 26th 2010, 04:31 AM
skeeter
Quote:

Originally Posted by Tessarina
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.

if $a$ and $b$ are the lengths of two adjacent sides of a parallelogram and $\theta$ is the included angle between those two sides, then the area of the parallelogram is given by

$
A = a \cdot b \cdot \sin{\theta}
$
• Aug 20th 2010, 04:43 AM
umangarora
../------------|--/
/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
--------------
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
• Aug 21st 2010, 10:37 PM
Educated
The area of a triangle is:
$A = \frac{1}{2}a * b * \sin{C}$

From your parallelogram, there are 2 identical triangles if you divide your parallelogram in half.
From your information, a = 4, b = 6, C = $54^{\circ}$.

There you can work out the area of a triangle, then double it to find the area of your parallelogram.