Question

How do you find the area of parallelogram with 2 sides and angle?

Answer 1

Hint: The formula for the area of a parallelogram is $ab\sin \theta$ where a and b are length of 2 consecutive sides of the parallelogram. We can find the area of the parallelogram using this formula.

Complete step by step answer:
Let’s draw the figure of a parallelogram

In the above figure, ABCD is a parallelogram and AE is the height of the parallelogram. DF is the height drawn from D.
Length of AE = length DF
And angle ABE = angle DCF
So triangle ABE and triangle DCF are congruent
So the area of ABE= area of DCF
Area of parallelogram ABCD = area of triangle ABE + area of trapezium AECD
We can replace ABE with the area of DCF in the above equation.
Area of parallelogram ABCD= area of triangle DCF + area of trapezium AECD

We can see in the figure area of triangle DCF + area of trapezium AECD is equal to the area of rectangle AEFD
We can write
Area of parallelogram ABCD = area of rectangle AEFD
We know that area of rectangle = length $\times$ breath
So area of rectangle AEFD= $EF\times AE$
 $\Rightarrow EF=EC+CF$
CF is equal to BE from congruence of ABE and ACF
 $\Rightarrow EF=EC+BE$
 $\Rightarrow EF=BC$
Area of AEFD= $BC\times AE$
Area of parallelogram ABCD= $BC\times AE$
In triangle ABE, $AE=AB\times \sin B$
Area of parallelogram ABCD= $AB\times BC\times \sin B$
 $\sin B=\sin A$ in ABCD as $A+B={180}^{\circ }$
AB and BC are consecutive side of ABCD
So if 2 sides and one angle is given parallelogram the area will be product of 2 sides and sin of given angle.

Note:
If the vector form of 2 consecutive sides are given in a parallelogram then the area of parallelogram is $|\overrightarrow{a}\times \overrightarrow{b}|$ where $\overrightarrow{a}$ and $\overrightarrow{b}$ are given vectors of sides. If all sides are same in a parallelogram it will became a rhombus area of rhombus is ${\displaystyle \frac{1}{2}}\times$ product of diagonals.