What is the area of a parallelogram if the base is 15in. the side is 10in. and the angle between them is 60 degrees?
The area of the parallelogram has to be determined which has a base of 15 inches and the length of the side is 10 inches. The angle between the base and the side is 60 degrees.
This parallelogram can be divided into 3 parts; two right triangles with perpendicular sides 5 inches and `5*sqrt 3` inches and a rectangle with sides `5*sqrt 3` and 10 inches.
The combined area of both the triangles is `2*(1/2)*5*5*sqrt 3 = 25*sqrt 3` and the area of the rectangle is `50*sqrt 3`
Adding the two, the area of the parallelogram is `75*sqrt 3`
The required area of the given parallelogram is `75*sqrt 3 ` square inches.
Let ABCD be a parallelogram with base AB. and parallell side are (AB, DC) and (AD, BC)
Given :Base AB = 15in. and the other side AD= BC = 10in. nad the angle BAD = 60 degrees
If we draw perpendicular DE from the vertex D on the base BC, Or perpendicllar CF from the other vertes C on the base AB then DE or CF will be the height(h) of the parallelogram. and DE = CF. We also get right angled triangle ADE or CBF. Let us take the Rt. angled triangle ADE.
sin60 = DE/AD [ Using trigonometrical ratio of sin(A) = P/H ] => (√3)/2 = h/10
=> h = (1.732/2) * 10 = 0.866 *10 = 8.66
Using the formula for Area of the parallelogram(A)=base*height
We get, A = AB*h = 15*8.66= 129.90 sq.inches
The area of the given parallelogram = 129.90 sq.inches <--Answer