Answer to Question #134213 in Geometry for Nirvana

Question

the longer diagonal of a parallelogram measures 62 cm and makes an angle of 30 degrees with the base. Find the area of the parallelogram if the diagonals intersect at an angle of 70 degrees?

Answer 1

solution

given data

length of diagonal (a)=62 cm

angle with base($\theta$ ) =30⁰

diagonal are intersecting at 70⁰ angle

Since the angle of intersection of the diagonals is 70 degrees then the other angle of intersection of the diagonals is 110 degrees.

180 = 70 + θ

where θ = the other angle of intersection

θ = 110 degree

diagram for this question can be drawn like this

by applying the sine law in the triangle AOD

$\frac{sin40^0}{31}=\frac{sin30^0}{x}$

therefore a half of the shorter diagonal x is

$x=24.11cm$

therefore total length of shorter diagonal (b)=2x=48.22cm

area of parallelogram can be given as

$area=\frac{absin\theta}{2}$

by substituting value

$area=\frac{61\times48.22\times sin110^0}{2}$

$area=1404.96cm^2$

$area=0.14m^2$

therefore area of parallelogram is $0.14m^2.$