parallelogram

[Agatha]

d₁=40, d₂=74, one of the lines (a or b) are 51.i assume it's "a" since d₁ is 40.what is area of parallelogram?
i used this: d₁²+d₂² = 2(a²+b²) to get b²=d₁²+d₂²/2 - a² ... i don't get the integer so i don't think it0s correct...is it?
i get for b=30,6 and for area of one triangle 611,5 and area for parallelogram 1223.
the solution in book 1224!!!help!!!

 

[pappus]

d₁=40, d₂=74, one of the lines (a or b) are 51.i assume it's "a" since d₁ is 40.what is area of parallelogram?
i used this: d₁²+d₂² = 2(a²+b²) to get b²=d₁²+d₂²/2 - a² ... i don't get the integer so i don't think it0s correct...is it?
i get for b=30,6 and for area of one triangle 611,5 and area for parallelogram 1223.
the solution in book 1224!!!help!!!

1. Your result contains some minor rounding errors.

2. You probably know that the area of the parallelogarm s calculated by:

\(\displaystyle A = a \cdot \underbrace{d_1 \cdot \sin(\alpha)}_{\text{length of height}}\)

It is not necessary to calculate the value of \(\displaystyle \alpha\) but only the \(\displaystyle \sin(\alpha)\). I've got

\(\displaystyle \sin(\alpha) = 0.6\)

So the area is

\(\displaystyle A = 51 \cdot 40 \cdot 0.6\)

 

[Agatha]

i have to do it without trigonomety,so i thought i'll use heron's formula for one of triangles...since both are the same,area of parallelogram is 2*area of triangles...is it not?

 

[Agatha]

thanks denis!