The diagonals of a parallelogram have lengths of $15.6 \text{cm}$ and $17.2 \text{cm}$. They intersect at an angle of $120$. Find the area of the parallelogram.
The part I find most confusing is the intersection angle. I thought diagonals intersect at a right angle.
This comes from an 8th-grade school math textbook.
On
Recall that the area of the parallelogram is $\frac{1}{2}d_1 d_2\sin\theta$, where $d_1$ and $d_2$ are the lengths of the diagonals, and $\theta$ is the angle between the diagonals.
As regards the intersection angle take a look here: Diagonals of parallelogram intersect at $90^\circ$ if and only if figure is rhombus
The area is $\mathcal{A}=\dfrac{1}{2} 15.6\times 17.2 \,\dfrac{\sqrt{3}}{2}\approx 232.372$cm$^2$
The answer is 116 cm square. The answer would be 232 cm square if there was no halving required.