Area by the sine: the triangle and the parallelogram

The area of a triangle is half the product of two sides and the sine of the angle between them (from grade 10).

• $\displaystyle a$ and $\displaystyle b$ – two sides; $\displaystyle \gamma$ – the angle between them.
• The reason: the height to the side $\displaystyle a$ is $\displaystyle b\sin\gamma$, and the area is half the side times the height.

Theorem: The area of a parallelogram is the product of two adjacent sides and the sine of the angle between them.

Given: $\displaystyle ABCD$ is a parallelogram, $\displaystyle {AB}={a}$, $\displaystyle {AD}={b}$, $\displaystyle {\angle DAB}={\alpha}$.
To prove: $\displaystyle {A_{ABCD}}={ab\sin\alpha}$.

The proof below draws the diagonal $\displaystyle BD$ and proves that it cuts the parallelogram into two congruent triangles:

Congruent triangles have equal areas, so the parallelogram’s area is twice the area of the triangle $\displaystyle ABD$:

Any angle of the parallelogram will do: the angle next to $\displaystyle \alpha$ is $\displaystyle 180^{\circ}-\alpha$, and its sine is the same.

When the angle is $\displaystyle 30^{\circ}$, $\displaystyle 45^{\circ}$ or $\displaystyle 60^{\circ}$ (or $\displaystyle 120^{\circ}$, $\displaystyle 135^{\circ}$, $\displaystyle 150^{\circ}$), the area comes out exact: $\displaystyle {\sin30^{\circ}}={\frac{1}{2}}$, $\displaystyle {\sin45^{\circ}}={\frac{\sqrt{2}}{2}}$, $\displaystyle {\sin60^{\circ}}={\frac{\sqrt{3}}{2}}$.

• A parallelogram with sides $\displaystyle 6$ and $\displaystyle 10$ and the angle $\displaystyle 30^{\circ}$ between them: $\displaystyle {A}={6\cdot10\cdot\frac{1}{2}}={30}$.
• A rhombus with side $\displaystyle 8$ and an angle of $\displaystyle 120^{\circ}$: $\displaystyle {A}={8\cdot8\cdot\frac{\sqrt{3}}{2}}={32\sqrt{3}}$.

In the triangle $\displaystyle ABC$, $\displaystyle {AB}={8}$, $\displaystyle {AC}={5}$ and $\displaystyle {\angle A}={150^{\circ}}$.
Find the area of the triangle.

The sides of a parallelogram are $\displaystyle 7$ and $\displaystyle 4$, and the angle between them is $\displaystyle 45^{\circ}$.
Find the area of the parallelogram. Leave the answer with a square root.

The area of a polygon can be found by adding and subtracting triangles. In a polygon inscribed in a circle it is handy to join the centre to the vertices:

• Each triangle made is isosceles: two of its sides are radii, $\displaystyle R$ and $\displaystyle R$.
• The area of each is $\displaystyle \frac{1}{2}R^{2}\sin\theta$, where $\displaystyle \theta$ is its central angle.
• When the centre is outside the polygon, one triangle lies outside it – and its area is subtracted.

The trapezoid $\displaystyle ABCD$ is inscribed in a circle.
The point $\displaystyle O(6,2)$ is the centre of the circle, and the point $\displaystyle A(-2,2)$ is on the circle.
$\displaystyle {\angle AOB}={\angle COD}={36.2^{\circ}}$ and $\displaystyle {\angle BOC}={71.2^{\circ}}$.
Round to two decimal places.

The circle $\displaystyle {(x+1)^{2}+(y-3)^{2}}={25}$ has its centre at $\displaystyle M$.
The line $\displaystyle {y}={7}$ meets the circle at the points $\displaystyle A$ and $\displaystyle B$; $\displaystyle A$ is to the right of $\displaystyle B$.

Find the coordinates of the points $\displaystyle A$ and $\displaystyle B$.

The line $\displaystyle BM$ meets the $\displaystyle x$-axis at the point $\displaystyle C$. Find the equation of the line $\displaystyle BM$ and the coordinates of $\displaystyle C$: write the two equations – the line’s and the $\displaystyle x$-axis’ ($\displaystyle {y}={0}$) – and solve.