## How do you find the area of a quadrilateral inscribed?

Example: In a circular grassy plot, a quadrilateral shape with its corners touching the boundary of the plot is to be paved with bricks. Find the area of the quadrilateral when the sides of the quadrilateral are 36 m, 77 m, 75 m and 40 m. The area of the cyclic quadrilateral =√78×37×39×74=39×37×2=2886 square meters.

## What is always true about inscribed quadrilaterals?

All the four vertices of a quadrilateral inscribed in a circle lie on the circumference of the circle. The measure of an exterior angle is equal to the measure of the opposite interior angle.

**What conclusions can you make about the angles of a quadrilateral inscribed in a circle?**

Inscribed Quadrilateral Theorem If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary.

### How to find the central angle of an inscribed quadrilateral?

If a, b, c, and d are the inscribed quadrilateral’s internal angles, then a + b = 180˚ and c + d = 180˚. a + b = 180˚. Join the vertices of the quadrilateral to the center of the circle. Recall the inscribed angle theorem (the central angle = 2 x inscribed angle). a + b = 180 o. Hence proved!

### Are the opposite angles of this inscribed quadrilateral supplementary?

So, it looks like at least for this case that these angles, these opposite angles of this inscribed quadrilateral, it looks like they are supplementary. So, an interesting question is are they always going to be supplementary?

**What are the properties of a cyclic quadrilateral?**

1 All the four vertices of a quadrilateral inscribed in a circle lie on the circumference of the circle. 2 The sum of two opposite angles in a cyclic quadrilateral is equal to 180 degrees (supplementary angles) 3 The measure of an exterior angle is equal to the measure of the opposite interior angle.

## How to find the opposite angles of a quadrilateral PQRS?

In the above diagram, quadrilateral PQRS is inscribed in a circle. Then, its opposite angles are supplementary. Subtract 216° from each side. In (1), substitute m∠P = 72°.