Geometry and Trigonometry - Cyclic quadrilaterals-extended

In a cyclic quadrilateral $ABCD$, $\angle A = (2x + 15)^{\circ}$ and $\angle C = (3x - 10)^{\circ}$. Find the value of $x$ and the measure of $\angle A$.

Since $ABCD$ is a cyclic quadrilateral, opposite angles sum to $180^{\circ}$. $(2x + 15) + (3x - 10) = 180$ $5x + 5 = 180$ $5x = 175$ $x = 35$ Now, calculate $\angle A$: $\angle A = 2(35) + 15 = 70 + 15 = 85^{\circ}$

We use the property that opposite angles in a cyclic quadrilateral are supplementary. By setting up a linear equation for $x$, we solve for the variable and substitute it back into the expression for $\angle A$.

A cyclic quadrilateral $PQRS$ has side lengths $PQ = 3$, $QR = 4$, $RS = 5$, and $SP = 6$. Calculate the area of the quadrilateral.

First, find the semi-perimeter $s$: $s = \frac{3 + 4 + 5 + 6}{2} = \frac{18}{2} = 9$ Using Brahmagupta's Formula: $Area = \sqrt{(9-3)(9-4)(9-5)(9-6)}$ $Area = \sqrt{6 \cdot 5 \cdot 4 \cdot 3}$ $Area = \sqrt{360}$ $Area = 6\sqrt{10} \approx 18.97 \text{ units}^2$

Brahmagupta's formula is specifically used for the area of cyclic quadrilaterals when all four side lengths are known. We first calculate the semi-perimeter and then apply the square root of the product of the differences between the semi-perimeter and each side.

In cyclic quadrilateral $WXYZ$, the exterior angle at vertex $W$ is $112^{\circ}$. Find the measure of the interior angle $\angle Y$.

By the Exterior Angle Property of cyclic quadrilaterals, the exterior angle is equal to the interior opposite angle. $\text{Exterior } \angle W = \angle Y$ $112^{\circ} = \angle Y$ Therefore, $\angle Y = 112^{\circ}$.

This property is a direct consequence of the fact that the exterior angle and the adjacent interior angle sum to $180^{\circ}$ (straight line), and the adjacent interior angle and the opposite interior angle also sum to $180^{\circ}$ (cyclic property).

1. Opposite angles in a cyclic quadrilateral sum to $180^{\circ}$. Therefore: $\angle K + \angle M = 180^{\circ}$
2. Substitute the expressions: $(4y + 20) + (y + 10) = 180$ $5y + 30 = 180$
3. Solve for $y$: $5y = 150$ $y = 30$
4. Calculate $\angle K$: $\angle K = 4(30) + 20 = 140^{\circ}$
5. To find $\angle L$, note that without more information about $\angle N$, we can only state: $\angle L = 180^{\circ} - \angle N$

This problem uses the fundamental property that opposite angles of a cyclic quadrilateral are supplementary. By forming an equation with the algebraic expressions provided, we can solve for the unknown variable.

1. Using Ptolemy's Theorem: $AC \cdot BD = (AB \cdot CD) + (BC \cdot AD)$
2. Substitute the given side lengths: $AC \cdot BD = (5 \cdot 5) + (8 \cdot 8)$ $AC \cdot BD = 25 + 64 = 89$
3. Since the opposite sides are equal ($AB=CD$ and $BC=DA$), $ABCD$ is a rectangle (or an isosceles trapezoid that happens to be a rectangle in this symmetry). In a rectangle, $AC = BD$.
4. Therefore: $AC^2 = 89$ $AC = BD = \sqrt{89} \approx 9.43 \text{ cm}$

Ptolemy's theorem provides a direct relationship between the sides and diagonals. Because the opposite sides are equal, we can deduce properties about the diagonals through symmetry.