Quadrilaterals - More Quadrilaterals with Parallel Opposite Sides

In a parallelogram $ABCD$, the measure of $\angle A = 75^\circ$. Find the measures of the remaining angles $\angle B$, $\angle C$, and $\angle D$.

In a parallelogram, opposite angles are equal and adjacent angles are supplementary. Given $\angle A = 75^\circ$.

1. $\angle C = \angle A$ (Opposite angles) $\implies \angle C = 75^\circ$.
2. $\angle A + \angle B = 180^\circ$ (Adjacent angles are supplementary). $\begin{array}{r} 180^\circ \\ - 75^\circ \\ \hline 105^\circ \end{array}$ So, $\angle B = 105^\circ$.
3. $\angle D = \angle B$ (Opposite angles) $\implies \angle D = 105^\circ$.

The solution uses the property that consecutive angles in a parallelogram add up to $180^\circ$ and opposite angles are congruent.

The diagonals of a rhombus are $12\text{ cm}$ and $16\text{ cm}$. Find the length of one side of the rhombus.

The diagonals of a rhombus bisect each other at $90^\circ$. Let the diagonals be $d_1 = 12\text{ cm}$ and $d_2 = 16\text{ cm}$.

1. Half-lengths of the diagonals are: $\frac{12}{2} = 6\text{ cm}$ and $\frac{16}{2} = 8\text{ cm}$.
2. In the right-angled triangle formed by the half-diagonals and the side ($s$), use Pythagoras theorem: $s^2 = 6^2 + 8^2$ $s^2 = 36 + 64$ $s^2 = 100$ $s = \sqrt{100} = 10\text{ cm}$

Because the diagonals of a rhombus are perpendicular bisectors, they divide the rhombus into four congruent right-angled triangles where the side of the rhombus is the hypotenuse.

In rectangle $RENT$, the diagonals intersect at $O$. If $OT = 3x + 1$ and $OR = 2x + 4$, find the value of $x$.

In a rectangle, the diagonals are equal in length and they bisect each other. This means all four segments from the center to the vertices are equal: $OT = OR = OE = ON$. Set $OT = OR$: $3x + 1 = 2x + 4$ Subtract $2x$ from both sides: $x + 1 = 4$ Subtract $1$ from both sides: $x = 3$

Since diagonals of a rectangle are equal and bisect each other, the distance from the intersection point to any vertex is the same.

In a square, the diagonals are equal and bisect each other. Therefore, $AC = BD$. Since $O$ is the midpoint of both diagonals, $OA = OB = OC = OD$. Given $OA = 3x - 5$ and $OB = 10$, we set them equal: $3x - 5 = 10$ $3x = 10 + 5$ $3x = 15$ $x = 5$

Because a square is a special type of rectangle, its diagonals are equal in length. Because it is also a parallelogram, those diagonals bisect each other, meaning all four half-diagonal segments are equal.

In rhombus $PQRS$, $PQ = PS$ (all sides are equal). Thus, $\triangle PQS$ is an isosceles triangle. Since $PQRS$ is a parallelogram, opposite angles are equal: $\angle PSR = \angle PQR = 120^\circ$. Adjacent angles are supplementary: $\angle SPQ = 180^\circ - 120^\circ = 60^\circ$. In $\triangle PQS$: $\angle PSQ + \angle PQS + \angle SPQ = 180^\circ$ Since $PQ=PS$, $\angle PSQ = \angle PQS$. $2 \times \angle PSQ + 60^\circ = 180^\circ$ $2 \times \angle PSQ = 120^\circ$ $\angle PSQ = 60^\circ$

We use the property that a rhombus has equal sides to form an isosceles triangle with the diagonal. Then we apply the angle sum property of triangles and the supplementary property of adjacent angles in a parallelogram.