Understanding Quadrilaterals - Properties of Parallelogram, Rhombus, Rectangle, and Square

In a parallelogram $ABCD$, the measure of $\angle A$ is $(3x - 20)^\circ$ and the measure of $\angle B$ is $(x + 40)^\circ$. Find the measure of all the angles of the parallelogram.

Step 1: Identify the relationship between $\angle A$ and $\angle B$. In a parallelogram, adjacent angles are supplementary. $\angle A + \angle B = 180^\circ$ Step 2: Substitute the given expressions: $(3x - 20) + (x + 40) = 180$ $4x + 20 = 180$ Step 3: Solve for $x$: $4x = 180 - 20$ $4x = 160$ $x = 40$ Step 4: Calculate individual angles: $\angle A = 3(40) - 20 = 120 - 20 = 100^\circ$ $\angle B = 40 + 40 = 80^\circ$ Step 5: Use the property that opposite angles are equal: $\angle C = \angle A = 100^\circ$ $\angle D = \angle B = 80^\circ$

This solution uses the property that adjacent angles in a parallelogram add up to $180^\circ$ to form a linear equation. Once $x$ is found, we apply the property that opposite angles are congruent.

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

Step 1: Recall that diagonals of a rhombus bisect each other at right angles ($90^\circ$). Let the diagonals be $d_1 = 16 \text{ cm}$ and $d_2 = 12 \text{ cm}$. Step 2: The segments of the diagonals from the center to the vertices are half the total length: $\text{Half-diagonal 1} = \frac{16}{2} = 8 \text{ cm}$ $\text{Half-diagonal 2} = \frac{12}{2} = 6 \text{ cm}$ Step 3: These segments form a right-angled triangle with the side of the rhombus ($s$) as the hypotenuse. Apply Pythagoras Theorem: $s^2 = 8^2 + 6^2$ $s^2 = 64 + 36$ $s^2 = 100$ Step 4: Take the square root: $s = \sqrt{100} = 10 \text{ cm}$

This problem uses the property that rhombus diagonals are perpendicular bisectors. By considering one of the four small right-angled triangles formed inside the rhombus, we use the Pythagorean theorem to find the side length.

1. In a rectangle, diagonals are equal: $PR = QS$.
2. Diagonals also bisect each other, so $OP = OR = \frac{1}{2}PR$ and $OQ = OS = \frac{1}{2}QS$.
3. Since $PR = QS$, their halves are also equal: $OP = OS$.
4. Equating the given expressions: $2x + 4 = 3x + 1$ $4 - 1 = 3x - 2x$ $x = 3$
5. Now, find $OP$: $OP = 2(3) + 4 = 6 + 4 = 10$
6. Since $O$ is the midpoint of $PR$: $PR = 2 \times OP = 2 \times 10 = 20$ Therefore, $x = 3$ and $PR = 20$ units.

This problem uses the property that diagonals of a rectangle are equal and bisect each other, making the distance from the intersection to any vertex identical.

1. In square $ABCD$, $\angle DAB = 90^\circ$ because all angles of a square are right angles.
2. In a square, all sides are equal, so $AD = AB$.
3. The diagonal $AC$ acts as a transversal for the square. In $\triangle ADC$, we have $AD = DC$ (sides of a square).
4. This makes $\triangle ADC$ an isosceles right-angled triangle.
5. Therefore, $\angle CAD = \angle ACD$.
6. In $\triangle ADC$: $\angle CAD + \angle ACD + \angle D = 180^\circ$ $2 \times \angle CAD + 90^\circ = 180^\circ$ $2 \times \angle CAD = 90^\circ$ $\angle CAD = 45^\circ$

This demonstrates that the diagonal of a square bisects the vertex angle.