Geometry and Trigonometry - Properties of quadrilaterals

In a quadrilateral $ABCD$, the interior angles are given as $x$, $2x$, $3x$, and $4x$. Calculate the value of $x$ and the size of the largest angle.

The sum of the interior angles of a quadrilateral is $360^\circ$. $x + 2x + 3x + 4x = 360^\circ$ $10x = 360^\circ$ $x = \frac{360^\circ}{10} = 36^\circ$ The largest angle is $4x$: $4 \times 36^\circ = 144^\circ$

We use the angle sum property of quadrilaterals to set up a linear equation and solve for the unknown variable $x$.

A trapezium has parallel sides of length $12\text{ cm}$ and $18\text{ cm}$. If the perpendicular distance between these sides is $10\text{ cm}$, find the area of the trapezium.

Using the area formula for a trapezium: $A = \frac{1}{2}(a + b)h$ $A = \frac{1}{2}(12 + 18) \times 10$ $A = \frac{1}{2}(30) \times 10$ $A = 15 \times 10 = 150\text{ cm}^2$

Substitute the given lengths of the parallel sides ($a$ and $b$) and the height ($h$) into the area formula.

In a rhombus $PQRS$, the diagonals $PR$ and $QS$ have lengths $16\text{ cm}$ and $12\text{ cm}$ respectively. Find the length of one side of the rhombus.

The diagonals of a rhombus bisect each other at $90^\circ$. Let the intersection be $O$. $PO = \frac{1}{2}PR = \frac{16}{2} = 8\text{ cm}$ $QO = \frac{1}{2}QS = \frac{12}{2} = 6\text{ cm}$ In right-angled triangle $\triangle POQ$, using Pythagoras' theorem: $PQ^2 = PO^2 + QO^2$ $PQ^2 = 8^2 + 6^2$ $PQ^2 = 64 + 36 = 100$ $PQ = \sqrt{100} = 10\text{ cm}$

We use the property that diagonals of a rhombus are perpendicular bisectors of each other to create a right-angled triangle and then apply the Pythagorean theorem.

1. In a parallelogram, consecutive angles are supplementary (add up to $180^\circ$). $\angle QRS = 180^\circ - \angle PQR = 180^\circ - 115^\circ = 65^\circ$
2. Opposite angles are equal. $\angle RSP = \angle PQR = 115^\circ$ $\angle SPQ = \angle QRS = 65^\circ$

Final values: $\angle QRS = 65^\circ$, $\angle RSP = 115^\circ$, $\angle SPQ = 65^\circ$.

We use the properties of parallelograms: adjacent angles are supplementary because they are co-interior angles between parallel lines, and opposite angles are congruent.

1. In a kite, diagonals intersect at $90^\circ$. This creates four right-angled triangles.
2. In $\triangle KXI$, use Pythagoras' theorem: $KI^2 = KX^2 + IX^2$ $KI^2 = 4^2 + 3^2 = 16 + 9 = 25$ $KI = \sqrt{25} = 5\text{ cm}$
3. In $\triangle TXI$, use Pythagoras' theorem: $IT^2 = XT^2 + IX^2$ $IT^2 = 10^2 + 3^2 = 100 + 9 = 109$ $IT = \sqrt{109} \approx 10.44\text{ cm}$

Side $KI = 5\text{ cm}$, Side $IT = \sqrt{109}\text{ cm}$.

The diagonals of a kite are perpendicular. By treating the intersection as the origin of four right-angled triangles, we can find the side lengths using the Pythagorean theorem.