Quadrilaterals - Kite and Trapezium

Find the area of a trapezium whose parallel sides are $12\text{ cm}$ and $20\text{ cm}$ and the distance between them is $8\text{ cm}$.

Given: $a = 12\text{ cm}$, $b = 20\text{ cm}$, and $h = 8\text{ cm}$. Using the formula: $\text{Area} = \frac{1}{2} \times (a + b) \times h$ $\text{Area} = \frac{1}{2} \times (12 + 20) \times 8$ $\text{Area} = \frac{1}{2} \times 32 \times 8$ $\text{Area} = 16 \times 8 = 128\text{ cm}^2$

To find the area of a trapezium, add the lengths of the parallel sides, multiply by the height (perpendicular distance), and then divide by 2.

In a kite $PQRS$, the lengths of the diagonals are $10\text{ cm}$ and $24\text{ cm}$. Calculate its area.

Given: $d_1 = 10\text{ cm}$ and $d_2 = 24\text{ cm}$. Using the area formula for a kite: $\text{Area} = \frac{1}{2} \times d_1 \times d_2$ $\text{Area} = \frac{1}{2} \times 10 \times 24$ $\text{Area} = 5 \times 24 = 120\text{ cm}^2$

The area of a kite is half the product of the lengths of its diagonals.

In an isosceles trapezium $ABCD$, if $\angle A = 70^\circ$ and $AB \parallel DC$, find the remaining angles.

Since $ABCD$ is an isosceles trapezium with $AB \parallel DC$: 1. Base angles are equal, so $\angle B = \angle A = 70^\circ$. 2. Adjacent angles between parallel sides are supplementary: $\angle D = 180^\circ - 70^\circ = 110^\circ$ 3. Similarly, $\angle C = 110^\circ$. The calculation for the total sum is: $\begin{array}{r} 70 \\ 70 \\ 110 \\ + 110 \\ \hline 360 \end{array}$

In an isosceles trapezium, angles sharing the same base are equal, and the sum of angles on a non-parallel side is $180^\circ$.

$\text{Area} = \frac{1}{2} \times (a + b) \times h$ $480 = \frac{1}{2} \times (20 + b) \times 15$ $480 \times 2 = 15(20 + b)$ $960 = 300 + 15b$ $660 = 15b$ $b = \frac{660}{15} = 44\text{ cm}$

We use the area formula for a trapezium. Given Area ($480$), height ($15$), and one side ($20$), we solve the linear equation for the unknown side $b$.

$\text{Area of Kite} = \frac{1}{2} \times d_1 \times d_2$ $72 = \frac{1}{2} \times 12 \times d_2$ $72 = 6 \times d_2$ $d_2 = \frac{72}{6} = 12\text{ cm}$

The area of a kite depends on the product of its diagonals. By substituting the given area and one diagonal into the formula, we find the second diagonal.