Mensuration - Area of Trapezium and General Quadrilaterals

Find the area of a trapezium whose parallel sides are $14\ cm$ and $20\ cm$, and the perpendicular distance between them is $8\ cm$.

1. Identify given values: $a = 14\ cm$, $b = 20\ cm$, $h = 8\ cm$
2. Apply formula: $Area = \frac{1}{2} \times (a + b) \times h$
3. Substitute: $Area = \frac{1}{2} \times (14 + 20) \times 8$
4. Calculate: $Area = \frac{1}{2} \times 34 \times 8 = 17 \times 8 = 136\ cm^2$

We use the standard trapezium area formula by adding the parallel sides, multiplying by the height, and dividing by 2.

The diagonal of a quadrilateral is $30\ cm$ and the perpendiculars dropped on it from the opposite vertices are $10.5\ cm$ and $12.5\ cm$. Find the area.

1. Identify given values: $d = 30\ cm$, $h_1 = 10.5\ cm$, $h_2 = 12.5\ cm$
2. Apply formula: $Area = \frac{1}{2} \times d \times (h_1 + h_2)$
3. Substitute: $Area = \frac{1}{2} \times 30 \times (10.5 + 12.5)$
4. Calculate sum: $10.5 + 12.5 = 23$
5. Final calculation: $Area = 15 \times 23 = 345\ cm^2$

To find the area of a general quadrilateral, we treat it as two triangles sharing the same diagonal base and sum their individual areas.

1. Area of rhombus = $base \times altitude$ Area = $6\ cm \times 4\ cm = 24\ cm^2$
2. Also, Area of rhombus = $\frac{1}{2} \times d_1 \times d_2$ $24 = \frac{1}{2} \times 8 \times d_2$ $24 = 4 \times d_2$ $d_2 = \frac{24}{4} = 6\ cm$

Since a rhombus is a parallelogram, we first use the base and altitude to find the area. Then, we use the diagonal formula for the area of a rhombus to solve for the unknown diagonal.

Let the length of the other parallel side be $b$. Area of trapezium = $\frac{1}{2} \times (a + b) \times h$ $34 = \frac{1}{2} \times (10 + b) \times 4$ $34 = 2 \times (10 + b)$ $\frac{34}{2} = 10 + b$ $17 = 10 + b$ $b = 17 - 10 = 7\ cm$

We substitute the known values (Area, $a$, and $h$) into the trapezium area formula and solve the linear equation for the unknown side $b$.