Will Approach 2 work for any type of trapezium?

Question

Homi · Accepted Answer

Understand the Question

Approach 2 works by splitting any trapezium into two familiar geometric figures: a parallelogram and a triangle.

By drawing a line from one upper vertex parallel to the opposite non-parallel side, any trapezium (whether scalene, right-angled, or isosceles) can be decomposed into these two shapes. Since standard area formulas apply to both shapes, adding their areas always yields the general area formula of a trapezium: $\dfrac{1}{2} h (a + b)$ .

Step 1 · Decompose the Trapezium

Consider trapezium $ABCD$ where side $AB \parallel DC$ , with lengths $AB = a$ , $DC = b$ , and perpendicular height $h$ . Question diagram

Draw a line $BG$ from vertex $B$ parallel to side $AD$ , with point $G$ lying on side $DC$ .

In quadrilateral $ABGD$ :

$AB \parallel DG$ (since $AB \parallel DC$ )
$AD \parallel BG$ (by construction)

Therefore, $ABGD$ is a parallelogram, and the remaining part is triangle $BGC$ .

Step 2 · Evaluate Total Area

The total area of trapezium $ABCD$ is the sum of the areas of parallelogram $ABGD$ and triangle $BGC$ :

$\text{Area of trapezium } ABCD = \text{Area of parallelogram } ABGD + \text{Area of } \Delta BGC$

Using standard formulas:

$\text{Area of parallelogram } ABGD = \text{base} \times \text{height} = a \times h$
$\text{Area of } \Delta BGC = \dfrac{1}{2} \times \text{base} \times \text{height} = \dfrac{1}{2} \times (b - a) \times h$

Summing the two areas:

\begin{aligned} \text{Area} &= ah + \dfrac{1}{2}(b - a)h \\[0.6em] &= h\left(a + \dfrac{b - a}{2}\right) \\[0.6em] &= h\left(\dfrac{2a + b - a}{2}\right) \\[0.6em] &= \dfrac{1}{2}h(a + b) \end{aligned}

Since this construction ( $BG \parallel AD$ ) and decomposition can be performed on any trapezium regardless of side lengths or angles, the method is universally valid.

Answer

Yes, Approach 2 will work for any type of trapezium.

Common Mistakes

Assuming Symmetry Restrictions: Believing this decomposition only applies to symmetric (isosceles) trapeziums. Drawing a line parallel to an opposite side is geometrically valid for any trapezium (scalene, right-angled, or isosceles).
Height Confusion: Forgetting that both the parallelogram and the triangle share the exact same perpendicular height $h$ , which is the distance between the two parallel sides.

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