Find the area of a trapezium whose parallel sides measure 25 cm and 13 cm, while the other two sides measure 15cm each.

Given:

• Parallel sides: $25$ cm and $13$ cm
• Non-parallel sides: $15$ cm each

Since both non-parallel sides are equal ($15$ cm each), this is an isosceles trapezium.

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When we drop perpendiculars from the shorter parallel side to the longer one, we create two identical right triangles on the sides.

The base of each triangle $= \dfrac{25 - 13}{2} = \dfrac{12}{2} = 6$ cm

Using Pythagoras theorem:

$\text{height}^2 + 6^2 = 15^2$

$\text{height}^2 + 36 = 225$

$\text{height}^2 = 189$

$\text{height} = \sqrt{189} = \sqrt{9 \times 21} = 3\sqrt{21}$ cm

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Area of trapezium $= \dfrac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}$

Area $= \dfrac{1}{2} \times (25 + 13) \times 3\sqrt{21}$

Area $= \dfrac{1}{2} \times 38 \times 3\sqrt{21}$

Area $= 19 \times 3\sqrt{21}$

Area $= 57\sqrt{21}$ cm²