In a trapezium $\mathrm{ABCD}, \mathrm{AB}$ is parallel to $\mathrm{DC}, \mathrm{BC}$ is perpendicular to DC and $\angle \mathrm{BAD}=45^{\circ}$. If $\mathrm{DC}=5 \mathrm{~cm}, \mathrm{BC}=4 \mathrm{~cm}$, the area of the trapezium in sq cm is

Given information:

ABCD is a trapezium with AB || DC

BC ⊥ DC (BC is perpendicular to DC)

∠BAD = 45°

DC = 5 cm, BC = 4 cm

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The key insight is using the 45° angle to find the length of AB.

Since BC ⊥ DC, if we drop a perpendicular from point A to line DC, let's call the foot of this perpendicular point E. This creates:

AE = BC = 4 cm (since ABCE forms a rectangle)

Triangle AED where ∠EAD = 45°

Since AB || DC and AE ⊥ DC, we have ∠BAE = 0°. Therefore, ∠EAD = ∠BAD = 45°.

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In triangle AED:

∠AED = 90° (AE ⊥ DC)

∠EAD = 45° (given)

∠ADE = 45° (since angles in triangle sum to 180°)

In a 45-45-90 triangle, the two legs are equal in length.

Since AE = 4 cm, we get ED = 4 cm as well.

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Now we can find AB:

AB = EC (since ABCE is a rectangle)

EC = DC + ED = 5 + 4 = 9 cm

Therefore, AB = 9 cm.

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Using the trapezium area formula:

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

$\text{Area} = \tfrac{1}{2} \times (AB + DC) \times BC$

$\text{Area} = \tfrac{1}{2} \times (9 + 5) \times 4$

$\text{Area} = \tfrac{1}{2} \times 14 \times 4 = 28 \text{ sq cm}$

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The reference solution breaks this into:

Rectangle area: 5 × 4 = 20 sq cm

Triangle area: $\tfrac{1}{2} \times 4 \times 4 = 8$ sq cm

Total: 20 + 8 = 28 sq cm

Answer: 28 sq cm