A trapezium $ABCD$ has side $AD$ parallel to $BC$, $\angle$BAD = $90°$, $BC = 3$cm and $AD = 8$cm. If the perimeter of this trapezium is $36$cm, then its area, in sq. cm, is

We need to find the area when we know some sides and the perimeter.

Trapezium ABCD with AD || BC (parallel sides)

∠BAD = 90° (right angle)

BC = 3 cm, AD = 8 cm

Perimeter = 36 cm

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Since ∠BAD = 90° and AD || BC, we can drop a perpendicular from C to meet AD at point E. This creates rectangle ABCE.

ABCE is a rectangle because:

BA ⊥ AD (given ∠BAD = 90°)

CE ⊥ AD (constructed perpendicular)

AD || BC (given)

So ABCE has all angles = 90°

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In rectangle ABCE:

BC = AE = 3 cm (opposite sides of rectangle are equal)

Since AD = 8 cm and AE = 3 cm:

ED = AD - AE = 8 - 3 = 5 cm

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Triangle CED is a right triangle (since ∠CED = 90°) with:

ED = 5 cm (one leg)

CE = height of trapezium = h (other leg)

CD = hypotenuse

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Let CE = AB = h (height of trapezium)

From the perimeter:

AB + BC + CD + AD = 36

h + 3 + CD + 8 = 36

h + CD = 25

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In right triangle CED:

$CE^2 + ED^2 = CD^2$

$h^2 + 5^2 = CD^2$

$h^2 + 25 = CD^2$

Since h + CD = 25, substitute CD = 25 - h:

$h^2 + 25 = (25 - h)^2$

$h^2 + 25 = 625 - 50h + h^2$

$25 = 625 - 50h$

$50h = 600$

$h = 12$

Therefore: CD = 25 - 12 = 13

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The triangle CED has sides 5, 12, 13 - this is a Pythagorean triplet since $5^2 + 12^2 = 25 + 144 = 169 = 13^2$

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

$= \frac{1}{2} \times (BC + AD) \times h$

$= \frac{1}{2} \times (3 + 8) \times 12$

$= \frac{1}{2} \times 11 \times 12$

$= 66$ cm²

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Answer: 66 cm²

When dealing with trapeziums with a right angle, creating a rectangle by dropping perpendiculars often simplifies the problem and reveals useful right triangles!