Question : The lengths of the two parallel sides of a trapezium are 28 cm and 40 cm. If the length of each of its other two sides is 12 cm, then the area (in cm2) of the trapezium is:
Option 1: $312\sqrt{5}$
Option 2: $408\sqrt{3}$
Option 3: $204\sqrt{3}$
Option 4: $504\sqrt{3}$
Correct Answer: $204\sqrt{3}$
Solution :
Let ABCD be a trapezium.
AB = 40 cm, CD = 28 cm, and AD = BC = 12 cm.
DE and CF are perpendicular to AB.
As per symmetry AE = FB = 6 cm
In the right-angled triangle CFB
$CF^2= BC^2-FB^2$
⇒ $CF^2=12^2-6^2$
⇒ $CF^2=144-36= 108$
⇒ $CF=6\sqrt3$
CF is the height of trapezium.
Now, the area of trapezium = $\frac{1}{2}\times$(sum of parallel sides) × height
= $\frac{1}{2}\times (40+28)\times 6\sqrt3$
= $\frac{1}{2}\times 68\times 6\sqrt3$
= $204\sqrt{3}$ cm$^2$
Hence, the correct answer is $204\sqrt{3}$.
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