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: $504\sqrt{3}$
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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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