Question

ABCD is an isosceles trapezium with AB parallel to DC, AD= BC= 12 cm, angle A=60° and DC= 16 cm. Taking √3= 1.732, find length of side AB and the area of the trapezium ABCD.

Answer 1

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Answer 2

Answer:

AB = 28 cm

Area = 207.84 cm²

Step-by-step explanation:

Construction: Draw Perpendicular DM and CN such that, DM⊥AB and CN⊥AB.

Then, cos 60° = $\frac{AM}{AD}=\frac{AM}{12}$

⇒ $\frac{1}{2}=\frac{AM}{12}$

⇒ AM = 6 cm

Similarly, BN = 6 cm

Given, DC = 16 cm

Since, DMNC is a square

∴ DC = MN

⇒ AB = AM + DC + NB

Thus, AB = 6 + 16 + 6 = 28 cm

For finding Area of Trapezoid firstly we have to find height (DM or CN)

DM² = AD² - AM²

⇒ DM² = 144 - 36

⇒ DM = 6√3 = 6 × 1.732 = 10.392

∴ Area = $\frac{1}{2}$ × (Sum of bases) × Height

⇒ Area = $\frac{1}{2}$ × (28 + 12) × 10.392

⇒ Area = 207.84 cm²