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In the figure given below, ABCF is an isosceles trapezium with AB || CF and BC = AF. AE and BD are two perpendicular drawn to the side CF. If perimeters of the trapezium ABCF, rectangle ABDE and triangle AEF are 28 cm, 20 cm and 12 cm respectively and AF : FE = 5 : 3, what is the area of the trapezium ABCF (in cm2)?

In the figure given below, ABCF is an isosceles trapezium with AB || CF and BC = AF. AE and BD are two perpendicular drawn to the side CF. If perimeters of the trapezium ABCF, rectangle ABDE and triangle AEF are 28 cm, 20 cm and 12 cm respectively and AF : FE = 5 : 3, what is the area of the trapezium ABCF (in cm2)? Correct Answer 36

Let AB = ED = a, AE = BD = b, AF = BC = c and FE = DC = d

⇒ Perimeter of the trapezium ABCF = AB + BC + DC + ED + EF + AF

⇒ 28 = 2a + 2c + 2d

⇒ 14 = a + c + d → 1

⇒ Perimeter of the rectangle ABDE = AB + BD + DE + AE

⇒ 20 = 2a + 2b

⇒ 10 = a + b → 2

⇒ Perimeter of triangle AEF = AE + EF + FA

⇒ 12 = c + b + d → 3

Equations 3 + 2 – 1

⇒ 8 = 2b

⇒ b = 4 cm

Substituting in equation 2,

⇒ a = 6 cm

Substituting in equation 3,

⇒ c + d = 8

Given c : d = 5 : 3

⇒ d = 3 cm and c = 5 cm

⇒ Area of trapezium = 1/2 × (sum of parallel sides) × Height

∴ Area of trapezium = 1/2 × (6 + 12) × 4 = 36 cm2

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