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A circle of radius 3.5 cm is drawn inside the trapezium ABCD as shown in the figure. AE and BF are perpendicular to side CD. If AD = BC = 25 cm, find the ratio of the area of trapezium ABCD to that of trapezium ABCE.

A circle of radius 3.5 cm is drawn inside the trapezium ABCD as shown in the figure. AE and BF are perpendicular to side CD. If AD = BC = 25 cm, find the ratio of the area of trapezium ABCD to that of trapezium ABCE. Correct Answer 31 : 19

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AB and EF are parallel, and AE and BF are perpendicular to CD

⇒ AE = BF = 2 × (radius of inscribed circle) = 2 × 3.5 = 7 cm

From the figure, we know that AB = EF = 2 × (radius of inscribed circle) = 7 cm

ADE is a right-angled triangle. Applying Pythagoras’ theorem,

(DE)2 = (AD)2 – (AE)2 = 625 – 49 = 576

⇒ DE = 24 cm

By RHS test, we know that ∆ADE is congruent to ∆BFC

⇒ FC = DE = 24 cm

CD = DE + EF + FC = 55 cm

CE = EF + FC = 31 cm

Area of trapezium ABCD = ½ × (AB + CD) × AE

Area of trapezium ABCE = ½ × (AB + CE) × AE

(Area of trapezium ABCD) / (Area of trapezium ABCE) = (AB + CD)/(AB + CE) = (55 + 7)/(31 + 7) = 62/38 = 31/19

∴ The ratio of the areas is 31 : 19

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