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The area of isosceles trapezium is 96 cm2 and the height is 11.11% of the sum of parallel sides if SR is 28.56% less than PQ as shown in the figure then find the length of diagonal SQ of the trapezium.

The area of isosceles trapezium is 96 cm2 and the height is 11.11% of the sum of parallel sides if SR is 28.56% less than PQ as shown in the figure then find the length of diagonal SQ of the trapezium. Correct Answer 2√183 cm

Given:

The area of trapezium is 96 cm2.

Formula Used:

Area of trapezium = 1/2 × (sum of two parallel sides) × height

Calculation:

∵ 28.56 % = 2/7

Let PQ be 7x

⇒ SR = 5x

11.11% = 1 / 9

⇒ Height of trapezium, h = sum of parallel side × 1/9

⇒ h = (7x + 5x) × 1/9 = 4x/3

Area of trapezium = 1/2 × (7x + 5x) × 4x / 3

⇒ 96 = 8x2

⇒ x = 2√3

∵ Given trapezium is isosceles

⇒ PT = QU = (7x - 5x)/2 = x

In ∆STQ, by Pythagoras theorem,

SQ² = ST² + TQ²

⇒ SQ2 = (4x/3)² + (6x)²

= 16x²/9 + 36x² = 340x²/9

⇒ SQ2 = 340 × 12/9

∴ SQ = 4√(85/3) cm

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