□ABCD is a trapezium, AB || CD and its area is 33 cm². AB = x, CD = 2x + 1, height AM = x - 4, BC = x - 2 and AD = BC. Find the lengths of all sides of □ABCD.

In the adjoining fig. □ABCD is a trapezium AB ∥ CD and its area is 33 cm². From the information given in the figure find the lengths of all sides of the □ABCD. Fill in the empty boxes to get the solution.

Figure: AB \(= x\), DC \(= 2x + 1\), seg AM ⊥ DC with M on DC and AM \(= x - 4\), BC \(= x - 2\), and AD and BC are marked equal.

Solution: A(□ABCD) \(= \frac{1}{2}(AB + CD) \times \square\); \(33 = \frac{1}{2}(x + 2x + 1) \times \square\); ∴ \(\square = (3x + 1) \times \square\); ∴ \(3x^2 + \square - \square = 0\); ∴ \(3x(\ldots) + 10(\ldots) = 0\); ∴ \((3x + 10)(\ldots) = 0\); ∴ \(x = -\frac{10}{3}\) or \(x = \square\); AB = ..., CD = ..., AD = BC = ...

Doubtly Asked 24 Oct 2026 Save WhatsApp Report

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Answer: AB = 7 cm, CD = 15 cm, AD = BC = 5 cm

\[A = \frac{1}{2}(AB + CD) \times AM\]\[33 = \frac{1}{2}(x + 2x + 1)(x - 4)\]\[66 = (3x + 1)(x - 4)\]\[3x^2 - 11x - 4 - 66 = 0\]\[3x^2 - 11x - 70 = 0\]\[3x(x - 7) + 10(x - 7) = 0\]\[(3x + 10)(x - 7) = 0\]

\(x = -\frac{10}{3}\) is rejected (length cannot be negative), so \(x = 7\).

AB \(= 7\) cm, CD \(= 2(7) + 1 = 15\) cm, AD = BC \(= 7 - 2 = 5\) cm (height AM \(= 3\) cm).

Boxes in order: AM, \((x - 4)\), 66, \((x - 4)\), \(-11x\), 70, \((x - 7)\), \((x - 7)\), \(x - 7\), 7, 7.

Doubtly AI · 24 Oct 2026 AI-verified · Oct 2026 Report

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□ABCD is a trapezium, AB || CD and its area is 33 cm². AB = x, CD = 2x + 1, height AM = x - 4, BC = x - 2 and AD = BC. Find the lengths of all sides of □ABCD.

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