Class 10 Maths Periodic Test 2
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Class 10 Maths Periodic Test 2
This periodic test covers three core chapters: Triangles (similarity, Basic Proportionality Theorem, criteria for similarity), Coordinate Geometry (distance formula, section formula, midpoint, collinearity), and Introduc...
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Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of a triangle PQR (see figure). Show that triangle ABC is similar to triangle PQR.
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Use SSS similarity: since AB/PQ = BC/QR = AD/PM, and D, M are midpoints, we get AB/PQ = BC/QR = AC/PR, hence △ABC ~ △PQR.
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In the given figure, DE || AC and DF || AE. Prove that BF/FE = BE/EC.
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Apply Basic Proportionality Theorem in △BAE (DF∥AE gives BF/FE = BD/DA) and in △BAC (DE∥AC gives BE/EC = BD/DA); hence BF/FE = BE/EC.
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Prove the following trigonometric identity: cosθ/(1 + sinθ) + (1 + sinθ)/cosθ = 2secθ
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Take LCM: [cos²θ + (1+sinθ)²] / [cosθ(1+sinθ)] = [cos²θ + 1 + 2sinθ + sin²θ] / [cosθ(1+sinθ)] = [2 + 2sinθ] / [cosθ(1+sinθ)] = 2/cosθ = 2secθ.
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State 'Basic Proportionality Theorem' and use it to prove the following: In a quadrilateral ABCD, diagonals AC and BD intersect each other at O such that AO/BO = CO/DO as shown in the given figure. Prove that ABCD is a trapezium.
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BPT: In a triangle, a line parallel to one side divides the other two sides proportionally. Given AO/BO = CO/DO ⇒ AO/CO = BO/DO. Consider △AOB and △COD: ∠AOB = ∠COD (vertically opposite) and AO/CO = BO/DO, so △AOB ~ △COD (SAS similarity). Hence ∠OAB = ∠OCD, so AB ∥ CD (alternate angles equal). Therefore ABCD is a trape...
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