Similarity — Maharashtra (SSC) Class 10 Mathematics Solutions (Free)

Free step-by-step Maharashtra (SSC) Class 10 Mathematics solutions for "Similarity" — important questions with detailed answers, download PDF for board exam preparation.

Similarity — Maharashtra (SSC) Class 10 Mathematics Solutions (Free)

Free step-by-step Maharashtra (SSC) Class 10 Mathematics solutions for "Similarity" — important questions with detailed answers, download PDF for board exam preparation.

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TL;DR: Free step-by-step Maharashtra (SSC) Class 10 Mathematics solutions for "Similarity" — important questions with detailed answers, download PDF for boar…

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Similarity — Textbook Questions Answered (7)

Q1: Triangles ABC and DEF are similar. If AB = 6 cm, BC = 8 cm, and DE = 9 cm, find EF.

Step 1: Understand similarity. If triangles ABC and DEF are similar, corresponding sides are proportional. Step 2: Identify corresponding sides. AB corresponds to DE, BC corresponds to EF, AC corresponds to DF. Step 3: Set up proportion. AB/DE = BC/EF. Step 4: Substitute known values. 6/9 = 8/EF. Step 5: Cross multiply. 6 * EF = 8 * 9. Step 6: Solve for EF. 6 * EF = 72, so EF = 72/6 = 12 cm. Step 7: Verify ratio. AB/DE = 6/9 = 2/3, BC/EF = 8/12 = 2/3. Ratios equal, confirming similarity. Answer: EF = 12 cm.

Q2: In triangle ABC, DE is parallel to BC. If AD = 4 cm, DB = 6 cm, and EC = 9 cm, find AE.

Step 1: Apply Basic Proportionality Theorem (BPT). If DE || BC, then AD/DB = AE/EC. Step 2: Substitute known values. 4/6 = AE/9. Step 3: Cross multiply. 4 * 9 = 6 * AE. Step 4: Solve for AE. 36 = 6 * AE, so AE = 36/6 = 6 cm. Step 5: Verify. AD/DB = 4/6 = 2/3, AE/EC = 6/9 = 2/3. Ratios equal, confirming DE || BC. Answer: AE = 6 cm.

Q3: Two similar triangles have areas 25 cm^2 and 100 cm^2. If a side of smaller triangle is 5 cm, find corresponding side of larger triangle.

Step 1: Recall area relationship for similar triangles. Ratio of areas = (Ratio of corresponding sides)^2. Step 2: Set up ratio. Area1/Area2 = (Side1/Side2)^2. Step 3: Substitute. 25/100 = (5/Side2)^2. Step 4: Simplify. 1/4 = (5/Side2)^2. Step 5: Take square root. 1/2 = 5/Side2. Step 6: Solve for Side2. Side2 = 5 * 2 = 10 cm. Step 7: Verify. (5/10)^2 = (1/2)^2 = 1/4. Area ratio = 25/100 = 1/4. Correct. Answer: Corresponding side of larger triangle is 10 cm.

Q4: State and prove the Basic Proportionality Theorem (Thales Theorem).

Statement: If a line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides those sides in the same ratio. Proof: Consider triangle ABC with DE parallel to BC, where D is on AB and E is on AC. Step 1: To prove: AD/DB = AE/EC. Step 2: Construct CF parallel to DE, meeting extended BA at F. Step 3: Since DE || BC and CF || DE, then CF || BC. Step 4: Since BCED is a parallelogram (opposite sides parallel), DE = BC and BD = EC. Step 5: Consider triangles ADE and AFC. Since DE || CF, angle ADE = angle ACF (corresponding angles). Step 6: By alternate segment theorem, angle AED = angle AFC. Step 7: Therefore, triangles ADE and AFC are similar (AA criterion). Step 8: By similarity, AD/AF = AE/AC. Step 9: But AF = AD + DF = AD + DB (since BCED is parallelogram), so AF = AD + DB. Step 10: Substituting: AD/(AD + DB) = AE/AC. Step 11: This leads to AD/DB = AE/EC (through cross multiplication and algebraic manipulation).

Q5: Triangles PQR and XYZ are similar with ratio of areas 16:25. If PQ = 12 cm, find XY.

Step 1: Use area ratio for similar triangles. Area(PQR)/Area(XYZ) = (PQ/XY)^2. Step 2: Substitute known values. 16/25 = (12/XY)^2. Step 3: Take square root of both sides. sqrt(16/25) = 12/XY. Step 4: Simplify. 4/5 = 12/XY. Step 5: Cross multiply. 4 * XY = 5 * 12. Step 6: Solve. 4 * XY = 60, so XY = 60/4 = 15 cm. Step 7: Verify. (12/15)^2 = (4/5)^2 = 16/25. Area ratio confirmed. Answer: XY = 15 cm.

Q6: In triangle ABC, a line DE parallel to BC intersects AB at D and AC at E. If AD:DB = 3:2 and AC = 20 cm, find AE.

Step 1: Apply BPT. AD/DB = AE/EC. Step 2: Given AD:DB = 3:2, so let AD = 3x and DB = 2x. Step 3: Then AB = AD + DB = 3x + 2x = 5x. Step 4: Apply proportion. 3x/2x = AE/EC, which simplifies to 3/2 = AE/EC. Step 5: From 3/2 = AE/EC, we get AE = (3/2)EC. Step 6: Also, AE + EC = AC = 20. Step 7: Substitute. (3/2)EC + EC = 20. Step 8: Combine. (3/2 + 1)EC = 20, so (5/2)EC = 20. Step 9: Solve. EC = 20 * 2/5 = 8 cm. Step 10: Find AE. AE = 20 - 8 = 12 cm. Step 11: Verify. AE/EC = 12/8 = 3/2, AD/DB = 3/2. Equal, confirming BPT. Answer: AE = 12 cm.

Q7: Two similar triangles have perimeters 60 cm and 80 cm. If the area of smaller triangle is 150 cm^2, find area of larger triangle.

Step 1: Find ratio of similarity from perimeters. Ratio = Perimeter(smaller)/Perimeter(larger) = 60/80 = 3/4. Step 2: For similar figures, ratio of areas = (Ratio of corresponding linear dimensions)^2. Step 3: Ratio of areas = (3/4)^2 = 9/16. Step 4: Set up equation. Area(smaller)/Area(larger) = 9/16. Step 5: Substitute. 150/Area(larger) = 9/16. Step 6: Cross multiply. 150 * 16 = 9 * Area(larger). Step 7: Calculate. 2400 = 9 * Area(larger). Step 8: Solve. Area(larger) = 2400/9 = 266.67 cm^2 (or 800/3 cm^2). Answer: Area of larger triangle is 800/3 cm^2 or approximately 266.67 cm^2.

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