Similar Triangles — Telangana (SSC) Class 10 Mathematics Solutions (Free)

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

Similar Triangles — Telangana (SSC) Class 10 Mathematics Solutions (Free)

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

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

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Similar Triangles — Textbook Questions Answered (8)

Q1: If triangle ABC is similar to triangle PQR with AB/PQ = 2/3, and area of ABC = 36 cm², find area of PQR.

Step 1: Ratio of corresponding sides = AB/PQ = 2/3 Step 2: Ratio of areas = (ratio of sides)² Step 3: Area of ABC/Area of PQR = (2/3)² = 4/9 Step 4: 36/Area of PQR = 4/9 Step 5: Area of PQR = 36 × 9/4 = 81 cm² Final Answer: Area of PQR = 81 cm²

Q2: In triangle ABC, D and E are points on AB and AC respectively such that DE ∥ BC. If AD = 3 cm, DB = 2 cm, AE = 4.5 cm, find EC.

Step 1: Since DE ∥ BC, by Basic Proportionality Theorem (Thales) Step 2: AD/DB = AE/EC Step 3: 3/2 = 4.5/EC Step 4: EC = (4.5 × 2)/3 = 9/3 = 3 cm Final Answer: EC = 3 cm

Q3: Prove that if a line is drawn parallel to one side of a triangle, it divides the other two sides proportionally.

Step 1: Let triangle ABC with line DE drawn parallel to BC, D on AB, E on AC. Step 2: We need to prove AD/DB = AE/EC Step 3: Draw EF ∥ AB (F on BC) Step 4: ADEF is a parallelogram, so AD = EF and AE = DF Step 5: In triangle ABC with parallel lines, triangles ADE and ABC are similar Step 6: By AA similarity, ∠ADE = ∠ABC and ∠AED = ∠ACB Step 7: Therefore AD/AB = AE/AC Step 8: This gives AD/DB = AE/EC (rearranging proportions) Final Answer: Line parallel to one side divides other sides proportionally

Q4: Two triangles have sides 3, 4, 5 and 6, 8, 10. Prove they are similar.

Step 1: Check if corresponding sides are proportional Step 2: Ratio = 6/3 = 2, 8/4 = 2, 10/5 = 2 Step 3: All ratios equal 2 Step 4: By SSS (Side-Side-Side) similarity criterion, the triangles are similar Step 5: Scale factor = 2:1 Final Answer: Triangles are similar by SSS with scale factor 2:1

Q5: If the areas of two similar triangles are 25 cm² and 36 cm², find the ratio of their corresponding sides.

Step 1: For similar triangles, Area ratio = (side ratio)² Step 2: Area₁/Area₂ = 25/36 Step 3: (side₁/side₂)² = 25/36 Step 4: side₁/side₂ = √(25/36) = 5/6 Final Answer: Ratio of corresponding sides = 5:6

Q6: In right triangle ABC with right angle at C, if a perpendicular from C meets AB at D, and AD = 4 cm, DB = 9 cm, find CD.

Step 1: In a right triangle, when perpendicular is drawn from right angle to hypotenuse: Step 2: CD² = AD × DB (geometric mean relation) Step 3: CD² = 4 × 9 = 36 Step 4: CD = 6 cm Final Answer: CD = 6 cm

Q7: In triangle ABC, angle A = angle P, angle B = angle Q. Prove triangle ABC is similar to triangle PQR.

Step 1: Given: ∠A = ∠P and ∠B = ∠Q Step 2: Sum of angles in any triangle = 180° Step 3: In triangle ABC: ∠A + ∠B + ∠C = 180° Step 4: In triangle PQR: ∠P + ∠Q + ∠R = 180° Step 5: Since ∠A = ∠P and ∠B = ∠Q Step 6: ∠C = 180° - ∠A - ∠B = 180° - ∠P - ∠Q = ∠R Step 7: All three angles are equal, so by AAA criterion, triangles are similar Final Answer: Triangle ABC ~ Triangle PQR (by AA or AAA similarity)

Q8: In triangle ABC, a line parallel to BC intersects AB at P and AC at Q. If AP = 2 cm, PB = 3 cm, and AQ = 4 cm, find QC.

Step 1: Since PQ ∥ BC, by Basic Proportionality Theorem Step 2: AP/PB = AQ/QC Step 3: 2/3 = 4/QC Step 4: QC = (4 × 3)/2 = 12/2 = 6 cm Final Answer: QC = 6 cm

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