Similarity Of Triangles — Important Questions

5 cm

Step 1: Given right triangle ABC with right angle at B, AB = 3 cm, BC = 4 cm. Step 2: By Pythagoras theorem: AC² = AB² + BC². Step 3: Substitute values: AC² = 3² + 4² = 9 + 16 = 25. Step 4: Taking square root: AC = √25 = 5 cm. Step 5: This is the famous 3-4-5 right triangle.

24 cm

Step 1: Given that sides are in ratio 3:4 and perimeter of smaller triangle = 18 cm. Step 2: For similar triangles, the ratio of perimeters equals the ratio of corresponding sides. Step 3: Let perimeter of larger triangle = P. Step 4: So 18/P = 3/4. Step 5: Cross multiply: 3P = 18 × 4 = 72, therefore P = 24 cm.

Right triangle

Step 1: Given PQ = 8 cm, QR = 6 cm, PR = 10 cm. Step 2: Check if it satisfies Pythagoras theorem: largest side² = sum of squares of other two sides. Step 3: PR² = 10² = 100. Step 4: PQ² + QR² = 8² + 6² = 64 + 36 = 100. Step 5: Since PR² = PQ² + QR², by converse of Pythagoras theorem, triangle PQR is a right triangle with right angle opposite to PR.

12 cm

Step 1: Given triangles ABC ~ DEF, AB = 6 cm, BC = 8 cm, DE = 9 cm. Step 2: For similar triangles, corresponding sides are proportional: AB/DE = BC/EF. Step 3: Substitute values: 6/9 = 8/EF. Step 4: Simplify: 2/3 = 8/EF. Step 5: Cross multiply: 2 × EF = 3 × 8 = 24, so EF = 12 cm.