Geometry

13 m

Step 1: Let AB = 6 m (shorter pole) and CD = 11 m (taller pole), with AC = 12 m (distance between feet). Step 2: Draw BE perpendicular to CD. Then DE = CD - CE = 11 - 6 = 5 m (difference in heights), and BE = AC = 12 m (horizontal distance). Step 3: In right triangle BED: BD² = BE² + DE² = 12² + 5² = 144 + 25 = 169. Step 4: BD = √169 = 13 m. Common mistake: Students add heights (6 + 11 = 17) or use the full height of taller pole instead of the difference in heights (5 m). The key insight is to form a right triangle using the difference in heights and the horizontal distance.

6 cm

Step 1: When a perpendicular is drawn from the right angle vertex to the hypotenuse, the altitude is the geometric mean of the two segments of the hypotenuse. Here PS ⊥ QR. Step 2: Using the geometric mean relation (from similar triangles): PS² = QS × SR. Step 3: PS² = 4 × 9 = 36. Step 4: PS = √36 = 6 cm. This result comes from the similarity: △PQS ~ △RSP (AA), giving PS/QS = SR/PS, so PS² = QS × SR. Common mistake: Students add QS and SR and take half (giving 6.5), or use PS = QS + SR = 13 directly.

8 cm

Step 1: For similar triangles, the ratio of corresponding sides equals the ratio of perimeters. Step 2: Ratio of perimeters = 30:20 = 3:2. Step 3: If one side of first triangle = 12 cm, let corresponding side of second = x cm. Then 12/x = 3/2. Step 4: 3x = 24, so x = 8 cm. Common mistake: Students use ratio of areas instead of ratio of perimeters, computing √(30/20) = √(3/2), giving an incorrect answer. For perimeters, the ratio is direct (not squared).

13 cm

Step 1: The radius OA is perpendicular to the tangent PA at the point of contact A (radius ⊥ tangent). Step 2: Triangle OAP is right-angled at A. Step 3: Using Pythagoras theorem: OP² = OA² + PA² = 5² + 12² = 25 + 144 = 169. Step 4: OP = √169 = 13 cm. Common mistake: Students sometimes use OP² = PA² - OA², reversing the formula. Since OP is the hypotenuse (longest side), OP² = OA² + PA².