Triangles

6 cm

Step 1: Given DE || BC, AD = 5 cm, AB = 15 cm, AC = 18 cm. We need to find AE. Step 2: By Basic Proportionality Theorem, DE || BC implies AD/AB = AE/AC. Step 3: Substituting known values: 5/15 = AE/18. Step 4: Simplify the left side: 1/3 = AE/18. Step 5: Cross multiply: AE = 18 × 1/3 = 6 cm.

13 m

Step 1: Draw perpendiculars from the top of shorter pole to both the ground and the taller pole. Step 2: This forms a right triangle with horizontal distance 12 m and vertical distance (11-6) = 5 m. Step 3: The distance between tops is the hypotenuse of this right triangle. Step 4: Using Pythagoras theorem: distance² = 12² + 5² = 144 + 25 = 169. Step 5: Therefore, distance = √169 = 13 m.

3:4

Step 1: Given PQ = 12 cm, QR = 16 cm, PR = 20 cm. The angle bisector from Q meets PR at point S. Step 2: By Angle Bisector Theorem, the angle bisector divides the opposite side in the ratio of the other two sides. Step 3: PS/SR = PQ/QR (sides adjacent to the bisected angle). Step 4: Substituting: PS/SR = 12/16 = 3/4. Step 5: Therefore, the angle bisector divides PR in the ratio 3:4.

4:9

Step 1: Given triangles ABC ~ PQR with AB/PQ = 4/9. Step 2: In similar triangles, all corresponding sides are in the same ratio. Step 3: Therefore, AB/PQ = BC/QR = CA/RP = 4/9. Step 4: Perimeter of ABC/Perimeter of PQR = (AB + BC + CA)/(PQ + QR + RP). Step 5: This equals (AB + BC + CA)/(PQ + QR + RP) = 4/9, since each corresponding side has the same ratio.