in Fig. 1, DE 11 BC, AD 1 cm and BD-2 cm.What is the ratio of the ar (

1.	in Fig. 1, DE 11 BC, AD 1 cm and BD-2 cm.What is the ratio of the ar (AABC) to thear (Δ ADE) ?Fig. 1
Answer» Since DE \|\| BC, by Thales theorem AD/DB = AE/EC Therefore, both AD/DB and AE/EC = 1: 2 So AD:AB = 1:3 or AB:AD = 3:1 The triangles are similar by SAS axiom. Hence the areas of the triangles are in the ratio of squares of corresponding sides. Therefore, the ratio of area of Δ ABC to the area of Δ ADE = 3^2:1^2 = 9:1

in Fig. 1, DE 11 BC, AD 1 cm and BD-2 cm.What is the ratio of the ar (AABC) to thear (Δ ADE) ?Fig. 1

Answer»

Since DE || BC, by Thales theorem AD/DB = AE/EC Therefore, both AD/DB and AE/EC = 1: 2 So AD:AB = 1:3 or AB:AD = 3:1

The triangles are similar by SAS axiom. Hence the areas of the triangles are in the ratio of squares of corresponding sides.

Therefore, the ratio of area of Δ ABC to the area of Δ ADE = 3^2:1^2 = 9:1