Question

(iii) AABC and ADEF are similar and BC and EF are corresponding sides.
Areas of AABC and ADEF are 25 cm² and 16 cm² respectively. If BC =
3.75 cm then length of EF will be-
(a) 2.4 cm (b) 3 cm (c) 5 cm (d) 4 cm​

Answer 1

(i) Given : ∆ ABC ∼ ∆DEF ,area of (ΔABC) = 16 cm², area (ΔDEF) = 25 cm² and BC = 2.3 cm.

We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides.

ar(ΔABC)/ar(ΔDEF) =(BC/EF)²

16/25 = (2.3/EF)²

√16/25 = (2.3/EF)

⅘ = 2.3/EF

4EF = 2.3 × 5

EF = 11.5/4

EF = 2.875 cm

Hence, the length of EF is 2.875 cm.