Question

Question 1 In figure.6.17. (i) and (ii), DE || BC. Find EC in (i) and AD in (ii). (i) (ii)

Class 10 - Math - Triangles Page 128

Answer 1

(i) In △ ABC, DE∥BC (Given)

∴ AD/DB = AE/EC [By using Basic proportionality theorem]

⇒ 1.5/3 = 1/EC

⇒ Σ EC = 3/1.5

EC = 3×10/15 = 2 cm
Hence, EC = 2 cm.

(ii) In △ ABC, DE∥BC (Given)
∴ AD/DB = AE/EC [By using Basic proportionality theorem]
⇒ AD/7.2 = 1.8/5.4
⇒ AD = 1.8×7.2/5.4 = 18/10 × 72/10 × 10/54 = 24/10
⇒ AD = 2.4
Hence, AD = 2.4 cm.

Answer 2

Hi

bhai

answer is here

I ) ∆ ABC s DE || BC

.°. AD/BD = AE/EC ( from formula )

=> 1.5/3 = 1/EC

=> EC × 1.5 = 3 × 1

=> EC = 3/1.5 = 30/15 = 2 cm

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ii) ∆ABC s DE || BC

.°. AD/BD = AE/EC

=>AD/7.2 = 1.8/5.4

=> AD/7.2 = 18/54

=> AD = 18/54 × 7.2

=> AD = 18/54 × 72/10 = 24/10 = 2.4 cm

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