Question

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

Answer 1

$\huge \underline \mathbb {SOLUTION:-}$

(i) Given, in ∆ ABC, DE∥BC

Therefore:

AD/DB = AE/EC [Using Basic proportionality theorem]

➠ 1.5/3 = 1/EC

➠ EC = 3/1.5

EC = 3×10/15

➠ EC = 2 cm

• Hence, EC = 2 cm.

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(ii) Given, in ∆ ABC, DE∥BC

Therefore:

AD/DB = AE/EC [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 cm

• Hence, AD = 2.4 cm.

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Answer 2

(i) Given, in △ ABC, DE∥BC

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

⇒1.5/3 = 1/EC

⇒EC = 3/1.5

EC = 3×10/15 = 2 cm

Hence, EC = 2 cm.

(ii) Given, in △ ABC, DE∥BC

∴ AD/DB = AE/EC [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.