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Given that ΔDEF ~ ΔABC. If the area of ΔABC is 9 cm2 and area of ΔDEF is 12cm2 , BC = 2.1 cm, then the length of EF is:
1. \(\frac{8\sqrt3}{5}cm\)
2. \(\frac{3\sqrt7}{5}cm\)
3. \(\frac{4\sqrt7}{5}cm\)
4. \(\frac{7\sqrt3}{5}cm\)

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Correct Answer - Option 4 : \(\frac{7\sqrt3}{5}cm\)

Given :-

 ΔDEF ~ ΔABC

Area of ΔABC is 9 cm2 and ΔDEF IS 12 cm

BC = 2.1 cm

Concept :-

Relation among areas, side, altitude and perimeters of similar triangle

(area of ΔABC/area of ΔDEF) = (AB/DE)2 = (BC/EF)2 = (AC/DF)2

Calculation :-

Let EF = x cm

⇒ (9/12) = (2.1/x)2

⇒ x2 = (2.1 × 2.1 × 12)/9

⇒ x2 = (7 × 7 × 3)/25

⇒ x = √((7 × 7 × 3)/25)

⇒ x = (7√3)/5

⇒ EF = (7√3)/5 cm

∴ EF is (7√3)/5 cm

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