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**Q.1. Let ΔABC ~ ΔDEF and their areas be, respectively, 64 cm ^{2} and 121 cm^{2}. If EF = 15.4 cm, find BC.**

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AB || DC and AB = 2 CD

In ΔAOB and ΔCOD,

∠AOB = ∠COD [V.O.A.]

∠1 = ∠2 [Alternate angles]

∴ ΔAOB ~ ΔCOD

⇒ ar ΔAOB: ar ΔCOD = 4 : 1

∠AOM = ∠DON [V.O.A.]

∠AMO = ∠DNO [Each 90°]

∴ ΔAOM ~ ΔDON [AA]

⇒

[Corresponding sides of similar triangles]

and ar ΔABC = ar ΔDEF

To Prove; ΔAABC ≌ ΔDEF

∴ ΔABC ≌ ΔDEF [By SSS congruency rule]

Proof: D and E are mid-points of sides BC and CA respectively

∴

[Line segment joining the mid-points of two sides of triangle is parallel to the third side and half of it.]

Similarly

Hence ar ΔDEF : ar ΔABC = 1 : 4

[Corresponding angles of similar triangle]

∴ ΔABP ~ ΔDEQ [SAS]

∴

From equation (i) and (ii)**Q.7. Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of its diagonals.****Sol.** Let the side of the square ABCD be a

ΔPAD ~ ΔQAC [AA similarity each angle = 60°]**Q.8. ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Ratio of the areas of triangles ABC and BDE is****(A) 2:1 ****(B) 1:2 ****(C) 4:1 ****(D) 1:4****Sol.** **Justification:** Let AB = BC = CA = a

D is the mid point of BC

ΔABC ~ ΔBDE

[Given that they are equilateral triangles]**Q.9. Sides of two similar triangles are in the ratio 4:9. Areas of these triangles are in the ratio****(A) 2:3****(B) 4:9****(C) 81:16****(D) 16:81****Sol.** **Justification:** Areas of two similar triangles are in the ratio of the squares of their corresponding sides.

∴ Ratio of areas of triangle

or 16:81

Correct answer is (d)

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