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

â‡’

â‡’

â‡’

AB || DC and AB = 2 CD

In Î”AOB and Î”COD,

âˆ AOB = âˆ COD [V.O.A.]

âˆ 1 = âˆ 2 [Alternate angles]

âˆ´ Î”AOB ~ Î”COD

â‡’ ar Î”AOB: ar Î”COD = 4 : 1

âˆ AOM = âˆ DON [V.O.A.]

âˆ AMO = âˆ DNO [Each 90Â°]

âˆ´ Î”AOM ~ Î”DON [AA]

â‡’

[Corresponding sides of similar triangles]

and ar Î”ABC = ar Î”DEF

To Prove; Î”AABC â‰Œ Î”DEF

âˆ´ Î”ABC â‰Œ Î”DEF [By SSS congruency rule]

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

âˆ´

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

Similarly

Hence ar Î”DEF : ar Î”ABC = 1 : 4

[Corresponding angles of similar triangle]

âˆ´ Î”ABP ~ Î”DEQ [SAS]

âˆ´

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

Î”PAD ~ Î”QAC [AA similarity each angle = 60Â°]**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

Î”ABC ~ Î”BDE

[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.

âˆ´ Ratio of areas of triangle

or 16:81

Correct answer is (d)

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