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Consider the triangle shown in the figure where BC = 12 cm, Db = 9 cm, CD = 6 cm and What is the ratio of the perimeter of the triangle ADC to that of the triangle BDC ?

- a)7:9
- b)8:9
- c)6:9
- d)5:9
- e)None of these

Correct answer is option 'A'. Can you explain this answer?

Here, ∠ACB = c+180-(2c-b) = 180-(b+c)

So, We can say that Δ BCD and &delta ABC will be similar.

According to property of similarity,

AB/12 = 12/9

Hence,

AB = 16

AC/6 = 12/9

AC = 8

Hence, AD = 7 and AC = 8

Now,

Perimeter of Delta; ADC / Perimeter of &delta BDC,

= (6+7+8)/(9+6+12)

= 21/27 = 7/9.

So, We can say that Δ BCD and &delta ABC will be similar.

According to property of similarity,

AB/12 = 12/9

Hence,

AB = 16

AC/6 = 12/9

AC = 8

Hence, AD = 7 and AC = 8

Now,

Perimeter of Delta; ADC / Perimeter of &delta BDC,

= (6+7+8)/(9+6+12)

= 21/27 = 7/9.

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Here, ∠ACB = c+180-(2c-b) = 180-(b+c)So, We can say that Δ BCD and &delta ABC will be similar.According to property of similarity,AB/12 = 12/9Hence,AB = 16AC/6 = 12/9AC = 8Hence, AD = 7 and AC = 8Now,Perimeter of Delta; ADC / Perimeter of &delta BDC,= (6+7+8)/(9+6+12)= 21/27 = 7/9.