Question

The vertices of a triangle abc are a(3,0) ,b(0,6) ,c(6,9) and de divides ab and ac in the same ratio 1:2. prove that area of triangle abc=9(area of triangle ade)

Answer 1

Arun5445 Points

Dear Student

 

By question D divides AB internally in the ratio 1 : 2; hence, the co-ordinates of D are ((1 ∙ 0 + 2 ∙ 3)/(1 + 2), (1 ∙ 6 + 2 ∙ 0)/(1 + 2)) = (6/3, 6/3) = (2, 2). 

Again, E divides AC internally in the ratio 1 : 2; hence, the co-ordinates of E are 

((1 ∙ 6 + 2 ∙ 3)/(1 + 2), (1 ∙ 9 + 2 ∙ 0)/(1 + 2)) = (12/3, 9/3) = (4, 3). 

Now, the area of the triangle ABC

= ½ |(18 + 0 + 0) - (0 + 36 + 27)| sq. units. 

= ½ |18 - 63| sq. units. 

= 45/2 sq. units. 

And the area of the triangle ADE 

= ½ |( 6 + 6 + 0) - (0 + 8 + 9)| sq. units. 

= ½ |12 - 17| sq. units.

= 5/2 sq. units. 

therefore, area of the ∆ ABC 

= 45/2 sq. units = 9 ∙ 5/2 sq. units. 

= 9 ∙ area of the ∆ ADE. Proved. 
 

Regards

S.K

Answer 2

Step-by-step explanation:

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