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`        The vertices of a triangle ABC are (3,0) (0,6)&(6,9). The line DE divides AB and AC in the ratio of 1:2.prove that∆ABC=9∆ADE.`
2 years ago

Arun
23832 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.  RegardsArun (askIITians forum expert)
```
2 years ago
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• Mind Map
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