Question

The vertices of a triangle ABC are A(3, 0), B(0, 6) and C(6, 9) and DE divides AB and AC in the ratio 1 : 2. Prove that ABC = 9 ADE.

Answer 1

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.

Hope it helps

Answer 2

Step-by-step explanation:

22 क्वेश्चन हिंदी दिवस ए बी इन द इन द रेश्यो वन डे यू टू हेल्प द कोऑर्डिनेटर ऑफ द ईयर 1.0 प्लस 2.31 प्लस 2 प्लस 6 प्लस 2.01 प्लस 2 इक्वल टो 1 बाय 3 माइनस इज इक्वल टू टू टू टू टू टू टू टू