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Answer 1

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1/16 * area of ΔABC

Step-by-step explanation:

Given vertices are A(4,6), B(1,5), C(7,2).

We know that Area of triangle having vertices (x₁,y₁), (x₂,y₂) and (x₃,y₃) is :

A = 1/2[x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)]

∴ Area of triangle ABC:

A = 1/2[4(5 - 2) + 1(2 - 6) + 7(6 - 5)]

  = 1/2[20 - 8 - 4 + 7]

  = 1/2[15]

  = 15/2 sq.units.

In ΔADE and ΔABC,

AD/AB = AE/AC = 1/4.

∠DAE = ∠BAC.

Hence ΔADE ~ ΔABC.

∴ Area of ΔACE/Area of ΔABC:

⇒ ΔADE/ΔABC = (AD/AB)²

⇒ ΔADE/(15/2)  = (1/4)²

⇒ ΔADE = 15/32

∴ Area of ΔADE : Area of ΔABC = 15/32 : 15/2

= 1/16

Therefore, Area of ΔADE = (1/16) * Area of ΔABC.

Hope it helps!

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Answer 2

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