THE VERTICES OF TRIANGLE ABC ARE A(4,6) , B(1,5) , C(7,2). A LINE IS DRAWN TO INTERSECT SIDES AB AND AC AT D AND E RESPECTIVELY SUCH THAT AD/AB=AE/AC=1/4. CALCULATE THE AREA OF TRIANGLE ADE AND COMPARE IT WITH AREA OF TRANGLE ABC.
Answers
Answer:
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!
Answer:
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!