SSC CGL Tier II 25 Oct 2015 Paper 1

Question : 11 of 100

Marks: +1, -0

In triangle ABC, DE || BC where D is a point on AB and E is a point on AC. DE divides the area of ΔABC into two equal parts. Then DB : AB is equal to

Solution:

(3)

DE || BCArea of ΔADE = Area of quadrilateral BDEC
⇒ Area of D ABC = 2 × Area of ΔADE
In ΔADE and D ABC,
∠D = ∠B ; ∠E = ∠C
∴ DADE ~ DABC
∴

{ { \Area \of } \Δ{\ABC}}/{ { \Area \of } \Δ {\ADE}}= { {AB}^{2}}/{ {AD}^{2}}

⇒{ {AB}^{2}}/{ {AD}^{2}}=2 ⇒{AB}=√{2} {AD}

⇒ A B=√{2}(A B-D B)

⇒√{2} A B-A B=√{2} D B

⇒ A B(√{2}-1)=√{2} D B

⇒{D B}/{A B}= {√{2}-1}/{√{2}}

Go to Question:

Prev Question

Next Question