Question

n the figure, ABCD is a quadrilateral. F is a point on AD such that = 2.1 cm and = 4.9. E and G are points on AC and AB respectively such that EF∥ CD and GE∥BC. Find ( ∆)/( ∆)

Answer 1

Answer:

Given:

1. A quadrilateral ABCD in which:
2. AF= 2.1 cm
3. FD=4.9 cm
4. EF ∥CD
5. GE∥BC

To find: $\frac{Ar(BCD)}{Ar(GEF)}$

Proof:

In Δ ABC, GE∥BC (Given)

∴ $\frac{AG}{GB} = \frac{AF}{FD}---(1)$  [If a line is drawn parallel to any one side of a triangle                                     to intersect the other two sides at distinct points, then                                     the two sides are divided in the same ratio]

Similarly, in Δ ACD,

$\frac{AE}{EC} = \frac{AF}{FD} -----(2)$

By (1) and (2),

$\frac{AG}{GB} = \frac{AE}{EC}$

∴ GF || BD                      [ By the converse of B.P.T, the above used theorem]

In Δ AGF and Δ ABD

1. ∠A= ∠A                                               (Common)
2. ∠AFG=∠ADB                                      (Corresponding angles are equal)

∴ Δ AGF ~ Δ ABD                           (By AA similarity criterion)

⇒$\frac{AF}{AD} = \frac{GF}{BD}$                                      (Similar triangles have proportional sides)

⇒$\frac{2.1}{7} =\frac{GF}{BD}$

⇒$\frac{3}{10} =\frac{GF}{BD}$

⇒ $\frac{BD}{GF}$ = $\frac{10}{3}$

Now,

$\frac{Ar(BCD)}{Ar(GEF)}$ = $(\frac{BD}{GF})^{2}$      [The ratio of the area of both triangles is proportional

                                 to the square of the ratio of their corresponding sides}    

⇒ $\frac{Ar(BCD)}{Ar(GEF)}$ = $(\frac{10}{3})^{2}$

⇒ $\frac{Ar(BCD)}{Ar(GEF)}$ = $(\frac{100}{9})$

⇒ $Ar(BCD) : Ar(GEF) = 100:9$

Hence,

$Ar(BCD) : Ar(GEF) = 100:9$

Proved.

P.F.A the rough figure.

Hope you got that.

Thank You.