Question

In the figure given above (not drawn to scale), AD || GE || BC, DE = 18 cm, EC = 3 cm, AD = 35 cm.

Find:

1. AF : FC 
2. length of EF 
3. area (trapezium ADEF) : area (ΔEFC)
4. BC : GF

Answer 1

a. In △ACD and △FCE,

⇒ ∠ACD = ∠FCE   ...(Common angles)

⇒ ∠ADC = ∠FEC   ...(Corresponding angles are equal)

∴ △ACD ~ △FCE   ...(By A.A. axiom)

We know that,

Ratio of corresponding sides of similar triangle are proportional.

∴ `(FC)/(AC) = (EC)/(CD)`

⇒ `(FC)/(AC) = (EC)/(EC + ED)`

⇒ `(FC)/(AC) = 3/(3 + 18)`

⇒ `(FC)/(AC) = 3/21`

⇒ `(FC)/(AC) = 1/7`

Let FC = x and AC = 7x.

From figure,

AF = AC – FC

= 7x – x

= 6x

∴ `(AF)/(FC) = (6x)/x`

= `6/1`

= 6 : 1

Hence, AF : FC = 6 : 1.

b. Since, △ACD ~ △FCE

∴ `(EF)/(AD) = (EC)/(CD)`

⇒ `(EF)/35 = 3/21`

⇒ `EF = 3/21 xx 35`

⇒ `EF = 105/21`

⇒ EF = 5 cm

Hence, EF = 5 cm.

c. We know that,

The ratio of the area of two similar triangles is equal to the square of the ratio of any pair of the corresponding sides of the similar triangles.

∴ `("Area of"  ΔADC)/("Area of"  ΔFCE) = ((DC)/(EC))^2`

⇒ `("Area of"  ΔADC)/("Area of"  ΔFCE) = (21/3)^2`

⇒ `("Area of"  ΔADC)/("Area of"  ΔFCE) = 441/9`

⇒ `("Area of"  ΔADC)/("Area of"  ΔFCE) = 49/1`

Let area of △ADC = 49a and area of △FCE = a.

Area of trapezium ADEF = Area of △ADC – Area of △FCE

= 49a – a

= 48a

∴ `("Area of trapezium"  ADEF)/("Area of" △FCE) = (48a)/a`

= `48/1`

Hence, area of trapezium ADEF : area of △EFC = 48 : 1.

d. In △AGF and △ABC,

⇒ ∠AGF = ∠ABC   ...(Corresponding angles are equal)

⇒ ∠GAF = ∠BAC   ...(Common angles)

∴ △AGF ~ △ABC   ...(By A.A. axiom)

We know that,

Ratio of corresponding sides of similar triangle are proportional.

∴ `(AC)/(AF) = (BC)/(GF)`

From part (a),

⇒ FC = x and AC = 7x

⇒ AF = AC – FC

= 7x – x

= 6x

⇒ `(BC)/(GF) = (7x)/(6x)`

⇒ `(BC)/(GF) = 7/6`

Hence, BC : GF = 7 : 6.