Directions (Q. 53 to 56): Study the following diagram carefully and answer the given questions:

In ΔABC, D and E are points on AB and AC respectively such that AD = a, DB = 3a, AE = b and EC = 3b. DQ ∥ EA and EP ∥ DA are drawn. QP is joined.

53. ΔADE is similar to which of the following triangles?

I. ΔABC

II. ΔDAQ

III. ΔADQ

IV. ΔEPA

V. ΔEAP

1. I, II and IV only

2. I, III and V only

3. I, II and V only

4. I, III and IV only

54. If DE = 2 cm, then BC is equal to:

1. 4 cm

2. 6 cm

3. 7 cm

4. 8 cm

55. The ratio of the perimeters of ΔADE and ΔABC is:

1. 1 : 2

2. 1 : 3

3. 1 : 4

4. 1 : 6

56. The ratio of the areas of ΔADE and trapezium DBCE is:

1. 1 : 8

2. 1 : 9

3. 1 : 15

4. 1 : 16

53. Given,

In ΔADE and ΔABC,

∠A is common.

$\dfrac{AD}{AB} = \dfrac{a}{a + 3a} = \dfrac{1}{4}$

$\dfrac{AE}{AC} = \dfrac{b}{b + 3b} = \dfrac{1}{4}$

By SAS Similarity of Triangles, ΔADE ∼ ΔABC.

In ΔADE and ΔDAQ,

∠DQA = ∠ADE [Corresponding angle]

∠QDA = ∠DAE [Alternate Interior Angle]

∴ ΔADE ∼ ΔDAQ by AA similarity

In ΔADE and ΔEPA,

∠EPA = ∠ADE [Corresponding angle]

∠PEA = ∠DAE [Alternate Interior Angle]

∴ ΔADE ∼ ΔEPA by AA similarity

Hence, option 1 is the correct option.

54. Given,

Since ΔADE ∼ ΔABC, corresponding sides are proportional.

$\dfrac{DE}{BC} = \dfrac{AD}{AB}$

Given AD = a and DB = 3a, so AB = 4a.

$\Rightarrow \dfrac{2}{BC} = \dfrac{a}{4a} \\[1em] \Rightarrow \dfrac{2}{BC} = \dfrac{1}{4}\\[1em] \Rightarrow BC = 8 \text{ cm.}$

Hence, option 4 is the correct option.

55. Given,

$\Rightarrow \dfrac{\text{ Perimeter(ΔADE)}}{\text{ Perimeter(ΔABC)}} = \dfrac{AD}{AB} \\[1em] = \dfrac{a}{4a} \\[1em] = \dfrac{1}{4} = 1:4.$

Hence, option 3 is the correct option.

56. Given,

$\Rightarrow \dfrac{\text{ Area(ΔABC)}}{\text{ Area(ΔADE)}} = \Big(\dfrac{AB}{AD}\Big)^2 \\[1em] \Rightarrow \dfrac{\text{ Area(ΔABC)}}{\text{ Area(ΔADE)}}= \Big(\dfrac{4a}{a}\Big)^2 \\[1em] \Rightarrow \dfrac{\text{ Area(ΔABC)}}{\text{ Area(ΔADE)}}= \Big(\dfrac{4}{1}\Big)^2 \\[1em] \Rightarrow \dfrac{\text{ Area(ΔABC)}}{\text{ Area(ΔADE)}} = 16 \\[1em] \Rightarrow \text{ Area(ΔABC)} = 16 (\text{ Area(ΔADE)}) \\[1em] \Rightarrow \text{ Area quad. DBCE} = \text{ Area(ΔABC)} - \text{ Area(ΔADE)} \\[1em] \Rightarrow \text{ Area quad. DBCE} = 16 (\text{ Area(ΔADE)}) - \text{ Area(ΔADE)} \\[1em] \Rightarrow \text{ Area quad. DBCE} = 15 (\text{ Area(ΔADE)}) \\[1em] \Rightarrow \dfrac{\text{ Area(ΔADE)}}{\text{ Area quad. DBCE}} = \dfrac{1}{15} \\[1em]$

Hence, option 3 is the correct option.