Directions (Q. 57 to 60): Study the given information and answer the questions that follow:

In the given figure ABCD is a trapezium in which DC is parallel to AB. AB = 16 cm and DC = 8 cm, OD = 5 cm, OB = (y + 3) cm, OA = 11 cm and OC = (x − 1) cm.

57. From the given figure name the pair of similar triangles :

1. ΔAOD, ΔOBC

2. ΔCOD, ΔAOB

3. ΔADB, ΔACB

4. ΔCOD, ΔCOB

58. The corresponding proportional sides with respect to the pair of similar triangles obtained above is :

1. $\dfrac{CD}{AB} = \dfrac{OC}{OA} = \dfrac{OD}{OB}$

2. $\dfrac{AD}{BC} = \dfrac{OC}{OA} = \dfrac{OD}{OB}$

3. $\dfrac{AD}{BC} = \dfrac{BD}{AC} = \dfrac{AB}{DC}$

4. $\dfrac{OD}{OB} = \dfrac{CD}{CB} = \dfrac{OC}{OA}$

59. The ratio of the sides of the pair of similar triangles is:

1. 1 : 3

2. 1 : 2

3. 2 : 3

4. 3 : 1

60. Using the ratio of sides of the pair of similar triangles, the values of x and y are respectively :

1. x = 4.6, y = 7

2. x = 7, y = 7

3. x = 6.5, y = 7

4. x = 6.5, y = 2

57. Given,

In ΔAOB and ΔCOD,

∠AOB = ∠COD [Vertically opposite angle are equal]

∠OAB = ∠OCD [Alternate interior angle are equal]

∴ ΔAOB ∼ ΔCOD (By A.A. axiom)

Hence, option 2 is the correct option.

58. Given,

ΔAOB ∼ ΔCOD.Since the triangles are similar, the ratios of the corresponding sides are equal,

$\dfrac{CD}{AB} = \dfrac{OC}{OA} = \dfrac{OD}{OB}$

Hence, option 1 is the correct option.

59. Given,

ΔAOB ∼ ΔCOD. Since the triangles are similar, the ratios the corresponding sides are equal,

Ratio = $\dfrac{CD}{AB} = \dfrac{8}{16} = \dfrac{1}{2}$ = 1 : 2.

Hence, option 2 is the correct option.

60. Given,

ΔAOB ∼ ΔCOD. Since the triangles are similar, the ratios the corresponding sides are equal,

$\dfrac{CD}{AD} = \dfrac{OC}{OA} = \dfrac{OD}{OB} = \dfrac{1}{2}$

Solving,

$\dfrac{OC}{OA} = \dfrac{1}{2}$

Substituting values we get :

$\Rightarrow \dfrac{(x - 1)}{11} = \dfrac{1}{2} \\[1em] \Rightarrow 2 \times (x - 1) = 11 \\[1em] \Rightarrow 2x - 2 = 11 \\[1em] \Rightarrow 2x = 11 + 2 \\[1em] \Rightarrow 2x = 13 \\[1em] \Rightarrow x = \dfrac{13}{2} \\[1em] \Rightarrow x = 6.5$

Solving,

$\dfrac{OD}{OB} = \dfrac{1}{2}$

Substituting values we get :

$\Rightarrow \dfrac{5}{(y + 3)} = \dfrac{1}{2} \\[1em] \Rightarrow 2 \times 5 = (y + 3) \\[1em] \Rightarrow 10 = y + 3 \\[1em] \Rightarrow y = 10 - 3 \\[1em] \Rightarrow y = 7.$

x = 6.5, y = 7

Hence, option 3 is the correct option.