Directions (Q. 41 to 44): These questions are based on the following information :

In the given figure, ABC is a triangle and PQ is a straight line meeting AB in P and AC in Q. It is given that AP = 1 cm, PB = 3 cm, AQ = 1.5 cm, QC = 4.5 cm and PQ = 2 cm.

41. The perimeter of ΔABC is:

1. 12 cm

2. 16 cm

3. 18 cm

4. 24 cm

42. The ratio of the areas of ΔAPQ and ΔABC is:

1. 1 : 3

2. 1 : 4

3. 1 : 9

4. 1 : 16

43. Which of the following holds true?

1. ΔAPQ ∼ ΔABC

2. ΔAQP ∼ ΔABC

3. ΔAPQ ∼ ΔACB

4. none of these

44. Which axiom of similarity applies in the above case?

1. AAA

2. AA

3. SSS

4. SAS

41. Given,

From figure,

AB = AP + PB = 1 + 3 = 4 cm

AC = AQ + QC = 1.5 + 4.5 = 6 cm

In ΔAPQ and ΔABC,

$\Rightarrow \dfrac{AP}{AB} = \dfrac{1}{4}$

$\Rightarrow \dfrac{AQ}{AC} = \dfrac{1.5}{6} = \dfrac{1}{4}$

∠PAQ = ∠BAC [Common angles]

ΔAPQ ∼ ΔABC [By SAS Similarity]

Since, corresponding sides of similar triangle are proportional to each other.

$\Rightarrow \dfrac{PQ}{BC} = \dfrac{AQ}{AC} \\[1em] \Rightarrow \dfrac{PQ}{BC} = \dfrac{1}{4} \\[1em] \Rightarrow \dfrac{2}{BC} = \dfrac{1}{4} \\[1em] \Rightarrow BC = 2 \times 4 \\[1em] \Rightarrow BC = 8 \text{ cm}.$

Perimeter = AB + AC + BC

= 4 + 6 + 8

= 18 cm.

Hence, option 3 is the correct option.

42. Given,

The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

$\Rightarrow \dfrac{\text{ Area(ΔAPQ)}}{\text{ Area(ΔABC)}} = \Big(\dfrac{AP}{AB}\Big)^2 \\[1em] = \Big(\dfrac{1}{4}\Big)^2 \\[1em] = \dfrac{1}{16} = 1 : 16.$

Hence, option 4 is the correct option.

43. Given,

In ΔAPQ and ΔABC,

$\dfrac{AP}{AB} = \dfrac{AQ}{AC} = \dfrac{1}{4}$

∠PAQ = ∠BAC [Common angles]

∴ ΔAPQ ∼ ΔABC by SAS Similarity.

Hence, option 1 is the correct option.

44. Given,

Two pairs of corresponding sides are proportional ($\dfrac{AP}{AB} = \dfrac{AQ}{AC}$) and icluded angle is also equal (angle A).

This is the S.A.S. (Side-Angle-Side) Similarity Axiom.

Hence, option 4 is the correct option.