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SSC CHSL 2023 PYQ — Question 60

SSC CHSL - 04 Aug 2023 - Shift 1 • Quantitative Aptitude • Q60

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In a triangle RST, M, and N are two points on RS and RT such that MN is parallel to the base ST of the triangle RST. If RM = 13\frac{1}{3} MS, and ST = 5.6 cm, what is the ratio of Area];of  Triangle  RMNArea  of  Trapezium  MNST\frac{Area]; of\; Triangle\; RMN}{Area\; of\; Trapezium\; MNST}

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Correct answer: C

Explanation

The correct answer is Option 3 i.e 115\frac{1}{15}.

For two similar triangles, the ratio of their areas will be in the ratio of the square of their corresponding sides

in ΔRST, 

⇒ ∠RMN = ∠RST (corresponding sides)

⇒ ∠R (common)

ΔRST ~ ΔRMN

⇒ RM/RS = MN/ST

⇒ 1/4 = MN/5.6

⇒ MN = 1.4

So, Area of ΔRST/Area ΔRMN = ST2/MN2

⇒ 5.62/1.42 = 16

Suppose, the area of ΔRST is 16 sq. units, and the area of ΔRMN is 1 sq. units.

Area of trapezium MNST = Area of ΔRST - Area of ΔRMN

⇒ 16 - 1 = 15

Area of trapezium MNST : Area of ΔRMN = 15 : 1 = 1/15

So, value of Area];of  Triangle  RMNArea  of  Trapezium  MNST\frac{Area]; of\; Triangle\; RMN}{Area\; of\; Trapezium\; MNST} = 1/15