In 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 = \(1 \over 3\) MS, and ST = 5.6 cm, what is the ratio of \(\rm {{ Area \space of \space Triangle \space RMN} \over {Area \space of \space Trapezium \space MNST}}?\)

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Question

This question was previously asked in

SSC CHSL Exam 2023 Tier-I Official Paper (Held On: 04 Aug, 2023 Shift 1)

Answer 1

Given:

In ΔRST, points M and N are the points on sides RS and RT, respectively.

RM = \(1 \over 3\) MS

ST = 5.6 cm

Concept Used:

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

Calculation:

F2 SSC Priya 13-2-24 D7

ΔRST, ∠RMN = ∠RST (Corresponding angles)

∠R is a common angle.

∴ ΔRST ∼ ΔRMN

⇒ RM/RS = MN/ST

⇒ 1/4 = MN/5.6

⇒ MN = 1.4

So, Area of ΔRST / Area ΔRMN = ST^{2 / 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

The answer is 1/15.

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