Comprehension

ABC is a triangle right-angled at A. Further,
AB = 8 cm, BC = 10 cm. D is the point on BC such that AD is perpendicular to BC.

What is ratio of area of triangle ADC to area of triangle ADB?

This question was previously asked in
UPSC CDS-I 2025 (Elementary Mathematics) Official Paper (Held On: 13 Apr, 2025)
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  1. 7 : 15
  2. 9 : 16
  3. 2 : 3
  4. 3 : 4

Answer (Detailed Solution Below)

Option 2 : 9 : 16
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Detailed Solution

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Given:

 

The triangle ABC is right-angled at A.

AB = 8 cm, BC = 10 cm,

D is perpendicular to BC.

Calculation:

AC2 = BC2 - AB2

AC = √(100 - 64)

AC = √36 = 6 cm  

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ΔADB ~ ΔADB

So, \(\frac{Ar\ \Delta ADC}{Ar\ \Delta ADB} = \frac{6^2}{8^2}\)

⇒ \(\frac{Ar\ \Delta ADC}{Ar\ \Delta ADB} = \frac{36}{64}\) 

⇒ \(\frac{Ar\ \Delta ADC}{Ar\ \Delta ADB} = \frac{9}{16}\) 

∴ The correct answer is option 2.

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