Question
Download Solution PDFComprehension
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?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFGiven:
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
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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