A) \[\frac{3}{2}\]
B) \[\frac{2}{3}\]
C) \[\frac{3}{5}\]
D) \[\frac{2}{5}\]
Correct Answer: D
Solution :
Given, AB = 5 cm, DB = 3 can |
\[\therefore \] \[AD=5-3=2\,\,\text{cm}\] |
In the figure we can see that both \[\Delta ADC\]and \[\Delta ABC\]the same height h. |
Area of a triangle \[=\frac{1}{2}\times \]Base \[\times \]Height |
When, height is constant. Area of triangle \[\propto \]base, |
\[\therefore \] \[\frac{\text{Area}\,\,\text{of}\,\,\Delta ADC}{\text{Area}\,\,\text{of}\,\,\Delta ABC}=\frac{AD}{AB}=\frac{2}{5}\] |
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