| In the given figure, \[\Delta MNP\tilde{\ }\Delta ABC\] such that \[\text{MN}=\text{6 cm},\] \[\text{AB}=8\,\text{cm}\] and area of \[\Delta MNP\] is \[\text{15 c}{{\text{m}}^{\text{2}}},\] then the area of \[\Delta ABC\] is: |
 |
A) \[26.66c{{m}^{2}}\]
B) \[16.66c{{m}^{2}}\]
C) \[20\,c{{m}^{2}}\]
D) \[None\,\, of\,\, these\]
Correct Answer: A
Solution :
| [a] \[\Delta MNP\tilde{\ }\Delta ABC\] |
| \[\therefore \,\,\frac{ar(\Delta MNP)}{ar(\Delta ABC)}=\frac{{{(MN)}^{2}}}{{{(AB)}^{2}}}\,\,\,\Rightarrow \,\,\,\frac{(15)}{ar\,(\Delta ABC)}=\frac{{{(6)}^{2}}}{{{(8)}^{2}}}\] |
| \[\Rightarrow \,\,\,\,\,\,\,ar(\Delta ABC)=\frac{15\times 64}{36}=26.66\,c{{m}^{2}}\] |
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