If \[\Delta ABC\tilde{\ }\Delta PQR,\] area of \[\Delta ABC=81c{{m}^{2}}\] and area of \[\Delta PQR=121c{{m}^{2}}\] and altitude \[\text{AD}=\text{9 cm},\]then \[PM=\] |
A) 10 cm
B) 11 cm
C) 12 cm
D) 15 cm
Correct Answer: B
Solution :
[b] \[\frac{ar(\Delta ABC)}{ar(\Delta PQR)}=\frac{A{{D}^{2}}}{P{{M}^{2}}}\] |
\[\therefore \,\,\,\,\,\,\,\,\,\,\frac{81}{121}=\frac{{{9}^{2}}}{P{{M}^{2}}}\,\,\,\Rightarrow \,\,\,PM=11\,cm.\] |
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