A) \[10\,m\]
B) \[11\,m\]
C) \[12\,m\]
D) \[13\,m\]
Correct Answer: D
Solution :
| Let the distance between P and Q is x m. |
| Given; \[QR=4m,\] \[RS=15m,\]and \[SP=8m\] |
| \[\therefore \,\,\,\,\,\,\,\,\,\,\,\,\,\,PR=PQ+QR=(4+x)m\] |
| In right \[\Delta RSP,\] \[{{(PR)}^{2}}={{(RS)}^{2}}+{{(PS)}^{2}}\] |
| (By Pythagoras theorem) |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,{{(x+4)}^{2}}={{(15)}^{2}}+{{(8)}^{2}}\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,{{x}^{2}}+16+8x=225+64\,\,\,\Rightarrow \,\,{{x}^{2}}+8x-273=0\] \[\Rightarrow \,\,\,\,\,{{x}^{2}}+21x-13x-273=0\] |
| (By spliting the middle term) |
| \[\Rightarrow \,\,\,\,\,x(x+21)-13(x+21)=0\] |
| \[\Rightarrow \,\,\,\,\,x+21=0\] or \[x-13=0\] |
| But, the distance can't be negative |
| \[\therefore \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x=13\] |
| \[\therefore \]The required distance of garden is\[\text{13 m}\]. |
| So, option [d] is correct. |
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