A) \[1632\,c{{m}^{2}},1886\,c{{m}^{2}}\]
B) \[1538\,c{{m}^{2}},1632\,c{{m}^{2}}\]
C) \[1632\,c{{m}^{2}},1868\,c{{m}^{2}}\]
D) \[1538\,c{{m}^{2}},1632\,c{{m}^{2}}\]
Correct Answer: C
Solution :
Area of rectangular tile \[=(50\times 70)c{{m}^{2}}=3500\,c{{m}^{2}}\] We have, \[a=25\,cm,\,b=17\,cm\]and \[c=26\,cm\] \[\therefore \]\[s=\frac{a+b+c}{2}=\left( \frac{25+17+6}{2} \right)\,cm\] = 34 cm \[\therefore \]Area of 1 triangular tile \[=\sqrt{s(s-a)(s-b)(s-c)}\] \[=\sqrt{34(34-25)(34-17)(34-26)}\,c{{m}^{2}}\] \[=\sqrt{34\times 9\times 17\times 8}\,c{{m}^{2}}=204\,c{{m}^{2}}\] \[\therefore \]Total area of 8 triangles \[=(204\times 8)c{{m}^{2}}\] \[=1632\,c{{m}^{2}}\] So, area of the design \[=1632\,c{{m}^{2}}\] Also, remaining area of the tile \[=(3500-1632)\,c{{m}^{2}}=1868\,c{{m}^{2}}\]
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