A) 3
B) 4
C) 2
D) 5
Correct Answer: C
Solution :
In\[\Delta ABC,\] \[a=9\,cm,\,b=15\,cm,\] \[c=12\,cm\] \[s=\frac{a+b+c}{2}\] \[=\frac{12+9+15}{2}=\frac{36}{2}=18\] Area of \[\Delta ABC=\sqrt{s(s-a)(s-b)(s-c)}\] \[=\sqrt{18(18-12)(18-9)(18-15)}\] \[=\sqrt{18\times 6\times 9\times 3}=54\,c{{m}^{2}}\] Area of parallelogram ABCD \[=2(Area\,of\,\Delta \Alpha \Beta C)\] \[=2\times 54=108\,c{{m}^{2}}=k\,c{{m}^{2}}\] (given) \[\Rightarrow \]\[k=108\] \[\therefore \]The value of \[\frac{k-100}{4}=\frac{108-100}{4}=2\]You need to login to perform this action.
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