question_answer

The area of a parallelogram ABCD in which AB = 12 cm, BC = 9 cm and diagonal AC = 15 cm is k cm2. Find the value of \[\frac{k-100}{4}.\]

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 \[\frac{k-100}{4}=\frac{108-100}{4}=2\]