A) 16cm
B) 18cm
C) 17cm
D) data insufficient
Correct Answer: B
Solution :
| Suppose hypotenuse of the triangle is c and other sides are a and b, obviously. |
| \[c=\sqrt{{{a}^{2}}+{{b}^{2}}}\] ...(i) |
| We have, |
| \[a+b+c=40\] |
| And \[\frac{1}{2}ab=40\Rightarrow \,ab=80\] |
| or \[40-\left( a+b \right)=\sqrt{{{a}^{2}}+{{b}^{2}}}\] |
| \[\Rightarrow \,\,\,c=40-\left( a+b \right)\] |
| and \[ab=80\] |
| or \[40-\left( a+b \right)=\sqrt{{{a}^{2}}+{{b}^{2}}}\] [from Eq. (i)] |
| By squaring the above equation both sides |
| \[\Rightarrow \,\,{{\left( a+b \right)}^{2}}-2\times 40\left( a+b \right)+1600={{a}^{2}}+{{b}^{2}}\] |
| \[\Rightarrow \,{{a}^{2}}+{{b}^{2}}+2\times 80-80\left( a+b \right)+1600={{a}^{2}}+{{b}^{2}}\] |
| \[\Rightarrow \,\,80\left[ \left( a+b \right)-2 \right]=1600\] |
| \[\Rightarrow \,\,\,a+b=20+2=22\] |
| \[\therefore \,\,\,c=40-\left( a+b \right)=40-22=18\,\,cm\] |
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