A) 6
B) 8
C) 10
D) 12
Correct Answer: A
Solution :
[a]Let the side of right angled triangle are \[a-d,\]\[a,a+d.\] | |
\[\therefore \]\[{{(a-d)}^{2}}+{{a}^{2}}={{(a+d)}^{2}}\] | |
\[\Rightarrow \] \[{{a}^{2}}={{(a+d)}^{2}}-{{(a-d)}^{2}}\] | |
\[\Rightarrow \] \[{{a}^{2}}=4ad\] | |
\[\Rightarrow \] \[a=4d\] | ? (i) |
Area of triangle \[=\frac{1}{2}a\,(a-d)=24\] | |
\[\Rightarrow \]\[a\,(a-d)=48\] | ? (ii) |
From Eqs. (i) and (ii) we get | |
\[a=8,\,\,d=2\] | |
\[\because \]Smallest side is \[a-d=8-2=6\] |
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