A) 5 units
B) 12 units
C) 11 units
D) \[\left( 7+\sqrt{5} \right)\]units
Correct Answer: B
Solution :
We plot the vertices of a triangle i.e., (0, 4), (0, 0) and (3, 0) on the paper shown as given below |
Now, perimeter of \[\Delta AOB\]= Sum of the length of all its sides |
\[=\text{ }d\left( AO \right)+d\left( OB \right)+d\left( AB \right)\] |
\[\because\] Distance between the points \[\left( {{x}_{1}},\,{{y}_{1}} \right)\]and \[\left( {{x}_{2}},\,{{y}_{2}} \right)\] is |
\[d=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}\] |
= Distance between A(0, 4) and O(0, 0) |
+ Distance between O(0, 0) and B(3, 0) |
+ Distance between A(0, 4) and B (3, 0) |
\[=\sqrt{{{\left( 0-0 \right)}^{2}}+{{\left( 0-4 \right)}^{2}}}+\sqrt{{{\left( 3-0 \right)}^{2}}+{{\left( 0-0 \right)}^{2}}}\] |
\[+\sqrt{{{\left( 3-0 \right)}^{2}}+{{\left( 0-4 \right)}^{2}}}\] |
\[=\sqrt{0+16}+\sqrt{9+0}+\sqrt{{{\left( 3 \right)}^{2}}+{{\left( 4 \right)}^{2}}}\] |
\[=4+3+\sqrt{9+16}\] |
\[=7+\sqrt{25}=7+5=12\] |
Hence, the required perimeter of triangle is 12 units. |
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