question_answer

The perimeter of a triangle with vertices (0, 4), (0, 0) and (3, 0) is

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.