Question

Question 6
The perimeter of a triangle with vetices (0,4), (0,0) and (3,0) is:
(A) 5
(B) 12
(C) 11
(D) $7+\sqrt{5}$

Answer 1

We plot the vertice of a triangles 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 = d(AO)+d(OB)+d(AB)
$\because \text{Distance between the points}(x_1,y_1)and(x_2,y_2),d=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}\therefore \text{Distance between A(0,4) and O(0,0)}$
$+\text{Distance between O(0,0)}\text{and B(3,0)}$
$+\text{Distance between A(0,4), and B(3,0)}=\sqrt{(0-0{)}^2+(0-4{)}^2}+\sqrt{(3-0{)}^2+(0-0{)}^2}$
$+\sqrt{(3-0{)}^2+(0-4{)}^2}=\sqrt{0+16}+\sqrt{9+0}+\sqrt{(3{)}^2+(4{)}^2}=4+3+\sqrt{9+16}=7+\sqrt{25}=7+5=12\text{Hence, the required perimeter of triangle is 12.}$