# The Perimeter of a Triangle with Vertices (0, 4), (0, 0) and (3, 0) is - Mathematics

MCQ

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

#### Options

• 7 + $\sqrt{5}$

• 5

• 10

• 12

#### Solution

Let A(0, 4), O(0, 0) and B(3, 0) be the vertices of ∆AOB.
Using distance formula, we get
OA = $\sqrt{\left( 0 - 0 \right)^2 + \left( 4 - 0 \right)^2} = \sqrt{16} = 4$  units

OB = $\sqrt{\left( 3 - 0 \right)^2 + \left( 0 - 0 \right)^2} = \sqrt{9} = 3$  units

AB =  $\sqrt{\left( 3 - 0 \right)^2 + \left( 0 - 4 \right)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$  units

∴ Perimeter of ∆AOB = OA + OB + AB = 4 + 3 + 5 = 12 units
Thus, the required perimeter of the triangle is 12 units.

Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Class 10 Maths
Chapter 6 Co-Ordinate Geometry
Q 48 | Page 66