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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

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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.

 

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APPEARS IN

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