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Home » Choose the correct answer from the given four options in the following questions: The perimeter of a triangle with vertices (0, 4), (0, 0) and (3, 0) is (A) 5 (B) 12 (C) 11 (D) 7+ √5
Choose the correct answer from the given four options in the following questions: The perimeter of a triangle with vertices (0, 4), (0, 0) and (3, 0) is (A) 5 (B) 12 (C) 11 (D) 7+ √5
CBSE | Class 10 | Coordinate Geometry | Exercise 7.1 | Maths | NCERT Exemplar
Choose the correct answer from the given four options in the following questions: The perimeter of a triangle with vertices (0, 4), (0, 0) and (3, 0) is (A) 5 (B) 12 (C) 11 (D) 7+ √5

Solution:

Option (B) 12 is the correct answer.

(0, 4), (0, 0) and (3, 0) are the vertices of a triangle.

The perimeter of triangle AOB = Sum of the length of all its sides:

= distance between (OA+OB+AB)

The distance between the points (x1,y1) and (x2, y2) can be calculated using the formula,

d~=\text{ }\surd \text{ }({{({{x}_{2}}~\text{ }{{x}_{1}})}^{2}}~+\text{ }{{({{y}_{2}}~\text{ }{{y}_{1}})}^{2}})

NCERT Exemplar Class 10 Maths Chapter 7 Ex. 7.1 Question 6

We need to find:

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

As a result, the required perimeter of triangle is 12

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