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:

Consider ABC as a triangle having A(0, 4), B(0, 0) and C(3,0)

Let us make use of the distance formula

\(Distance=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}\)

Here

\(AB=\sqrt{(0-0)^{2}+(0-4)^{2}}\)

AB = √16 = 4

\(BC=\sqrt{(3-0)^{2}+(0-0)^{2}}\)

BC = √9 = 3

\(CA=\sqrt{(0-3)^{2}+(4-0)^{2}}\)

CA = √25 = 5

The formula to find the perimeter is

Perimeter of triangle ABC = AB + BC + CA

Substituting the values

= 4 + 3 + 5

= 12

Therefore, the perimeter of the triangle is 12.

✦ Try This: The perimeter of a triangle with vertices (0, 3), (0, 0) and (4, 0) is

☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 7

NCERT Exemplar Class 10 Maths Exercise 7.1 Problem 6

The perimeter of a triangle with vertices (0, 4), (0, 0) and (3, 0) is a. 5, b. 12, c. 11, d. 7+ √5

Summary:

The perimeter of a triangle with vertices (0, 4), (0, 0) and (3, 0) is 12

☛ Related Questions:

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