# Calculate the perimeter of the triangle whose vertices are (1,2) (-1,4) and * (-2,3).

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To find the perimeter we need the distances between the points (1,2) and (-1,4), and the points (-1, 4) and (-2,3) and the points (-2,3) and (1,2).

The distance between (x1, y1) and (x2, y2) is given by sqrt[(x1 - x2)^2 + (y1 - y2)^2]

The distance between (1,2) and (-1,4)

=> sqrt [(1 +1)^2 + (2 - 4)^2]

=> sqrt ( 2^2 + 2^2)

=> sqrt 8

The distance between (-1, 4) and (-2,3)

=> sqrt [(-1 + 2)^2 + (4 - 3)^2]

=> sqrt (1^2 + 1^2)

=> sqrt 2

The distance between (-2,3) and (1,2)

=> sqrt [(1 + 2)^2 + (2 - 3)^2]

=> sqrt (3^2 + 1^2)

=> sqrt 10

**Therefore the perimeter is sqrt 10 + sqrt 8 + sqrt 2.**

Let ABC be a triangle.

Given the vertices's of a triangle are:

A(1,2) , B(-1,4), and C(-2,3).

We need to find the perimeter of the triangle.

First we need to find the length of the sides.

==> AB = sqrt[( 1+1)^2 + ( 2-4)^2)

= sqrt( 2^2 + 2^2 )

= sqrt( 4+4)

= sqrt8

**==> AB = 2sqrt2.**

==> AC = sqrt( 1+ 2)^2 + ( 2-3)^2

= sqrt( 3^2 + 1^2)

= sqrt(9+ 1)

= sqrt10.

**==> AC= sqrt10.**

==> BC = sqrt( -1+2)^2 +(4-3)^2]

= sqrt( 1^2 + 1^2 )

= sqrt2

==>** BC = sqrt2.**

**==> The perimeter of the triangle is:**

**P = AB + AC + BC**

** = 2sqrt2 + sqrt10 + sqrt2**

** = 7.4 units. ( approx.)**

To determine the perimeter of a triangle, we have to know the lengths of the sides of the triangle.

We don't know the sides but we know the vertices of the triangle.

We'll calculate the lengths of the sides, using the distance formula:

We'll compute the length a:

a = sqrt[(-1-1)^2 + (4-2)^2]

a = sqrt(4 + 4)

a = 2sqrt2

We'll compute the length b:

b = sqrt[(-2+1)^2 + (3-4)^2]

b = sqrt(1+1)

b = sqrt2

We'll compute the length c:

c = sqrt[(-2-1)^2 + (3-2)^2]

c = sqrt(9 + 1)

c = sqrt10

The perimeter of the triangle is the sum of the lengths of the sides a,b,c:

P = a+b+c

P = 2sqrt2 + sqrt2 + sqrt10

We'll combine like terms:

**P = 3sqrt2 + sqrt10**

We'll factorize by sqrt2:

**P = sqrt2(3 + sqrt5) units**