Question

find the perimeter of the triangle whose vertices are (3,2),(7,2) and (7,5)

Answer 1

Answer:

The Perimeter of a ΔABC is 12 units .

Step-by-step explanation:

Formula

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

As given

vertices of a triangle (3,2),(7,2) and (7,5).

Name the vertex A(3,2), B(7,2) and C(7,5).

Take the vertiex A(3,2) and B(7,2).

$AB = \sqrt{(7 - 3)^{2} +(2 - 2)^{2}}$

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

$AB = \sqrt{16}$

$\sqrt{16} = 4$

AB = 4 units

Take the vertiex B(7,2) and C(7,5)

$BC = \sqrt{(7 - 7)^{2} +(5 - 2)^{2}}$

$BC = \sqrt{(3)^{2}}$

$BC = \sqrt{9}$

$\sqrt{9} = 3$

BC = 3 units

Take the vertiex C(7,5) and  A(3,2).

$CA = \sqrt{(3 - 7)^{2} +(2 - 5)^{2}}$

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

$CA = \sqrt{16 + 9}$

$CA = \sqrt{25}$

$\sqrt{25} = 5$

CA =5 units

Thus

Perimeter of a ΔABC = AB + BC + CA

                                   = 4 + 3 + 5

                                   = 12 units

Therefore the Perimeter of a ΔABC is 12 units .

Answer 2

12 units

correct hai koi baat Nehi bilkul correct hai bhai