Question

A triangle has vertices (1,4), (1,-2), and (-3,-2). what is the approximate perimeter of the triangle

Answer 1

Answer:

• Perimeter of Traingle is 10 + 2√5

Step-by-step explanation:

Let,

• A = (1,4)
• B = (1,-2)
• C = (-3,-2)

To find the perimeter of the traingele with vertices of A (1,4), B(1, -2) and C(-3,-2). We have to first find the distance between each pair of points, which will give length of the sides.

Using distance formula,

$\dashrightarrow \sf \: \: AB = \sqrt{( x_{2} - x_{1})^2 + (y_{2} - y_{1})^2}$

$\dashrightarrow \sf \: \: AB = \sqrt{{ {(1 - 1)}^{2} } + {( - 2 - 4)}^{2} }$

$\dashrightarrow \sf \: \: AB = \sqrt{0 + {( - 6)}^{2} }$

$\dashrightarrow \sf \: \: AB = \sqrt{36}$

$\pink{ \dashrightarrow\sf \: \: {\underline{AB = 6 }}}$

$\dashrightarrow \sf \: \: BC = \sqrt{{ {( - 3 - 1)}^{2} } + {( - 2 + 2)}^{2} }$

$\dashrightarrow \sf \: \:BC = \sqrt{ {( - 4)}^{2} + 0}$

$\dashrightarrow \sf \: \:BC = \sqrt{16}$

$\pink{ \dashrightarrow \sf \: \:{ \underline{ BC = 4 }}}$

${\dashrightarrow { \sf \: \: {AC = \sqrt{{ {( - 3 - 1)}^{2} } + {(2 - 4)}^{2}}}} }$

$\dashrightarrow \sf \: \:AC = \sqrt{ {( - 4)}^{2} + {( - 2)}^{2} }$

$\dashrightarrow\sf \: \: AC = \sqrt{16 + 4}$

$\dashrightarrow \sf \: \:AC = \sqrt{20}$

$\pink{ \dashrightarrow \sf \: \: { \underline{AC = 2 \sqrt{5} }}}$

Now,

$\dashrightarrow\sf \: \: Perimeter = AB + BC + AC$

$\dashrightarrow\sf \: \: 6 + 4 + 2√5$

$\dashrightarrow\sf \: \: {\pink{ \underline{ \boxed { \pmb{ \sf{ \: 10 + 2√5 \: \: }}}}}}$

Hence,

• Perimeter of Traingle is 10 + 2√5