Question

2. Perimeter of A ABC whose vertices are A = (0,0), B = (6,0), C = (0,8) is
A) 44 units B) 24 units
C) 12 units D) 34 units​

Answer 1

Given :

•The vertices of a given triangles are :

$\bullet\bf{A=(0,0)}$

$\bullet\bf{B=(6,0)}$

$\bullet\bf{C=(0,8)}$

To Find :

•The Perimeter of ΔABC=?

Formula :

Distance Formula :

$\\ \red\bigstar\boxed{\bf{\sqrt{(y_{2}-y_{1})^{2}+(x_{2}-x_{1})^{2}} }}$

Solution :

1. Find length of each sides of Δ ABC

Let A=(0,0) ,B=(6,0) , C=(0,8)

Use distance formula :

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

$\implies\boxed{\sf{AB=6 cm}}$

$\\ \implies\sf{BC=\sqrt{(8-0)^{2}+(0-6)^{2}}}$

$\implies\boxed{\sf{BC=10cm}}$

$\\ \implies\sf{AC=\sqrt{(8-0)^{2}+(0-0)^{2}}}$

$\implies\boxed{\sf{AC=8cm}}$

Now,

$\\ \green \bigstar \implies \sf \: perimeter \: of \: \triangle = 6 + 10 + 8$

$\implies \sf \: perimeter \: of \: \triangle = 24cm$

$\\ \implies \boxed{ \bf{perimeter \: of \: \triangle = 24cm}}$