Answer:
The perimeter of ΔABC = 12 units
Step-by-step explanation:
From the given triangle ABC on a coordinate plane, we can observe that:
The length of AB is the distance between A(1, 1) and B(4, 1)
so using the distance formula
[tex]\mathrm{Compute\:the\:distance\:between\:}\left(x_1,\:y_1\right),\:\left(x_2,\:y_2\right):\quad \sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}[/tex]
[tex]\mathrm{The\:distance\:between\:}\left(1,\:1\right)\mathrm{\:and\:}\left(4,\:1\right)\mathrm{\:is\:}[/tex]
[tex]=\sqrt{\left(4-1\right)^2+\left(1-1\right)^2}[/tex]
[tex]=\sqrt{3^2+0}[/tex]
[tex]=\sqrt{3^2}[/tex]
[tex]\mathrm{Apply\:radical\:rule\:}\sqrt[n]{a^n}=a,\:\quad \mathrm{\:assuming\:}a\ge 0[/tex]
[tex]=3[/tex]
- So the length of AB = 3 units
similarly the length of BC is the distance between B(4, 1) and C(4, 5)
[tex]\mathrm{The\:distance\:between\:}\left(4,\:1\right)\mathrm{\:and\:}\left(4,\:5\right)\mathrm{\:is\:}[/tex]
[tex]=\sqrt{\left(4-4\right)^2+\left(5-1\right)^2}[/tex]
[tex]=\sqrt{0+4^2}[/tex]
[tex]=\sqrt{4^2}[/tex]
[tex]\mathrm{Apply\:radical\:rule\:}\sqrt[n]{a^n}=a,\:\quad \mathrm{\:assuming\:}a\ge 0[/tex]
[tex]=4[/tex]
- So the length of BC = 4 units
- As the length of of hypotenuse AC = 5 units
Therefore, the perimeter of the triangle ABC can be obtained using the formula:
P = AB + BC + AC
= 3 + 4 + 5
= 12 units
Therefore, the perimeter of ΔABC = 12 units
