Find the perimeter of the triangle whose vertices are (−3,−5), (2,−5), and (2,7)
Answers
Step-by-step explanation:
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}}Distance Formula=
(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=
(7−3)
2
+(2−2)
2
AB = \sqrt{(4)^{2} +(0)^{2}}AB=
(4)
2
+(0)
2
AB = \sqrt{16}AB=
16
\sqrt{16} = 4
16
=4
AB = 4 units
Take the vertiex B(7,2) and C(7,5)
BC = \sqrt{(7 - 7)^{2} +(5 - 2)^{2}}BC=
(7−7)
2
+(5−2)
2
BC = \sqrt{(3)^{2}}BC=
(3)
2
BC = \sqrt{9}BC=
9
\sqrt{9} = 3
9
=3
BC = 3 units
Take the vertiex C(7,5) and A(3,2).
CA = \sqrt{(3 - 7)^{2} +(2 - 5)^{2}}CA=
(3−7)
2
+(2−5)
2
CA = \sqrt{(-4)^{2} +(-3)^{2}}CA=
(−4)
2
+(−3)
2
CA = \sqrt{16 + 9}CA=
16+9
CA = \sqrt{25}CA=
25
\sqrt{25} = 5
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 .