find the perimeter of a triangle whose vertices have the coordinates. (3,10) (5,2) and (4,12)
Answers
Answer:
We know that the distance between the two points (x_1,y_1)(x
1
,y
1
) and (x_2,y_2)(x
2
,y
2
) is
d=\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 } }d=
(x
2
−x
1
)
2
+(y
2
−y
1
)
2
Let the given vertices be A=(3,10)A=(3,10), B=(5,2)B=(5,2) and C=(14,12)C=(14,12)
We first find the distance between A=(3,10)A=(3,10) and B=(5,2)B=(5,2) as follows:
AB=\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 } } =\sqrt { { (5-3) }^{ 2 }+{ (2-10) }^{ 2 } } =\sqrt { { 2 }^{ 2 }+{ (-8) }^{ 2 } } =\sqrt { 4+64 } =\sqrt { 68 }AB=
(x
2
−x
1
)
2
+(y
2
−y
1
)
2
=
(5−3)
2
+(2−10)
2
=
2
2
+(−8)
2
=
4+64
=
68
Similarly, the distance between B=(5,2)B=(5,2) and C=(14,12)C=(14,12) is:
BC=\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 } } =\sqrt { { (14-5) }^{ 2 }+{ (12-2) }^{ 2 } } =\sqrt { { 9 }^{ 2 }+{ 10 }^{ 2 } } =\sqrt { 81+100 } =\sqrt { 181 }BC=
(x
2
−x
1
)
2
+(y
2
−y
1
)
2
=
(14−5)
2
+(12−2)
2
=
9
2
+10
2
=
81+100
=
181
Now, the distance between C=(14,12)C=(14,12) and A=(3,10)A=(3,10) is:
CA=\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 } } =\sqrt { { (14-3) }^{ 2 }+{ (12-10) }^{ 2 } } =\sqrt { 11^{ 2 }+{ 2 }^{ 2 } } =\sqrt { 121+4 } =\sqrt { 125 }CA=
(x
2
−x
1
)
2
+(y
2
−y
1
)
2
=
(14−3)
2
+(12−10)
2
=
11
2
+2
2
=
121+4
=
125
Since the perimeter PP of a triangle ABCABC is AB+BC+CAAB+BC+CA, therefore,
P=\sqrt { 68 } +\sqrt { 181 } +\sqrt { 125 }P=
68
+
181
+
125
Hence, the perimeter of the triangle is \sqrt { 68 } +\sqrt { 181 } +\sqrt { 125 }
68
+
181
+
125
units