Math, asked by lakshmibai7338, 11 months ago

find the perimeter of a triangle whose vertices have the coordinates. (3,10) (5,2) and (4,12)​

Answers

Answered by 18shreya2004mehta
1

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

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