Area of a triangle

Geometry Level 3

Find the perimeter of a triangle with vertices at ( 8 , 3 ) , ( 5 , 5 ) (8,3) , (5,5) and ( 3 , 2 ) (3,2) .

Give your answer to 2 decimal places.

8.32 14.31 10.32 12.31

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1 solution

Consider the diagram on the above.

Using the distance formula, we have

a = ( 5 − 3 ) 2 + ( 5 − 2 ) 2 = 4 + 9 = 13 a=\sqrt{(5-3)^2+(5-2)^2}=\sqrt{4+9}=\sqrt{13}

b = ( 8 − 5 ) 2 + ( 3 − 5 ) 2 = 9 + 4 = 13 b=\sqrt{(8-5)^2+(3-5)^2}=\sqrt{9+4}=\sqrt{13}

c = ( 8 − 3 ) 2 + ( 3 − 2 ) 2 = 25 + 1 = 26 c=\sqrt{(8-3)^2+(3-2)^2}=\sqrt{25+1}=\sqrt{26}

P = 2 13 + 26 = P=2\sqrt{13}+\sqrt{26}= 12.31 \boxed{12.31}

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