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Triangle Inequality Theorem

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QuestionClick to flip

The points $\left( 0,\dfrac { 8 }{ 3 } \right),(1,3)$ and $(82,30)$ are the vertices of:

AnswerClick to flip back

an equilateral triangle

💡 Explanation:

According to the problem :

$AB^2=(0-1)^2+(\dfrac{8}{3}-3)^2$

$=1+\dfrac{1}{9}=\dfrac{10}{9}=1.11$

Similarly,

$BC^2=(82-1)^2+(30-3)^2=7290$

and

$AC^2=(82-0)^2+(30-\dfrac{8}{3})^2=7471.11$

Therefore,

$AB^2+BC^2<AC^2$

Hence the answer is acute-angled triangle.