Question

Can you draw a triangle with vertices (1, 5), (5, 8) and (13, 14) ? Give reason.m​

Answer 1

It is not possible to draw a triangle with vertices (1, 5), (5, 8) and (13, 14)

Given :

The points (1, 5), (5, 8) and (13, 14)

To find :

Is it possible to draw a triangle with vertices (1, 5), (5, 8) and (13, 14)

Solution :

Step 1 of 2 :

Calculate the sides of the triangle

Let the given vertices are A(1, 5) , B(5, 8) and C(13, 14)

AB

$\displaystyle \sf = \sqrt{ {(5 - 1)}^{2} + {(8 - 5)}^{2} } \: \: unit$

$\displaystyle \sf = \sqrt{ {(4)}^{2} + {(3)}^{2} } \: \: unit$

$\displaystyle \sf = \sqrt{16 + 9} \: \: unit$

$\displaystyle \sf = \sqrt{25 } \: \: unit$

$\displaystyle \sf = 5 \: \: unit$

BC

$\displaystyle \sf = \sqrt{ {(13 - 5)}^{2} + {(14 - 8)}^{2} } \: \: unit$

$\displaystyle \sf = \sqrt{ {(8)}^{2} + {(6)}^{2} } \: \: unit$

$\displaystyle \sf = \sqrt{64 + 36} \: \: unit$

$\displaystyle \sf = \sqrt{100 } \: \: unit$

$\displaystyle \sf = 10 \: \: unit$

AC

$\displaystyle \sf = \sqrt{ {(13 - 1)}^{2} + {(14 - 5)}^{2} } \: \: unit$

$\displaystyle \sf = \sqrt{ {(12)}^{2} + {(9)}^{2} } \: \: unit$

$\displaystyle \sf = \sqrt{144 + 81} \: \: unit$

$\displaystyle \sf = \sqrt{225 } \: \: unit$

$\displaystyle \sf = 15 \: \: unit$

Step 2 of 2 :

Find is it possible to draw a triangle or not

AB = 5 unit

BC = 10 unit

AC = 15 unit

∴ AB + BC = AC

Thus sum of two sides is equal to third side

But in order to draw a triangle sum of two sides must be greater than the third side

Hence it is not possible to draw a triangle with vertices (1, 5), (5, 8) and (13, 14)

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