Question

Find the lengths of the medians of the triangle
whose vertices are (5,6), (3,8) and (-1,2).​

Answer 1

Answer:

Step-by-step explanation:

Using the distance formula, AC=\sqrt{(3+1)^{2}+(5-1)^{2}}

AC=\sqrt{16+16}=\sqrt{32}=4\sqrt{2}

And BC=\sqrt{(3-5)^{2}+(5+3)^{2}}

=2\sqrt{17}

Now, BD=DC

Therefore, BD+CD=BC

2CD=BC

CD=\frac{\sqrt{17}}{2}

Now, in triangle ACD,

(AC)^{2}=(AB)^{2}+(CD)^{2}

(4\sqrt{2}) ^{2}=(AD)^{2}+(\sqrt{17})^{2}

32=(AD)^{2}+17

(AD)^{2}=15

AD=\sqrt{15}cm

therefore, the length of the median of the triangle is \sqrt{15}cm

Answer 2

The length of the median of the triangle is sqrt of 15 cm.
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