Question

find length of median of triangle whose vertices are (5,6) (3,8) and (-1,2)

Answer 1

Answer:

$[tex]<svg width="250" height="250" viewBox="0 0 100 100">\ \textless \ br /\ \textgreater \ \ \textless \ br /\ \textgreater \ <path fill="pink" d="M92.71,7.27L92.71,7.27c-9.71-9.69-25.46-9.69-35.18,0L50,14.79l-7.54-7.52C32.75-2.42,17-2.42,7.29,7.27v0 c-9.71,9.69-9.71,25.41,0,35.1L50,85l42.71-42.63C102.43,32.68,102.43,16.96,92.71,7.27z"></path>\ \textless \ br /\ \textgreater \ \ \textless \ br /\ \textgreater \ <animateTransform \ \textless \ br /\ \textgreater \ attributeName="transform" \ \textless \ br /\ \textgreater \ type="scale" \ \textless \ br /\ \textgreater \ values="1; 1.5; 1.25; 1.5; 1.5; 1;" \ \textless \ br /\ \textgreater \ dur="2s" \ \textless \ br /\ \textgreater \ repeatCount="40"> \ \textless \ br /\ \textgreater \ </animateTransform>\ \textless \ br /\ \textgreater \ \ \textless \ br /\ \textgreater \ </svg>$

$<svg width="250" height="250" viewBox="0 0 100 100">\ \textless \ br /\ \textgreater \ \ \textless \ br /\ \textgreater \ <path fill="red" d="M92.($

Answer 2

Step-by-step explanation:

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

(3+1)

2

+(5−1)

2

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

16+16

=

32

=4

2

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

(3−5)

2

+(5+3)

2

=2\sqrt{17}2

17

Now, BD=DC

Therefore, BD+CD=BC

2CD=BC

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

2

17

Now, in triangle ACD,

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

2

=(AB)

2

+(CD)

2

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

2

)

2

=(AD)

2

+(

17

)

2

32=(AD)^{2}+1732=(AD)

2

+17

(AD)^{2}=15(AD)

2

=15

AD=\sqrt{15}cmAD=

15

cm

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

15

cm