finding the length of the altitude of the rhombus the lengh of the side of a rhombus is given as $5\sqrt2$. if two of its opposites vertices have coordinates $(3,-4)$ and $(1,2)$. find the lengh of the altitude of the rhombus? i was trying this question , i got anwer $\sqrt{37}$ by using the distance formula, , i think $\sqrt{37}$ is not altitude of rhombus. im very confuse how to find altitude of rhombus, i dont know from where i have to start... if anbody help me,, i would be very thankful to him i have no idea and hint if any body pliz help me ,,,,i would be very thankful to him

Start finding the shorter diagonal $AC$

If we have $A(3,-4)$ and $C(1,2)$ then $AC=\sqrt{40}=2\sqrt{10}$

Call $O$ the midpoint of $AC$ and $B$ and $D$ the other vertices. From Pythagoras theorem $OB=\sqrt{BC^2-OC^2}=\sqrt{50-10}=\sqrt{40}=2\sqrt{10}$

The larger diagonal is $BD=4\sqrt{10}$ so the area of the rhombus is $\mathcal{A}=40$ and the height of the rhombus is $h=\dfrac{\mathcal{A}}{BC}=\dfrac{40}{5\sqrt{2}}=4\sqrt{2}$

$h=4\sqrt 2$

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