Question:
Diagonals of parallelogram are $6$ and $8$ cm. If one side is $5$ cm, then area of parallelogram is ?
My attempt:
Construct $AH$ perpendicular to base.
Let $DB = 6 cm$ and $AC = 6cm$. Since diagonals of parallelogram bisect each other $AO = OC = 4 cm$ and $OB = OD = 3cm$. But, I don't know where to proceed from where. I have an instinct that probably the AH segment will be used in the solution but I can't just connect to the solution.
Thus, I need help in this problem (hints are most welcome!)
$\endgroup$ 02 Answers
$\begingroup$Hint: $(3,4,5)$ is a Pythagorean triple.
If it wasn't a Pythagorean triple, you could use Heron's formula to get the area of triangle $OAD$, and then use the fact that triangles $OAD$ and $ODC$ have the same area: equal base ($AO=OC$) and equal height (perpendicular distance of $D$ from $AC$).
$\endgroup$ $\begingroup$In addition to the comments, I think this might help. The area of a parallelogram is $\frac {d_1d_2 \sin {\theta}}{2} .$ Here $d_1$ and $d_2$ are the lengths of the respective diagonals and $\theta$ is the angle between the two diagonals. Since $3-4-5$ is a right triangle, here $\theta=90°$. And hence our answer will be ${\frac {1}{2}}\cdot 8\cdot 6\cdot 1=24$ square units.
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