Answer:
The distance between 12m-sides is [tex]5\sqrt{2}m[/tex].
Step-by-step explanation:
It is given that the parallelogram has sides 10m and 12m and an angle of 45°.
Draw an altitude from one 12m side to another 12 m side as shown in below figure.
The opposite angles of parallelogram are same. Two angles are obtuse angles and two are acute angle.
Since angle C is acute angle therefore it must be 45 degree.
[tex]\sin\theta=\frac{perpendicular}{hypotenuse}[/tex]
[tex]\sin C=\frac{BE}{BC}[/tex]
[tex]\sin(45^{\circ})=\frac{d}{10}[/tex]
[tex]\frac{1}{\sqrt{2}}=\frac{d}{10}[/tex]
[tex]\frac{10}{\sqrt{2}}=d[/tex]
[tex]\frac{10}{\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}=d[/tex]
[tex]\frac{10\sqrt{2}}{2}=d[/tex]
[tex]5\sqrt{2}=d[/tex]
Therefore the distance between 12m-sides is [tex]5\sqrt{2}m[/tex].