Two opposite sides of a parallelogram each have a length of #3 #. If one corner of the parallelogram has an angle of #(3 pi)/4 # and the parallelogram's area is #24 #, how long are the other two sides?

Question

Two opposite sides of a parallelogram each have a length of #3  #.  If one corner of the parallelogram has an angle of #(3 pi)/4   # and the parallelogram's area is #24      #, how long are the other two sides?

Theodore Abad · Accepted Answer

#8sqrt2#

Let ABCD be the parallelogram as shown in the figure, with side AB=CD= 3. Let the other two sides AD and BC be equal to 'b'

Draw a perpendicular CE from C on to the extended side DE. If h is the height then DE would also be same as the height. Applying Pythagoras theorem to the rt. triangle CDE , #2h^2=9#. Hence #h=3/sqrt2#.

Area of ABCD would be base x height =bh=24.

This would give #b= 24/ h= 24/(3/sqrt2)= 8sqrt2#

Mason Adams · Answer

undefined Given that two opposite sides of the parallelogram have a length of 3 and one angle is $ \frac{3\pi}{4} $, with the area being 24, the lengths of the other two sides can be calculated using the formula for the area of a parallelogram:

$$ \text{Area} = \text{base} \times \text{height} $$

Since one side length is given as 3, the height of the parallelogram can be calculated using the formula for the area:

$$ \text{Area} = \text{base} \times \text{height} = 3 \times \text{height} $$

$$ 24 = 3 \times \text{height} $$

$$ \text{height} = \frac{24}{3} = 8 $$

Now, the height of the parallelogram is 8.

To find the lengths of the other two sides, we can use trigonometry since we have one angle and the height. We can use sine and cosine functions to find the lengths of those sides.

$$ \text{Length of side} = \text{height} \times \sin(\text{angle}) $$

$$ \text{Length of side} = 8 \times \sin\left(\frac{3\pi}{4}\right) $$

$$ \text{Length of side} = 8 \times \frac{\sqrt{2}}{2} $$

$$ \text{Length of side} = 8\sqrt{2} $$

So, the lengths of the other two sides of the parallelogram are $ 8\sqrt{2} $.