Two opposite sides of a parallelogram have lengths of #7 #. If one corner of the parallelogram has an angle of #pi/4 # and the parallelogram's area is #32 #, how long are the other two sides?

Question

Two opposite sides of a parallelogram have lengths of #7  #.  If one corner of the parallelogram has an angle of #pi/4   # and the parallelogram's area is #32      #, how long are the other two sides?

Wyatt Anderson · Accepted Answer

The other two sides #6.465# unit long.

Area of a parallelogram is given by #axxbxxsintheta#, where #a# and #b# are two sides of parallelogram and #theta# is angle included between them. As here one pair of sides are #7# units, included angle is #pi/4# and area is #32#, we have #7xxbxxsin(pi/4)=32# or #7xxbxx1/sqrt2=32# Hence #b=(32xxsqrt2)/7=(32xx1.4142)/7=6.465# or #b=6.465#

Savannah Aguilar · Answer

undefined Given that two opposite sides of a parallelogram have lengths of 7, and one angle of the parallelogram is $ \frac{\pi}{4} $ radians, and the area of the parallelogram is 32 square units, we can find the lengths of the other two sides.

The area of a parallelogram is given by the formula:

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

Let $ b $ be the length of one of the two sides adjacent to the angle $ \frac{\pi}{4} $, and let $ h $ be the perpendicular distance from this side to the opposite side.

From trigonometry, we know that $ h = b \sin(\frac{\pi}{4}) = \frac{b}{\sqrt{2}} $.

So, the area of the parallelogram can also be expressed as $ bh = 32 $.

We know that one side length is 7, so $ b = 7 $.

Substituting the values into the equation, we get:

$$ 7 \times \frac{7}{\sqrt{2}} = 32 $$

Solve for $ b $:

$$ \frac{49}{\sqrt{2}} = 32 $$

$$ \frac{49}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = 32 \times \frac{\sqrt{2}}{\sqrt{2}} $$

$$ b = \frac{49\sqrt{2}}{2} $$

So, the lengths of the other two sides are $ \frac{49\sqrt{2}}{2} $ and 7 units.