A parallelogram has sides A, B, C, and D. Sides A and B have a length of $7$ $7$ and sides C and D have a length of $4$ $4$ . If the angle between sides A and C is $\frac{7 π}{12}$ $(7 pi)/12$ , what is the area of the parallelogram?

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A parallelogram has sides A, B, C, and D. Sides A and B have a length of $7$ $7$ and sides C and D have a length of $4$ $4$ . If the angle between sides A and C is $\frac{7 π}{12}$ $(7 pi)/12$ , what is the area of the parallelogram?

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Answer 1 · 2016-02-05T17:53:41

The area is $7 \cdot (\sqrt{2} + \sqrt{6}) \approx 27.05 {units}^{2}$ $7 * (sqrt(2) + sqrt(6)) ~~ 27.05 " units"^2$ .

Explanation:

Let the angle between sides $A$ $A$ and $C$ $C$ be $α = \frac{7 π}{12}$ $alpha = (7pi)/12$ .

The formula to compute the area of the parallelogram is

$Area = A \cdot C \cdot sin (α)$ $"Area" = A * C * sin(alpha)$

$= 7 \cdot 4 \cdot sin (\frac{7 π}{12})$ $= 7 * 4 * sin((7pi)/12)$

$= 28 sin (\frac{7 π}{12})$ $= 28 sin((7pi)/12)$

So, the only thing left to do is compute $sin (\frac{7 π}{12})$ $sin((7pi)/12)$ .

Let me show how to do this without the calculator but with some basic knowledge of $sin$ $sin$ and $cos$ $cos$ functions:

$sin (\frac{7 π}{12}) = sin (\frac{π}{4} + \frac{π}{3})$ $sin((7pi)/12) = sin(pi/4 + pi/3)$

... use the formula $sin (x + y) = sin (x) \cdot cos (y) + cos (x) \cdot sin (y)$ $sin(x+y) = sin(x)*cos(y) + cos(x)*sin(y)$ ...

$= sin (\frac{π}{4}) \cdot cos (\frac{π}{3}) + cos (\frac{π}{4}) \cdot sin (\frac{π}{3})$ $= sin(pi/4) * cos(pi/3) + cos(pi/4) * sin(pi/3)$

$= \frac{1}{\sqrt{2}} \cdot \frac{1}{2} + \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2}$ $= 1/sqrt(2) * 1/2 + 1/sqrt(2) * sqrt(3)/2$

$= \frac{1 + \sqrt{3}}{2 \sqrt{2}}$ $= (1 + sqrt(3))/(2sqrt(2))$

$= \frac{\sqrt{2} + \sqrt{6}}{4}$ $= (sqrt(2) + sqrt(6))/4$

Thus, you have the area of

$Area = 28 sin (\frac{7 π}{12}) = 28 \cdot \frac{\sqrt{2} + \sqrt{6}}{4} = 7 \cdot (\sqrt{2} + \sqrt{6}) \approx 27.05 {units}^{2}$ $"Area" = 28 sin((7pi)/12) = 28 * (sqrt(2) + sqrt(6))/4 = 7 * (sqrt(2) + sqrt(6)) ~~ 27.05 " units"^2$

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A parallelogram has sides A, B, C, and D. Sides A and B have a length of $7$ $7$ and sides C and D have a length of $4$ $4$ . If the angle between sides A and C is $\frac{7 π}{12}$ $(7 pi)/12$ , what is the area of the parallelogram?

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A parallelogram has sides A, B, C, and D. Sides A and B have a length of 7 77 and sides C and D have a length of 4 4 4 . If the angle between sides A and C is (7 pi)/12 7π12(7 pi)/12 , what is the area of the parallelogram?

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A parallelogram has sides A, B, C, and D. Sides A and B have a length of $7$ $7$ and sides C and D have a length of $4$ $4$ . If the angle between sides A and C is $\frac{7 π}{12}$ $(7 pi)/12$ , what is the area of the parallelogram?