Question

A certain city block is in the form of a parallelogram. Two of its sides are each 42 m. long; the other two sides are each 22 m. in length. If the distance between the first pair of sides is 12 m., find the area of the land in the block, and the length of the diagonals.

Answer 1

Refer to the figure ABCD parallelogram.

In triangle ABE,

$sin\alpha = \frac{12}{22}$ and, $cos\alpha = \sqrt{1-sin^2\alpha} = \frac{18.44}{22}$

Now, In triangle ABD, applying cosine law,

$cos\alpha = \frac{22^2+42^2-x^2}{2\times 22 \times 42 }$

Putting value of $cos\alpha$, and solving for x diagonal BD, we get

$x = \sqrt{22^2+42^2-(2\times 22 \times 42 \times \frac{18.44}{22})} = 26.44 m$

Similarly for diagonal AC, $\theta = \angle ADC = 180 - \alpha$

Applying cosine law again, we get

$cos(\pi-\alpha) = \frac{22^2+42^2-y^2}{2\times 22 \times 42 }$

$y = \sqrt{22^2+42^2+(2\times 22 \times 42 \times \frac{18.44}{22})} = 61.62 m$

Area of the parallelogram is equal to the sum of the areas of the triangles ABD and BCD.

Area of parallelogram, $A = \frac{1}{2}(AD)(BE)+ \frac{1}{2}(BC)(BE) = \frac{1}{2}(12 \times 42)+ \frac{1}{2}(12 \times 42) = 504 m^2$