Question

To approximate the area of a lake, Cindy walks around the perimeter of the lake, taking the measurements shown in the illustration. Using this technique, what is the approximate area of the lake?

Answer 1

`A_t = A_1 +A_2 +A_3 = 1723.414+ 2287.462+4787.75= 8798.626 ft^2`

Explanation:

We assume that the plot attaced is the illustration for the problem.

For this case we need to begin finding the values for x and y using the cosine law given by:

`a^2 = b^2 + c^2 - a bc cos (a)`

We can begin finding the value of x like this:

`x^2 = 50^2 + 70^2 -2 (50)(70) cos 100 =8615.537`

`x= 92.820 ft`

And similar in order to find the value of y we got this:

`y^ 2 = 125^2 + 100^2 - 2(125) (100) cos 50=9555.310`

`y=97.751 ft`

We can find the area for the 3 triangls on this way, beggining from the left triangle:

`A_1 = (bh)/(2)= (50 * 70 sin (100))/(2)=1723.414 ft^2`

For the traingel in the most left part we have this:

`A_3 = (bh)/(2)= (125 * 100 sin (50))/(2)=4787.75 ft^2`

We obtain this value because if we find the angle a we got:

`sin a = (100)/(97.751) sin (50) = 0.784`

And then the angle a = 51.60 and for b = 180-51.60-50=78.402 degrees

Then we can create two equations in terms of h like this:

`(97.75)^2 = h^2 + (100-x)^2`

`(125)^2 = h^2 +x^2`

If we subtract both equation we got this:

`125^2 -97.75^2 = x^2 -(100-x)^2`

`200 x = 16069.742`

`x = 95.755`

And then the area is just
`A_3 = (100*95.755)/(2)= 4787.75 ft`

For the middle area we can begin finding the midsegment like this:

`r= (1)/(2) (50+92.820+97.75) = 120.285 ft`

And now we can find the area 2 like this:

`A_2 = 2287.462`

And then we just need to add the areas and we got:

`A_t = A_1 +A_2 +A_3 = 1723.414+ 2287.462+4787.75= 8798.626 ft^2`