Calculate the area of the above irregular object
1 answer:
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Answer:
27 miles.
Step-by-step explanation:
Here I attach the draw of the coordinates.
Tony traveled 3 segments. The first was from (12,6) to (12, 15), where, leting 12 constant, he moved from 6 to 15 in the ordinates axis, which implies 9 units. This is the section 1 in the draw.
Then he moved from point B to C. If you notice, this distance is the hypotenuse on the the triangle DBC. We can find this value using Pitagoras' theorem:
DB^2 + CD^2 = CB^2
With DB=15 and CD=8 (12 minus 4 = 8)
15^2 + 8^2 = 289
So CB^2=289
Applying sqr root:
CB = 17
So, the second section has a measure of 17 units.
Finally, the 3rd section is the hypotenuse of the DAC triangle and we can use Pitagoras to solve it:
CD^2 + AD^2 = CA^2
8^2 + 6^2 = CA^2
64 + 36 = 100
So, CA=10
In the 3r section we traveled 10 units.
So, in total he traveled 10 + 17 + 9 = 36 units
As every unit is 0.75 miles he traveled 36*0.75 miles:
36*0.75 = 27 miles
He traveled in total 27 miles!!
