Four congruent isosceles right triangles are cut from the 4 corners of a square with a side of 20 units. The length of one leg o
1 answer:
Answer:
368 sq. units.
Step-by-step explanation:
We have a square of side lengths 20 units and we cut four congruent isosceles right triangles from the corners of the square.
Now, the four isosceles right triangles have one leg equal to 4 units.
Therefore, the area of four triangles = sq. units.
Now, we have the area of the given square is (20 × 20) = 400 sq. units.
Therefore, the area of the remaining octagon will be (400 - 32) = 368 sq. units. (Answer)
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Check the picture below.
so notice, the sides AB and AC you can pretty much count them off the grid.
now, to get the CB side.
![\bf \textit{distance between 2 points}\\ \quad \\ \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) C&({{ -5}}\quad ,&{{ 1}})\quad % (c,d) B&({{ 3}}\quad ,&{{ -5}}) \end{array}\qquad % distance value d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2} \\\\\\ CB=\sqrt{[3-(-5)]^2+[-5-1]^2}\implies CB=\sqrt{(3+5)^2+(-5-1)^2} \\\\\\ CB=\sqrt{8^2+(-6)^2}\implies CB=\sqrt{100}\implies CB=10](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bdistance%20between%202%20points%7D%5C%5C%20%5Cquad%20%5C%5C%0A%5Cbegin%7Barray%7D%7Blllll%7D%0A%26x_1%26y_1%26x_2%26y_2%5C%5C%0A%25%20%20%28a%2Cb%29%0AC%26%28%7B%7B%20-5%7D%7D%5Cquad%20%2C%26%7B%7B%201%7D%7D%29%5Cquad%20%0A%25%20%20%28c%2Cd%29%0AB%26%28%7B%7B%203%7D%7D%5Cquad%20%2C%26%7B%7B%20-5%7D%7D%29%0A%5Cend%7Barray%7D%5Cqquad%20%0A%25%20%20distance%20value%0Ad%20%3D%20%5Csqrt%7B%28%7B%7B%20x_2%7D%7D-%7B%7B%20x_1%7D%7D%29%5E2%20%2B%20%28%7B%7B%20y_2%7D%7D-%7B%7B%20y_1%7D%7D%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0ACB%3D%5Csqrt%7B%5B3-%28-5%29%5D%5E2%2B%5B-5-1%5D%5E2%7D%5Cimplies%20CB%3D%5Csqrt%7B%283%2B5%29%5E2%2B%28-5-1%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0ACB%3D%5Csqrt%7B8%5E2%2B%28-6%29%5E2%7D%5Cimplies%20CB%3D%5Csqrt%7B100%7D%5Cimplies%20CB%3D10)
sum all three sides up, and that's the perimeter of the triangle.
so notice, the sides AB and AC you can pretty much count them off the grid.
now, to get the CB side.
sum all three sides up, and that's the perimeter of the triangle.