For the following polygons, using the correct notation for line segment lengths:
Do not perform any calculations.
For the following triangles:
For the following polygons:
For the following polygons:
The points A\left(2, 1\right), B\left(7, 3\right) and C\left(7, - 5 \right) are the vertices of a triangle. Determine the area of the triangle.
The points A\left(-2, -3\right), B\left(3, 4\right), C\left(-4,9 \right) and D\left(-9, 2 \right) are the vertices of a square. Determine the area of the square.
The points A\left(-2, -3\right), B\left(2, 5\right), C\left(10, 9\right) and D\left(6, 1 \right) are the vertices of a rhombus. Determine the perimeter of the rhombus as simplified radical value.
The hexagon ABCDEF has all sides congruent and vertices at A\left(-1, 3\right), B\left(4, 3\right), C\left(7, -1\right),D\left(4,-5\right), E\left(-1, -5\right) and F\left(-4, -1 \right). Determine:
Shantelle is putting up a fence around her chicken coop. She has posts at poles at A\left(0, 3\right), B\left(-1, -1\right), C\left(1, -3 \right) and D \left(6,-2 \right). Each unit on the grid represents 1 foot. She purchased 18 feet of fencing.
Gabriella and Joel are asked to calculate the area of the square below. Their solutions are shown below. Compare and contrast their solutions.
If you were helping a friend with this question, which approach would you take? Explain.
Gabriella's Solution
| 1 | \displaystyle XY | \displaystyle = | \displaystyle \sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2} | Distance formula |
| 2 | \displaystyle XY | \displaystyle = | \displaystyle \sqrt{\left(0--4\right)^2+\left(4-0\right)^2} | Substitution property of equality |
| 3 | \displaystyle XY | \displaystyle = | \displaystyle \sqrt{\left(4\right)^2+\left(4\right)^2} | Simplify |
| 4 | \displaystyle XY | \displaystyle = | \displaystyle 4\sqrt{2} | Simplify |
| 5 | \displaystyle \text{Area} | \displaystyle = | \displaystyle XY^2 | Area formula for square |
| 6 | \displaystyle \text{Area} | \displaystyle = | \displaystyle \left(4\sqrt{2}\right)^2 | Substitution property of equality |
| 7 | \displaystyle \text{Area} | \displaystyle = | \displaystyle 32 | Simplify |
Joel's Solution
| 1 | \displaystyle \text{Area} | \displaystyle = | \displaystyle \dfrac{YW \times XZ}{2} | Area formula for rhombus |
| 2 | \displaystyle \text{Area} | \displaystyle = | \displaystyle \dfrac{8 \times 8}{2} | Substitution property of equality |
| 3 | \displaystyle \text{Area} | \displaystyle = | \displaystyle \dfrac{64}{2} | Simplify |
| 4 | \displaystyle \text{Area} | \displaystyle = | \displaystyle 32 | Simplify |
The area of the triangle with vertices A\left(3, 0\right), B\left(8, 2\right) and C\left(8, - 6 \right) has been calculated.