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In this 2x3 grid, each lattice point is one unit away from its nearest neighbors. A total of 14 isosceles triangles(but not right triangles), each with an area of ½u squared have only two vertices that are one unit apart in the grid. how many such half-unit triangles have at least two vertices in an x-by-x grid?

Sagot :

The area of a shape is the amount of space it can occupy.

The number of isosceles triangle in the grid is: [tex]\mathbf{\frac{2x^2}{u^2}}[/tex]

The dimension of the grid is given as:

[tex]\mathbf{Dimension = x\ by\ x}[/tex]

Calculate the area

[tex]\mathbf{A_1 = x\ \times \ x}[/tex]

[tex]\mathbf{A_1 = x^2}[/tex]

The area of each isosceles triangle is given as:

[tex]\mathbf{A_2 = \frac 12u^2}[/tex]

The number (n) of the triangle in an x-by-x grid is:

[tex]\mathbf{n = A_1 \div A_2}[/tex]

So, we have:

[tex]\mathbf{n = x^2 \div \frac 12u^2}[/tex]

Express as products

[tex]\mathbf{n = x^2 \times \frac{2}{u^2}}[/tex]

[tex]\mathbf{n = \frac{2x^2}{u^2}}[/tex]

Hence, the number of isosceles triangle in the grid is: [tex]\mathbf{\frac{2x^2}{u^2}}[/tex]

Read more about areas at:

https://brainly.com/question/16418397