Question

1. find the value of x and y in each diagram.
2. pete was practicing finding the sum of the angles in polygons. below are his calculations, but he lost all his work on the shapes that went with them.

figure out which calculation goes with which shape, and show how to divide up each shape to match the calculation.
3(180) 5(180) - 360 2(360) + 180 5(180)

1. draw generic rectangles to match each calculation, and use your rectangles to find the area.

a. (2x - 3y)(3x - 2y)=
b. (x - y - 10)(y - x)=

Explanation:

2a.

Step1: Use angle - sum property of a triangle

In the larger triangle with angles \(x\), \(60^{\circ}\), and \(80^{\circ}\), the sum of interior angles of a triangle is \(180^{\circ}\). So \(x + 60+80=180\).
\[x=180-(60 + 80)=40^{\circ}\]

Step2: Use angle - sum property of a smaller triangle

In the smaller triangle with angles \(y\), \(30^{\circ}\), and the angle adjacent to \(x\) (which is also \(40^{\circ}\)), since the sum of interior angles of a triangle is \(180^{\circ}\), we have \(y+30 + 40=180\).
\[y=180-(30 + 40)=110^{\circ}\]

Step1: Use the property of exterior angles

The exterior - angle of a polygon is related to its interior angles. First, find the interior angles of the polygon. Let's consider the angles around the vertices.
The sum of the interior angles of a polygon and its exterior angles at each vertex is \(180^{\circ}\).
We know that the sum of the angles around a point is \(360^{\circ}\).
Let's find the missing interior angles.
The interior angle corresponding to the \(100^{\circ}\) exterior angle is \(180 - 100=80^{\circ}\), and the interior angle corresponding to the \(150^{\circ}\) exterior angle is \(180 - 150 = 30^{\circ}\)
The sum of the interior angles of the polygon is \(70+60 + 80+30+x=360\) (since the sum of interior angles of a quadrilateral is \(360^{\circ}\))
\[x=360-(70 + 60+80 + 30)=120^{\circ}\]

Step1: Recall the formula for the sum of interior angles of a polygon \(S=(n - 2)\times180^{\circ}\)

• For \(3(180)\):

We know that \(S=(n - 2)\times180=3\times180\), then \(n-2 = 3\), \(n = 5\), which is a pentagon. We can divide a pentagon into 3 non - overlapping triangles by drawing 2 non - intersecting diagonals from one vertex.

• For \(5(180)-360\):

\[S=(n - 2)\times180=5\times180-360\]
\[S = 900-360=540\]
\((n - 2)\times180=540\), \(n-2 = 3\), \(n=5\) (a pentagon). Another way to think about \(5(180)-360\) is that if we consider 5 triangles meeting at a point inside the pentagon, we subtract the \(360^{\circ}\) around that point.

• For \(2(360)+180\):

\[S=2\times360 + 180=720+180 = 900\]
\((n - 2)\times180=900\), \(n-2=5\), \(n = 7\), which is a heptagon. We can divide a heptagon into 5 non - overlapping triangles by drawing 4 non - intersecting diagonals from one vertex.

• For \(5(180)\):

\((n - 2)\times180=5\times180\), \(n-2 = 5\), \(n=7\), which is a heptagon. We can think of it as 5 non - overlapping triangles that make up the heptagon.

Answer:

\(x = 40^{\circ}\), \(y = 110^{\circ}\)

2b.