Question

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For problems 11 and 12, plot the points in the coordinate plane. Then find the perimeter and area of the polygon.

11. A(-2, 1), B(2, 5), C(5, 1) (round your answers to the nearest tenth.)
Perimeter:______________

Area: __________


12. A(-3, 5), B(1, 6), C(3, -2), D(-1, -3)

Perimeter:

Area:

Answer 1

11. Perimeter: 17.7 units

11. Area: 14 square units

12. Perimeter: 24.7 units

12. Area: 34 square units

Explanation:

Question 11

Given vertices of the polygon:

• A = (-2, 1)
• B = (2, 5)
• C = (5, 1)

Plot the given points in the coordinate plane (see attachment 1).

From inspection, we can see that the polygon is a triangle with the following dimensions:

• Base = 7 units
• Height = 4 units

`\boxed{\begin{minipage}{7.4 cm}\underline{Distance between two points}\\\\$d=√((x_2-x_1)^2+(y_2-y_1)^2)$\\\\\\where $(x_1,y_1)$ and $(x_2,y_2)$ are the two points.\\\end{minipage}}`

The length of AC is 7 units.

To find the length of AB and BC, use the distance formula.

`\begin{aligned}AB&=√((x_B-x_A)^2+(y_B-y_A)^2)\\&=√((2-(-2))^2+(5-1)^2)\\&=√((4)^2+(4)^2)\\&=√(16+16)\\&=√(32)\\&=4√(2)\;\sf units\end{aligned}`

`\begin{aligned}BC&=√((x_C-x_B)^2+(y_C-y_B)^2)\\&=√((5-2)^2+(1-5)^2)\\&=√((3)^2+(-4)^2)\\&=√(9+16)\\&=√(25)\\&=5\;\sf units\end{aligned}`

Therefore:

`\begin{aligned}\textsf{Perimeter}&=AB+BC+AC\\&=4√(2)+5+7\\&=12+4√(2)\\&=17.6568542...\\&=17.7\;\sf units\end{aligned}`

`\begin{aligned}\textsf{Area}&=(1)/(2)\cdot 7\cdot 4\\\\&=(7)/(2) \cdot 4\\\\&=(28)/(2)\\\\&=14\;\sf square\;units\end{aligned}`

Question 12

Given vertices of the polygon:

• A = (-3, 5)
• B = (1, 6)
• C = (3, -2)
• D = (-1, -3)

Plot the given points in the coordinate plane (see attachment 2).

From inspection, we can see that the polygon is a rectangle.

`\boxed{\begin{minipage}{7.4 cm}\underline{Distance between two points}\\\\$d=√((x_2-x_1)^2+(y_2-y_1)^2)$\\\\\\where $(x_1,y_1)$ and $(x_2,y_2)$ are the two points.\\\end{minipage}}`

Since AB = DC and BC = AD, find the length of AB and BC using the distance formula.

`\begin{aligned}AB&=√((x_B-x_A)^2+(y_B-y_A)^2)\\&=√((1-(-3))^2+(6-5)^2)\\&=√((4)^2+(1)^2)\\&=√(16+1)\\&=√(17)\;\sf units\end{aligned}`

`\begin{aligned}BC&=√((x_C-x_B)^2+(y_C-y_B)^2)\\&=√((3-1)^2+(-2-6)^2)\\&=√((2)^2+(-8)^2)\\&=√(4+64)\\&=√(68)\\&=2√(17)\;\sf units\end{aligned}`

Therefore:

`\begin{aligned}\textsf{Perimeter}&=AB+BC+CD+AD\\&=2(AB+BC)\\&=2(√(17)+2√(17))\\&=2(3√(17))\\&=6√(17)\\&=24.7386337...\\&=24.7\;\sf units\end{aligned}`

`\begin{aligned}\textsf{Area}&=AB\cdot BC\\&=√(17) \cdot 2√(17)\\&=34\;\sf square\;units\end{aligned}`

Answer 2

Question 11

Given vertices:

• A(-2, 1), B(2, 5), C(5, 1)

Plot the points (see attached).

AC is the base of the triangle:

• AC = 5 - (-2) = 7 units

Add BD, the height. It will help to find the length of the other two sides and the area:

• AD = 4 units and
• DC = 3 units (from the graph)
• BD = 5 - 1 = 4 units

Find AB and BC using Pythagorean:

• 
  `AB = √(4^2+4^2)=√(32) =5.7\ units`
• 
  `BC=√(4^2+3^2)=√(25)=5\ units`

Perimeter: 7 + 5.7 + 5 = 15.7 units

Area: 1/2*7*4 = 14 units²

Question 12

Given vertices:

• A(-3, 5), B(1, 6), C(3, -2), D(-1, -3)

Plot the points (see attached).

As we see this is a rectangle.

Find two adjacent sides using distance equation:

• 
  `AB = √((1 - (-3))^2+(6-5)^2) =√(16+1)=√(17)=4.1\ units`
• 
  `AD = √((-1 - (-3))^2+(-3-5)^2) =√(4+64)=√(68)=2√(17) =8.2\ units`

Perimeter: 2(4.1 + 8.2) = 2(12.3) = 24.6 units

Area: √17 * 2√17 = 2*17 = 34 units²