Question

Draw and classify the given vertices. Find the perimeter and area of the polygon to the nearest tenth.

Answer 1

QUESTION 1a

The given polygon has vertices $A(-2,3),B(3,1),C(-2,-1),D(-3,1)$.

We plot the points and connect them to obtain the figure as shown in the attachment.

The polygon has four sides and two pairs of adjacent sides equal.

Therefore Polygon ABCD can be classified as a quadrilateral, specifically a kite.

QUESTION 1b

We can find the perimeter by adding the length of all the sides of the kite

$Perimeter=|AB|+|BC|+|CD|+|AD|$

Or

$Perimeter=2|AB|+2|AD|$

Recall the distance formula

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

We use the distance formula to find the length of each side.

$|AB|=\sqrt{(3--2)^2+(1-3)^2}$

$|AB|=\sqrt{(3+2)^2+(1-3)^2}$

$|AB|=\sqrt{(5)^2+(-2)^2}$

$|AB|=\sqrt{25+4}$

$|AB|=\sqrt{29}=5.385$

Length of side AD

$|AD|=\sqrt{(-3--2)^2+(1-3)^2}$

$|AD|=\sqrt{(-3+2)^2+(1-3)^2}$

$|AD|=\sqrt{(-1)^2+(-2)^2}$

$|AD|=\sqrt{1+4}$

$|AD|=\sqrt{5}=2.24$

$Perimeter=2(5.1)+2(2.236)$

$Perimeter=10.77+4.472$

$Perimeter=15.242$

The perimeter of the kite to the nearest tenth is 15.2 units

QUESTION 1c

The area of kite ABCD is twice the area of ΔABD

$Area\:of\:ABD=\frac{1}{2}\times |BD| \times |AE|$

$Area\:of\:ABD=\frac{1}{2}\times 6 \times 2$

$Area\:of\:ABD=6\:square\units$

Therefore the are of the kite is

$=2\times 6=12$

The area of the kite is 12 square units.

QUESTION 2a

The vertices of the given polygon are

$P(-3,-4),Q(3,-3),R(3,-2),S(-3,2)$.

We plot all the four points as shown in the diagram in the attachment.

The polygon has one pair of opposite sides parallel and has four sides.

The polygon is a quadrilateral, specifically a trap-ezoid.

QUESTION 2b

The area of the trap-ezoid can be found using the formula

$Area=\frac{1}{2}(|RQ|+|PS|)\times |RU|$.

We use the absolute value method to find the length of RQ,PS and RU because they are vertical and horizontal lines.

$|RQ|=|-3--2|$

$|RQ|=|-3+2|$

$|RQ|=|-1|$

$|RQ|=1$

The length of PS is

$|PS|=|-4-2|$

$|PS|=|-6|$

$|PS|=6$

The length of RU

$|RU|=|-3-3|$

$|RU|=|-6|$

$|RU|=6$

The area of the trap-ezoid is

$Area=\frac{1}{2}(1+6)\times 6$.

$Area=(7)\times 3$.

$Area=21$.

Therefore the area of the trap-ezoid is 21 square units.

 

QUESTION 2c

The perimeter of the trap-ezoid

$=|PQ|+|RS|+|QR|+|PS|$

We use the distance formula to determine the length of RS and PQ.

$|RS|=\sqrt{(3--3)^2+(-2-2)^2}$

$|RS|=\sqrt{(3+3)^2+(-2-2)^2}$

$|RS|=\sqrt{(6)^2+(-4)^2}$

$|RS|=\sqrt{36+16}$

$|RS|=\sqrt{52}$

$|RS|=7.211$

We now calculate the length of PQ

$|PQ|=\sqrt{(3--3)^2+(-3--4)^2}$

$|PQ|=\sqrt{(3+3)^2+(-3+4)^2}$

$|PQ|=\sqrt{(6)^2+(1)^2}$

$|PQ|=\sqrt{36+1}$

$|PQ|=\sqrt{37}$

$|PQ|=6.083$

We already found that,

$|PS|=6$

and

$|RQ|=1$

We substitute all these values to get,

$Perimeter=6+1+7.211+6.083$

$Perimeter=20.294$

To the nearest tenth, the perimeter quadrilateral PQRS is 20.3 units.

QUESTION 3a

The given polygon has vertices

$E(-4,1),F(-2,3),G(-2,-4)$

We plot all the three points to the polygon shown in the diagram. See attachment.

The polygon has three unequal sides, therefore it is a triangle, specifically scalene triangle.

QUESTION 3b

We can calculate the area of this triangle using the formula,

$Area=\frac{1}{2} \times |FG| \times |EH|$ see attachment

We can use the absolute value method to find the length of FG and EH because they are vertical or horizontal lines.

$|FG|=|-4-3|$

$|FG|=|-7|$

$|FG|=7$

Now the length of EH is

$|EH|=|-4--2|$

$|EH|=|-4+2|$

$|EH|=|-2|$

$|EH|=2$

The area is

$Area=\frac{1}{2} \times 7 \times 2$

$Area=7$

Therefore the area of the triangle is 7 square units.

QUESTION 3c

The perimeter of the triangle can be found by adding the length of the three sides of the triangle.

$Perimeter=|EF|+|FG|+|GE|$

The length of EF can be found using the distance formula,

$|EF|=\sqrt{(-2--4)^2+(3-1)^2}$

$|EF|=\sqrt{(-2+4)^2+(3-1)^2}$

$|EF|=\sqrt{(2)^2+(2)^2}$

$|EF|=\sqrt{4+4}$

$|EF|=\sqrt{8}$

$|EF|=2.828$

The length of EG can also be found using the distance formula

$|EG|=\sqrt{(-4--2)^2+(1--4)^2}$

$|EG|=\sqrt{(-4+2)^2+(1+4)^2}$

$|EG|=\sqrt{(-2)^2+(5)^2}$

$|EG|=\sqrt{4+25}$

$|EG|=\sqrt{29}$

$|EG|=5.385}$

We found $|FG|=7$

The perimeter of the triangle is

$Perimeter=5.385+7+2.828$

$Perimeter=15.213$

Therefore the perimeter of the triangle is 15.2 units to the nearest tenth