Question

find the perimeter and area of the polygon with the given vertices. round to the nearest tenth, if necessary. p(-3,4),q(1,4),r(-3,-2),s(3,-2) the perimeter is about □ units. the area is □ square units. explain how you found your answer.

Explanation:

Step1: Identify the shape

Plot the points \( P(-3,4) \), \( Q(1,4) \), \( R(-3,-2) \), \( S(3,-2) \). We can see that \( PQ \) is horizontal (same y - coordinate), \( PR \) is vertical (same x - coordinate), \( RS \) is horizontal, and \( SQ \) (wait, actually, let's find the lengths of each side.

First, find the length of \( PQ \): The distance between \( P(-3,4) \) and \( Q(1,4) \). Since \( y \) - coordinates are the same, the distance is \( |x_2 - x_1|=\vert1-(-3)\vert = 4 \).

Step2: Find length of \( PR \)

The distance between \( P(-3,4) \) and \( R(-3,-2) \). Since \( x \) - coordinates are the same, the distance is \( |y_2 - y_1|=\vert-2 - 4\vert=6 \).

Step3: Find length of \( RS \)

The distance between \( R(-3,-2) \) and \( S(3,-2) \). Since \( y \) - coordinates are the same, the distance is \( |x_2 - x_1|=\vert3-(-3)\vert = 6 \).

Step4: Find length of \( SQ \)

Wait, no, the polygon is \( P - Q - S - R - P \)? Wait, let's check the coordinates again. Let's list the sides: \( PQ \), \( QS \), \( SR \), \( RP \). Wait, \( Q(1,4) \), \( S(3,-2) \): distance between \( Q(1,4) \) and \( S(3,-2) \) is \( \sqrt{(3 - 1)^2+(-2 - 4)^2}=\sqrt{2^2+(-6)^2}=\sqrt{4 + 36}=\sqrt{40}\approx6.3 \). Wait, no, maybe I made a mistake in the shape. Wait, \( P(-3,4) \), \( Q(1,4) \): horizontal line. \( Q(1,4) \), \( S(3,-2) \): line. \( S(3,-2) \), \( R(-3,-2) \): horizontal line. \( R(-3,-2) \), \( P(-3,4) \): vertical line. So the polygon is a quadrilateral with sides \( PQ = 4 \), \( QS=\sqrt{(3 - 1)^2+(-2 - 4)^2}=\sqrt{4 + 36}=\sqrt{40}\approx6.3 \), \( SR = 6 \), \( RP = 6 \). Wait, no, that can't be. Wait, maybe it's a trapezoid? Wait, \( PQ \) is from \( x=-3 \) to \( x = 1 \), \( y = 4 \); \( SR \) is from \( x=-3 \) to \( x = 3 \), \( y=-2 \); \( PR \) is from \( x=-3 \), \( y = 4 \) to \( x=-3 \), \( y=-2 \); and \( QS \) is from \( x = 1 \), \( y = 4 \) to \( x = 3 \), \( y=-2 \). Wait, actually, let's re - examine the coordinates.

Wait, \( P(-3,4) \), \( Q(1,4) \): length \( PQ=\vert1-(-3)\vert = 4 \) (horizontal line, \( y = 4 \)).

\( Q(1,4) \), \( S(3,-2) \): distance \( d_{QS}=\sqrt{(3 - 1)^2+(-2 - 4)^2}=\sqrt{4 + 36}=\sqrt{40}\approx6.3 \)

\( S(3,-2) \), \( R(-3,-2) \): length \( SR=\vert3-(-3)\vert = 6 \) (horizontal line, \( y=-2 \))

\( R(-3,-2) \), \( P(-3,4) \): length \( RP=\vert4-(-2)\vert = 6 \) (vertical line, \( x=-3 \))

Now, perimeter is \( PQ + QS+SR + RP=4+\sqrt{40}+6 + 6=16+\sqrt{40}\approx16 + 6.3246\approx22.3 \)

For area: The figure can be considered as a trapezoid. The two parallel sides (bases) are \( PQ = 4 \) and \( SR = 6 \), and the height is the vertical distance between the two horizontal lines \( y = 4 \) and \( y=-2 \), which is \( 4-(-2)=6 \). The formula for the area of a trapezoid is \( A=\frac{(b_1 + b_2)}{2}\times h \), where \( b_1 = 4 \), \( b_2 = 6 \), \( h = 6 \). So \( A=\frac{(4 + 6)}{2}\times6=\frac{10}{2}\times6 = 30 \). Wait, but let's check another way. The figure can also be seen as a rectangle plus a triangle? Wait, \( P(-3,4) \), \( Q(1,4) \), \( R(-3,-2) \), \( S(3,-2) \). The vertical side \( PR = 6 \), the horizontal side from \( P \) to \( Q \) is 4, from \( R \) to \( S \) is 6. The difference in the horizontal sides is \( 6 - 4 = 2 \), which is split equally on both sides? Wait, no, the trapezoid area formula: the two parallel sides are the horizontal sides \( PQ \) (length 4) and \( SR \) (length 6), and the height is the vertical distance between \( y = 4 \) and \( y=-2 \), which is 6. So \( A=\frac{(4 + 6)}{2}\times6=30 \).

Wait, let's re - check the perimeter. The sides are \( PQ = 4 \), \( QS=\sqrt{(3 - 1)^2+(-2 - 4)^2}=\sqrt{4 + 36}=\sqrt{40}\approx6.3 \), \( SR = 6 \), \( RP = 6 \). So perimeter \(=4 + 6.3+6 + 6=22.3 \) (rounded to the nearest tenth).

Step1 (Revised for perimeter):

Calculate length of \( PQ \):
Since \( P(-3,4) \) and \( Q(1,4) \) have the same \( y \) - coordinate, the distance \( PQ=\vert1-(-3)\vert=4 \).

Step2 (Revised for perimeter):

Calculate length of \( QS \):
Using distance formula \( d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2} \), for \( Q(1,4) \) and \( S(3,-2) \), \( d=\sqrt{(3 - 1)^2+(-2 - 4)^2}=\sqrt{4 + 36}=\sqrt{40}\approx6.3 \).

Step3 (Revised for perimeter):

Calculate length of \( SR \):
Since \( S(3,-2) \) and \( R(-3,-2) \) have the same \( y \) - coordinate, \( SR=\vert3-(-3)\vert = 6 \).

Step4 (Revised for perimeter):

Calculate length of \( RP \):
Since \( R(-3,-2) \) and \( P(-3,4) \) have the same \( x \) - coordinate, \( RP=\vert4-(-2)\vert = 6 \).

Step5 (Perimeter calculation):

Perimeter \(=PQ + QS+SR + RP=4+\sqrt{40}+6 + 6=16+\sqrt{40}\approx16 + 6.3246\approx22.3 \).

Step1 (Area calculation):

Identify the shape as a trapezoid with bases \( b_1 = PQ = 4 \) and \( b_2 = SR = 6 \), and height \( h \) (vertical distance between \( y = 4 \) and \( y=-2 \)) \( h=\vert4-(-2)\vert = 6 \).

Step2 (Area calculation):

Use trapezoid area formula \( A=\frac{(b_1 + b_2)}{2}\times h \).
Substitute \( b_1 = 4 \), \( b_2 = 6 \), \( h = 6 \): \( A=\frac{(4 + 6)}{2}\times6=\frac{10}{2}\times6 = 30 \).

Answer:

The perimeter is about \(\boldsymbol{22.3}\) units.
The area is \(\boldsymbol{30}\) square units.