McGraw Hill Glencoe Geometry, 2012
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McGraw Hill Glencoe Geometry, 2012 View details
6. Two-Dimensional Figures
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Exercise 27 Page 62

Draw the diagram on a coordinate plane.

Diagram:

Figure: Quadrilateral, square
Perimeter: 20
Area: 25

Practice makes perfect

Before we calculate the perimeter and area, let's draw a diagram of the figure with the given coordinates.

Drawing a Diagram

Let's start by plotting the points on a coordinate plane. We can see four sides, so this is a quadrilateral.

This quadrilateral appears to be a square, but we cannot assume this while we are calculating the perimeter and the area.

Finding the Perimeter

The perimeter is the sum of the length of the sides. To find the length of the sides, we can use the Distance Formula. Let's start with finding the distance between P and Q.
PQ=sqrt((x_2-x_1)^2+(y_2-y_1)^2)
PQ=sqrt(( 3-( - 1))^2+( 4- 1)^2)
â–Ľ
Simplify
PQ=sqrt((3+1)^2+(4-1)^2)
PQ=sqrt(4^2+3^2)
PQ=sqrt(16+9)
PQ=sqrt(25)
PQ=5
Similar calculations give us the lengths of the other sides.
Side Expression Value
QR sqrt((6-3)^2+(0-4)^2) 5
RS sqrt((2-6)^2+(- 3-0)^2) 5
SP sqrt((- 1-2)^2+(1-(- 3))^2) 5

We can add these lengths to find the perimeter. 5+5+5+5=20units

Finding the Area

Remember, although this quadrilateral appears to be a square, we cannot assume this. Instead, we can draw horizontal and vertical segments through the vertices to break this quadrilateral into smaller pieces. We can see a small square in the middle and four congruent right triangles when we do so.

The area of the 1 by 1 square in the middle is 1* 1=1 units^2. Counting grid units, we can find that the length of the legs of the right triangles are b=4 and h=3. Now, we can use the formula for area of a triangle to find the area of any one of these.
A=1/2bh
A=1/2( 4)( 3)
A=6
The area of the triangles is 6 units^2. Adding the area of the square in the middle to these, we get the area of the quadrilateral. 6+6+6+6+1=25units^2

Alternative Solution

Is this quadrilateral a square?

Draw a diagonal and take a look at only half of the quadrilateral. What do we know about this triangle?

We have already seen that PQ=QR=5. We can now use the Distance Formula to find PR.
PR=sqrt((x_2-x_1)^2+(y_2-y_1)^2)
PR=sqrt(( 6-( - 1))^2+( 0- 1)^2)
â–Ľ
Simplify
PR=sqrt((6+1)^2+(0-1)^2)
PR=sqrt(7^2+(- 1)^2)
PR=sqrt(49+1)
PR=sqrt(50)
To check if â–ł PQR is a right triangle, we can check whether the Pythagorean Theorem holds true for the known lengths of the sides.
a^2+b^2? =c^2
5^2+5^2? =(sqrt(50))^2
25+25? =50
50=50
Since a^2+b^2=c^2, we can conclude that â–ł PQR is indeed a right triangle with the right angle at Q. Similarly we can show that all angles of the quadrilateral PQRS are right angles. Since we already know that the sides are congruent, we have now proven that PQRS is a square.