PRECALCULUS:...COMMON CORE ED.-W/ACCESS
PRECALCULUS:...COMMON CORE ED.-W/ACCESS
10th Edition
ISBN: 9780134781839
Author: Demana
Publisher: PEARSON
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Chapter P.2, Problem 22E
Solution

Find the perimeter and area of the figure determined by the points.

Perimeter = 22

Area = 30

Given information:

The given points are (-2, 1), (-2, 6) (4, 6) (4, 1)

Concept used:

  • First, find the value for x on the x-axis.
  • Next, find the y-value on the y-axis.
  • The point should be plotted at the intersection of x and y.
  • Finally, plot the point on your graph at the appropriate spot.
  • Then join all the points to make a closed figure for calculating perimeter and area.

Calculation:

The given points are A(-2, 1), B(-2, 6), C(4, 6), D(4,1)

Plotting the points in the graph -

  PRECALCULUS:...COMMON CORE ED.-W/ACCESS, Chapter P.2, Problem 22E

Perimeter of enclosed figure ABCD is = AB+BC+CD+DA

Using distance formula −

A(-2, 1), B(-2, 6)

  AB=(x1−x2)+(y1−y2)=(−2−(−2))2+(1−6)2=(−2+2)2+(−5)2=0+25=25=5

AB = CD

So CD = 5

B(-2, 6), C(4, 6)

  BC=(x1−x2)+(y1−y2)=(−2−4)2+(6−6)2=(−6)2+(0)2=36=6

BC=DA

So DA=6

Perimeter of enclosed figure ABCD-

   AB+BC+CD+DA=5+6+5+6=22

Area of enclosed figure ABCD is −

As AB=CD and BC=DA

Therefore ABCD is a rectangle.

Area of rectangle ABCD = length⋅breadth

Length = DA =6

Breadth =AB 5

  Area=length⋅breadth=6⋅5=30

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Chapter P.2, Problem 21E
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Chapter P.2, Problem 23E

Chapter P Solutions

PRECALCULUS:...COMMON CORE ED.-W/ACCESS

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