PRECALCULUS:GRAPHICAL,...-NASTA ED.
PRECALCULUS:GRAPHICAL,...-NASTA ED.
10th Edition
ISBN: 9780134672090
Author: Demana
Publisher: PEARSON
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Chapter P.4, Problem 57E
Solution

Determine a and b so that figure ABCD is a parallelogram.

  PRECALCULUS:GRAPHICAL,...-NASTA ED., Chapter P.4, Problem 57E , additional homework tip 1

  a=6

  b=5

Given information:

ABCD is a parallelogram.

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:

ABCD is a parallelogram.

  PRECALCULUS:GRAPHICAL,...-NASTA ED., Chapter P.4, Problem 57E , additional homework tip 2

The opposite side of parallelogram are equal.

From figure b is equal to 5

  b=5

AB=CD and BC=AD

Using distance formula −

  AB=(x1−x2)+(y1−y2)=(0−2)2+(0−5)2=(−2)2+(−5)2=4+25=29

  CD=(x1−x2)+(y1−y2)=(4−a)2+(0−5)2=16+a2−8a+(−5)2=16+a2−8a+25=a2−8a+25+16=a2−8a+41

As AB=CD

                                    29=a2−8a+41on squaring both side                                        29=a2−8a+41 a2−8a+41−29=0             a2−8a+12=0     −−−−−−−equation 1

  AD=(x1−x2)+(y1−y2)=(0−4)2+(0−0)2=(−4)2+(0)2=16=4

  BC=(x1−x2)+(y1−y2)=(2−a)2+(5−5)2=4+a2−4a=a2−4a+4=a2−4a+4

BC=AD

  a2−4a+4=4on squaring both side a2−4a+4=16a2−4a+4−16=0a2−4a−12=0      −−−−−−−−equation  2

On comparing equation1 and equation2

   a2−8a+12=a2−4a−12            8a−4a=12+12                    4a=24                       a=244                        a=6

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Chapter P.4, Problem 56E
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Chapter P.4, Problem 58E

Chapter P Solutions

PRECALCULUS:GRAPHICAL,...-NASTA ED.

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