Monday, December 10, 2012

Line Intercept, Slope, Area/Geometry Questions

Check out Mathcounts here, the best competition math program for middle school students. Download this year's Mathcounts handbook here.

This question is similar to 2010 #3 Mathcouts Chapter Team problem. 

Solution I: Using the given slope, you can find the equation for AB ,
which is: y = -2 x + b, plug in (4, 7) and you have 7 = -2 x 4 + b so b = 15
The y intercept for AB is (0, 15)
Now you have four points, (0, 0), (0, 15), (4, 7) and (15, 0).
Using shoestring method to find area of any polygon, you have the area of polygon
AEDO = 82.5 square units.

Solution II:

After finding the y intercept for AB.

You break the area into three parts.

The side of the shaded region is 4 x 7 = 28.

The area of triangle ACE = 4 * (15-7) / 2 = 16

The area of triangle
EBD = (external height)7 * (15-4) / 2 = 38.5

So the total area is 82.5 square units.




Applicable problems: Answer key below.

#1:  The slope of AB is -3 and it intercepts with CD at point E, which is (3, 3). If the x intercepts of CD is (10,0), what is the area of AEDO?

#2: The slope of AB is 2 and it intercepts with CD at point E, which is (-2, 2 ). If the x intercepts of CD is (-7,0), what is the area of AEDO?

#3:  The slope of AB is -3 and it intercepts with CD at point E, which is ( 5, 8 ). If the x intercepts of CD is (14,0), what is the area of AEDO? -- Andrew's question.

#4: The slope of AB is \(\dfrac {-3} {2}\)and it intercepts with CD at point E, which is ( 2, 7 ). If the x intercepts of CD is (16,0), what is the area of AEDO? --Daniel's question.

#5: The slope of AB is -4/5 and it intercepts with CD at point E, which is (5, 2 ). If the x intercepts of CD is (12,0), what is the area of AEDO?












Answer key: 
#1: 33
#2: 13
#3: 113.5
#4: 66
#5: 27

2 comments:

  1. For question 4 the answer is satisfied only when the slope is the reciprocal of that given in the original question.

    ReplyDelete
  2. Thanks for the correction. I have changed the slope.
    Take care and cheers, Mrs. Lin

    ReplyDelete