Find the area of the shaded region between triangle ABC and triangle GHI, if the corresponding sides of the three triangles are parallel and 1 unit apart. Triangle DEF lies midway between the other two triangles. The lengths of the three sides of triangle DEF are 5, 6, and 7 units. (See Figure 1)
Solution. Students will initially try to find the area of triangles ABC and GHI, and then take the difference. Rather than pursue this method, let’s look at the problem from another point of view. Consider the figure as cut into three trapezoids (AGIC, AGHB, and BCIH) as shown in Figure 2
In each case, the altitude of the trapezoid is 2. The medians are 5, 6, and 7, respectively. Thus, we apply the formula for the area of a trapezoid, A =median X height:
Trapezoid AGIC = 5 . 2 = 10.
Trapezoid AGHB = 7 . 2 = 14.
Trapezoid BCIH = 6 . 2 = 12.
The total area of the shaded region is 10 + 14 + 12 = 36 square units.
What is the greatest value of the expression
ab + be + cd + ad
if a, b, c, and d have values 1, 2, 3, and 4, but not necessarily in that order?
We could list all possibilities for a, b, c, and d, and then for ab + bc + cd + ad, but this appears to be somewhat cumbersome. Let’s consider this problem from another point of view. Can we possibly factor the expression?
ab + bc + cd + ad = b(a + c) + d(c + a)
= b(a + c) + d(a + c)
Now consider all possibilities for the two factors:
(1 + 2)(3 + 4) = 3.7 = 21
(1 + 3) (2 + 4) = 4 . 6 = 24
(1 + 4)(2 + 3) = 5 . 5 = 25.
The largest sum is 25.
In Figure 2.1, ABCD is a square, and P and Q are the midpoints of the sides. What is the ratio of the area of triangle DPQ to the area of the square?
Solution. The usual solution to this problem begins by considering a square with side x, finding the areas of each of the three component right triangles whose sides are either x or , and then adding them and subtracting their sum from the whole square to get the area of triangle DPQ. We then express this area compared to the area of the entire square as a ratio.
We can make our task a bit easier, however, by employing one of our problem-solving strategies, namely, solve a simpler analogous problem. We assign a convenient length to the side of the original square, say 10, and proceed in a manner similar to that already described:
Area of square ABCD = (10)(10) = 100.
Area of triangle APD = (1/2) (AP)( AD) = (1/2)(5)(10) = 25.
Area of triangle PBQ = (1/2)(PB)(BQ) = (1/2)(5)(5) = 25/2.
Area of triangle QCD = (1/2)(CD)(CQ) = (1/2)(10)(5) = 25.
The sum of the areas of the three right triangles is 62 1/2. By subtracting this area sum from the area of the square, we obtain the area of triangle DPQ = 37.
Thus our required ratio is 37 ½ : 100 = 3/8.
Let’s examine this problem from another point of view. Select E and F as midpoints of CD and AD, respectively (Figure 4)
Now, because rectangle APED is one half the area of the original square, and diagonal DP divides it in half, the area of triangle APD is half the area of the rectangle APED and thus one quarter the area of the original square. Similarly, with F the midpoint of AD, the area of rectangle DCQF is one half the area ofthe original square, and triangle DQC is also one quarter the area of the original square. Furthermore, since PE and QF meet at G, the center of the original square, the area of square PBQG is one quarter that of the original square, and triangle PBQ (which is one half the area of square PBQG) is one eighth the area of the original square. We now have the following information:
- The area of triangle APD = 1/4 the area of square ABCD.
- The area of triangle QCD = 1/4 the area of square ABCD.
- The area of triangle PBQ = 1/8 the area of square ABCD.
- The sum of the area of the three triangles is 1/4 + 1/4 + 1/8 = 5/8.
Hence, the area of triangle DPQ is 3/8 the area of the original square.
Solve for x:
(x – 3)3 + (x – 7)3 = (2x – 10)3.
The traditional approach to solving this problem is to cube all three of the parenthetical expressions and solve the resulting equation as follows:
x 3 – 9x 2 + 27x – 27 + X 3 – 21x 2 + 147x – 343 = 8x 3 -120x 2 + 600x -1000
2x 3 – 30x 2 + 174x – 370 = 8x 3 – 120x 2 + 600x – 1000
0 = 6x 3 – 90x 2 + 426x – 630
0 = x 3 – 15x 2 + 71 x – 105
0 = (x – 7)( x – 5)( x – 3)
x = 3
x = 5
x = 7.
Of course, this solution requires a great deal of algebraic manipulation, and could easily lead to an error, making the expression unfactorable. Instead, let’s solve this problem by adopting a different point of view.
Notice that (x – 3) + (x – 7) = (2 x – 10). Suppose we let
a = x-3
b = x – 7.
a + b = 2x – 10.
Substituting in the original,
a 3 + b 3 = (a + b) 3
a 3 + b 3 = a 3 + 3a 2 b + 3ab 2 + b 3
0 = 3a 2 b + 3ab 2 = 3ab(a + b)
0 = (ab) (a + b).
a = 0,
b = 0,
If a = 0, then x – 3 = 0, x = 3.
If b = 0, then x – 7 = 0, x = 7.
If a + b = 0, then 2 x – 10 = 0 and x = 5.
How many different candy items can a vending machine offer if each item costs a different amount of money and the machine accepts, at most, one coin for each of the slots from among a penny, nickel, dime, or quarter?
The immediate approach to solving this problem would be to make a list of all possible combinations of the four coins in an organized manner. However, this could prove difficult (and confusing) for students if they are not systematic in the way they create this list. It is also quite easy to miss one or more possible coin combinations.
However, let’s look at this problem from another point of view. We can solve it by considering each of the coin slots separately. That is, for each item, a decision about inserting or not inserting a coin into each slot has to be made. Thus, one either puts a penny into the penny slot or not-two possibilities. One either puts a nickel into the nickel slot or not (again, two possibilities). One either puts a dime into the dime slot or not. Finally, one either puts a quarter into the quarter slot or not. Then, using the fundamental counting principle, there are 2 X 2 X 2 X 2 = 16 ways of inserting or not inserting coins into the machine with, at most, one coin per slot. However, one of these 16 options involves not putting any coins into the slots, and, because there are no free candy items in the machine, there are 16 – 1 or 15 ways.
Find the value of (x + y) if
123x + 321y = 345
321x + 123y = 543.
When students are confronted with two equations that contain two variables, they automatically revert to the process that has been taught as the method of solution; that is, to solve them simultaneously. Students begin by making the coefficients of either x or y the same:
(321)(123) x + (321)(321)y = (321)(345)
(123)(321) x + (123)(123)y = (543)(123)
39,483x + 103,041y = 110,745
39,483x + 15,129y = 66,789
87,912y = 43,956
Y = 0.5
(123) x + (321)(0.5) = 345
123x + 160.5 = 345
123x = 184.5
x = 1.5
( x + y) = 1.5 + 0.5 = 2.
Notice that solving this pair of equations simultaneously leads to a great deal of complicated arithmetic computation. Let’s examine this problem from another point of view. We are not really interested in the specific values for x and y; rather, we are interested in their sum. If we add the two equations, we obtain
444x + 444y = 888
x + y = 2.
The problem is quite easily solved. Students should not think that this is merely a “trick” problem, although it would appear so on the surface. This problem exemplifies the important skill of looking at what is to be found in a given problem, rather than beginning in the traditional way, given familiar stimuli. Students should begin to realize that there are problems (in everyday life, as well) the solutions to which can be much simpler with a bit of “foresight.”
*Problem Solving Class