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Aim: Perfecting students' skills for transferring knowledge of planimetry (metric dependencies in a right triangle, triangle's area and cosine rule) for solving systems of quadratic equations with tree or two unknowns (9th and 10th grade).

Example 1. If $tema3_EN_files\tema3_EN_MathML_0.jpg$ and $tema3_EN_files\tema3_EN_MathML_1.jpg$ are positive numbers and $tema3_EN_files\tema3_EN_MathML_2.jpg$ then without finding the values for $tema3_EN_files\tema3_EN_MathML_3.jpg$ and $tema3_EN_files\tema3_EN_MathML_4.jpg$ find the value of the expression $tema3_EN_files\tema3_EN_MathML_5.jpg$

Solution: If we have to solve this system it would be easy for anyone manage to find $tema3_EN_files\tema3_EN_MathML_6.jpg$ and $tema3_EN_files\tema3_EN_MathML_7.jpg$ but in this problem $tema3_EN_files\tema3_EN_MathML_8.jpg$ is to be found. The system can be written down in the form

$tema3_EN_files\tema3_EN_MathML_9.jpg$

Since $tema3_EN_files\tema3_EN_MathML_10.jpg$ and $tema3_EN_files\tema3_EN_MathML_11.jpg$ are positive, from the first equation $tema3_EN_files\tema3_EN_MathML_12.jpg$ and $tema3_EN_files\tema3_EN_MathML_13.jpg$ can be considered using a reverse Pythagorean theorem as the lengths respectively of the legs and the hypotenuse in a right triangle ABD $tema3_EN_files\tema3_EN_MathML_14.jpg$ (fig.1)

Figure 1.

If we consider the second equation of the system we can reach an analogical conclusion $tema3_EN_files\tema3_EN_MathML_15.jpg$ и $tema3_EN_files\tema3_EN_MathML_16.jpg$ are the respective lengths of the legs and hypotenuse in a triangle BCD $tema3_EN_files\tema3_EN_MathML_17.jpg$ (fig.1).

From the third equation we can make the conclusion that y is a number which is a geometrical average of $tema3_EN_files\tema3_EN_MathML_18.jpg$ and $tema3_EN_files\tema3_EN_MathML_19.jpg$ , and using the reverse theorem for proportional line segments in a right triangle the consequence is that $tema3_EN_files\tema3_EN_MathML_20.jpg$

Now we examine the expression $tema3_EN_files\tema3_EN_MathML_21.jpg$ , which can be represented as $tema3_EN_files\tema3_EN_MathML_22.jpg$ . Considering figure 1 $tema3_EN_files\tema3_EN_MathML_23.jpg$ but $tema3_EN_files\tema3_EN_MathML_24.jpg$

In this way we find out that $tema3_EN_files\tema3_EN_MathML_25.jpg$ i.e. $tema3_EN_files\tema3_EN_MathML_26.jpg$

For this problem the question could be to find $tema3_EN_files\tema3_EN_MathML_27.jpg$ .

Example 2. If $tema3_EN_files\tema3_EN_MathML_28.jpg$ and $tema3_EN_files\tema3_EN_MathML_29.jpg$ are positive numbers and

$tema3_EN_files\tema3_EN_MathML_30.jpg$

then without finding values for $tema3_EN_files\tema3_EN_MathML_31.jpg$ and $tema3_EN_files\tema3_EN_MathML_32.jpg$ , calculate the expression $tema3_EN_files\tema3_EN_MathML_33.jpg$ .

Solution: The given system we represent in the form

$tema3_EN_files\tema3_EN_MathML_34.jpg$

Since $tema3_EN_files\tema3_EN_MathML_35.jpg$ and $tema3_EN_files\tema3_EN_MathML_36.jpg$ , considering the second equation of the system and having in mind a reverse Pythagorean theorem the numbers $tema3_EN_files\tema3_EN_MathML_37.jpg$ and $tema3_EN_files\tema3_EN_MathML_38.jpg$ are the lengths respectively of the legs and hypotenuse in a right triangle AMC $tema3_EN_files\tema3_EN_MathML_39.jpg$ (fig.2).

If we consider the first equation the numbers $tema3_EN_files\tema3_EN_MathML_40.jpg$ , $tema3_EN_files\tema3_EN_MathML_41.jpg$ and $tema3_EN_files\tema3_EN_MathML_42.jpg$ are lengths of sides in a triangle AMB in which $tema3_EN_files\tema3_EN_MathML_43.jpg$ and using a reverse cosine rule. Analogically $tema3_EN_files\tema3_EN_MathML_44.jpg$ , $tema3_EN_files\tema3_EN_MathML_45.jpg$ and $tema3_EN_files\tema3_EN_MathML_46.jpg$ are lengths of sides in a triangle BMC where $tema3_EN_files\tema3_EN_MathML_47.jpg$ ( fig. 2).

Figure 2.

Since $tema3_EN_files\tema3_EN_MathML_48.jpg$ , then using a reverse Pythagorean theorem we can conclude that ABC is a right triangle and $tema3_EN_files\tema3_EN_MathML_49.jpg$ ACB = $tema3_EN_files\tema3_EN_MathML_50.jpg$

We will find the areas of the following three triangles: $tema3_EN_files\tema3_EN_MathML_51.jpg$ using which the area of $tema3_EN_files\tema3_EN_MathML_52.jpg$ ABC can be found.

$tema3_EN_files\tema3_EN_MathML_53.jpg$

$tema3_EN_files\tema3_EN_MathML_54.jpg$

$tema3_EN_files\tema3_EN_MathML_55.jpg$

$tema3_EN_files\tema3_EN_MathML_56.jpg$

This way we get $tema3_EN_files\tema3_EN_MathML_57.jpg$ $tema3_EN_files\tema3_EN_MathML_58.jpg$ , i.e. $tema3_EN_files\tema3_EN_MathML_59.jpg$ .

For individual work

Problem 3. If $tema3_EN_files\tema3_EN_MathML_72.jpg$ , $tema3_EN_files\tema3_EN_MathML_73.jpg$ , $tema3_EN_files\tema3_EN_MathML_74.jpg$ does the system have a solution:

$tema3_EN_files\tema3_EN_MathML_75.jpg$

Hint: If $tema3_EN_files\tema3_EN_MathML_76.jpg$ are solutions to the system, than the geometric interpretation of the system is presented in figure 3. Use the inequality between the sides of a triangle i.e. $tema3_EN_files\tema3_EN_MathML_77.jpg$

Figure 3.

Answer: No.

Problem 4. Calculate the value of the expression $tema3_EN_files\tema3_EN_MathML_78.jpg$ , if $tema3_EN_files\tema3_EN_MathML_79.jpg$ and

$tema3_EN_files\tema3_EN_MathML_80.jpg$

Hint: $tema3_EN_files\tema3_EN_MathML_81.jpg$ and $tema3_EN_files\tema3_EN_MathML_82.jpg$ , because if $tema3_EN_files\tema3_EN_MathML_83.jpg$ and $tema3_EN_files\tema3_EN_MathML_84.jpg$ , then $tema3_EN_files\tema3_EN_MathML_85.jpg$ . The given system we represent in the form

$tema3_EN_files\tema3_EN_MathML_86.jpg$

Look above at the solution of problem 1.

Problem 5. Solve the system $tema3_EN_files\tema3_EN_MathML_87.jpg$ .

Hint: It is not hard to prove that $tema3_EN_files\tema3_EN_MathML_88.jpg$ and $tema3_EN_files\tema3_EN_MathML_89.jpg$ are positive numbers and $tema3_EN_files\tema3_EN_MathML_90.jpg$ and $tema3_EN_files\tema3_EN_MathML_91.jpg$ are lengths of the legs and hypotenuse in a right triangle.

Answer: $tema3_EN_files\tema3_EN_MathML_92.jpg$ .