
Example  Problem 1:
A right triangle has a perimeter of 24 cm and a hypotenuse of 10 cm. Find the sides x and y, x > y, that make the right angle of the triangle.Solution to Problem 1:

We start by drawing a triangle with the given information

The perimeter of the triangle is 24, hence
x + y + 10 = 24

It is a right triangle, use Pythagoras theorem to obtain.
x^{2} + y^{2} = 10^{2}

Solve the equation x + y + 10 = 24 for y.
y = 14  x

Substitute y in the equation x^{2} + y^{2} = 10^{2} by the expression obtained above.
x^{2} + (14  x)^{2} = 10^{2}

Expand the square, group like terms and write the above equation with the right side equal to zero.
2x^{2} 28x + 96 = 0

Multiply all terms in the above equation by 1/2.
x^{2} 14x + 48 = 0

Find the discriminant of the above quadratic equation.
Discriminant Δ = b^{2}  4*a*c = 196  192 = 4

Use the quadratic formulas to solve the quadratic equation; two solutions
x1 = [ b + √Δ ] / (2 a) = [ 14 + 2 ] / 2 = 8
x2 = [ b  √Δ ] / (2 a) = [ 14  2 ] / 2 = 6

use the equation y = 14  x to find the corresponding value of y.
y1 = 14  8 = 6
y2 = 14  6 = 8

Taking into account the condition x > y, the sides that make the right angle of the triangle are: x = 8 cm and y = 6 cm.
Check answer:
Hypotenuse h = √ (x^{2} + y^{2})
= √ (8^{2} cm^{2} + 6^{2} cm^{2})
= √(64 cm^{2} + 36 cm^{2})
= 10 cm, it agrees with the given value.
Perimeter = y + x + hypotenuse
= 8 cm + 6 cm + 10 cm
= 24 cm, it agrees with the given value.
Matched Problem 1:
A rectangle has a perimeter of 60 m and an area of 200 m^{2}. Find the length x and width y, x > y, of the rectangle.
Detailed Solution.
Example  Problem 2:
The sum of the squares of two consecutive real numbers is 61. Find the numbers.Solution to Problem 2:

Let x and x+1 be the two consecutive numbers. The sum of the square of x and x + 1 is equal to 61.
x^{2} + (x + 1)^{2} = 61

Expand (x + 1)^{2}, group like terms and write the above equation with the right side equal to zero.
2x^{2} + 2x  60 = 0

Multiply all terms in the above equation by 1/2.
x^{2} + x  30 = 0

Find the discriminant of the above quadratic equation.
Discriminant Δ = b^{2}  4*a*c = 1 + 120 = 121

Use the quadratic formulas to solve the quadratic equation; two solutions
x1 = [  b + √Δ ] / 2*a = [ 1 + 11 ] / 2 = 5
x2 = [  b  √Δ ] / 2*a = [ 1  11 ] / 2 = 6

First solution to the problem
first number: x1 = 5
second number: x1 + 1 = 6

Second solution to the problem
first number: x2 = 6
second number: x2 + 1 = 5
Check answer:
first solution sum of squares: 5^{2} + 6^{2}
= 25 + 36 = 61
second solution sum of squares: (6)^{2} + (5)^{2}
= 36 + 25 = 61
The two solutions to the problem agree with the given information in the problem.
Matched Problem 2:
The sum of the squares of two consecutive even real numbers is 52. Find the numbers.
Detailed Solution.
More References and linksSolve Equations, Systems of Equations and Inequalities.
