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3 years ago

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In a right triangle, the hypotenuse is three inches more than twice the length of one leg. If the other leg is 12 inches, what i

s the lengths of the other leg and hypotenuse?

1 answer:

andrew-mc [135]3 years ago

6 0

Answer:

Length of hypothenuse = 13 inches

Length of other leg = 5 inches

Step-by-step explanation:

A triangle has three sides and three angles. A right angle triangle has 90 degrees as one of the angles. The remaining two angles are complimentary. This means that the sum if the other two angles is 90 degrees.

The longest side is the hypotenuse. The shorter sides are the opposite and adjacent.

In this triangle, the hypotenuse is three inches more than twice the length of one leg. Let the length of one leg be x inches.

Hypothenuse = 2x + 3

The other leg is 12 inches.

The diagram is illustrated in the attached photo.

Applying Pythagoras theorem,

Hypothenuse ^2 = adjacent^2 + opposite ^2

(2x+3)^2 = 12^2 + x^2

(2x+3)(2x+3)= 144 + x^2

4x^2 + 6x + 6x + 9 = 144 + x^2

4x^2 - x^2 + 6x + 6x + 9 - 144 = 0

3x^2 +12x - 135 = 0

3x^2 +27x - 15x - 135 = 0

3x(x+9)-15(x+9)=0

(3x -15)(x+9) = 0

3x -15 = 0 or x+9= 0

3x = 15 or x = -9

x = 15/3 = 5

x = 5inches because it cannot be negative.

Length of hypothenuse = 2x + 3

= 2×5 + 3 = 10 + 3 = 13 inches

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Determine all prime numbers a, b and c for which the expression a ^ 2 + b ^ 2 + c ^ 2 - 1 is a perfect square .

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Answer:

The family of all prime numbers such that $a^{2} + b^{2} + c^{2} -1$ is a perfect square is represented by the following solution:

$a$ is an arbitrary prime number. (1)

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$c$ is another arbitrary prime number. (3)

Step-by-step explanation:

From Algebra we know that a second order polynomial is a perfect square if and only if $(x+y)^{2} = x^{2} + 2\cdot x\cdot y + y^{2}$ . From statement, we must fulfill the following identity:

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By Associative and Commutative properties, we can reorganize the expression as follows:

$a^{2} + (b^{2}-1) + c^{2} = x^{2} + 2\cdot x \cdot y + y^{2}$ (1)

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From Number Theory, we remember that a number is prime if and only if is divisible both by 1 and by itself. Then, $a, b, c > 1$ . If $a$ , $b$ and $c$ are prime numbers, then $2\cdot a\cdot c$ must be an even composite number, which means that $a$ and $c$ can be either both odd numbers or a even number and a odd number. In the family of prime numbers, the only even number is 2.

In addition, $b$ must be a natural number, which means that:

$1 + 2\cdot a\cdot c \ge 4$

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But the lowest possible product made by two prime numbers is $2^{2} = 4$ . Hence, $a\cdot c \ge 4$ .

The family of all prime numbers such that $a^{2} + b^{2} + c^{2} -1$ is a perfect square is represented by the following solution:

$a$ is an arbitrary prime number. (1)

$b = \sqrt{1 + 2\cdot a \cdot c}$ (2)

$c$ is another arbitrary prime number. (3)

Example: $a = 2$ , $c = 2$

$b = \sqrt{1 + 2\cdot (2)\cdot (2)}$

$b = 3$

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