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The lengths of the sides of a right-angled triangle are given by the expressions $x - 1$, $4x$, and $5x - 9$, as shown in the diagram - Leaving Cert Mathematics - Question 5 - 2016

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The lengths of the sides of a right-angled triangle are given by the expressions $x - 1$, $4x$, and $5x - 9$, as shown in the diagram. Find the value of $x$. Verify... show full transcript

Worked Solution & Example Answer:The lengths of the sides of a right-angled triangle are given by the expressions $x - 1$, $4x$, and $5x - 9$, as shown in the diagram - Leaving Cert Mathematics - Question 5 - 2016

Step 1

Find the value of $x$

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Answer

To find the value of xx, we will set up the equation based on the Pythagorean theorem, which states that for a right triangle with sides aa, bb, and hypotenuse cc:

a2+b2=c2a^2 + b^2 = c^2

We assign the sides as follows:

  • a=x1a = x - 1
  • b=4xb = 4x
  • c=5x9c = 5x - 9

Substituting into the equation:

(x1)2+(4x)2=(5x9)2(x - 1)^2 + (4x)^2 = (5x - 9)^2

Expanding each term gives:

(x22x+1)+(16x2)=(25x290x+81)(x^2 - 2x + 1) + (16x^2) = (25x^2 - 90x + 81)

Combining like terms results in:

17x22x+1=25x290x+8117x^2 - 2x + 1 = 25x^2 - 90x + 81

Rearranging the equation yields:

0=8x288x+800 = 8x^2 - 88x + 80

Dividing through by 8 simplifies this to:

0=x211x+100 = x^2 - 11x + 10

Factoring gives:

(x1)(x10)=0(x - 1)(x - 10) = 0

Therefore, the solutions are: x=1 or x=10x = 1 \text{ or } x = 10

Step 2

Verify, with this value of $x$, that the lengths of the triangle above form a pythagorean triple.

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Answer

For x=1x = 1:

  • Side lengths are:
    • 11=01 - 1 = 0 (not valid)

For x=10x = 10:

  • Side lengths are:
    • 101=910 - 1 = 9
    • 4(10)=404(10) = 40
    • 5(10)9=415(10) - 9 = 41

Now we check:

92+402=4129^2 + 40^2 = 41^2

Calculating each term:

  • 92=819^2 = 81
  • 402=160040^2 = 1600
  • 412=168141^2 = 1681

Thus, 81+1600=168181 + 1600 = 1681

This confirms that the sides indeed form a Pythagorean triple.

Step 3

Show that $f(x) = 3x - 2$, where $x \, \text{is} \, \text{real}$, is an injective function.

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Answer

To show that f(x)f(x) is injective, we need to demonstrate that: f(a)=f(b)a=bf(a) = f(b) \Rightarrow a = b

Let: f(a)=3a2 and f(b)=3b2f(a) = 3a - 2 \text{ and } f(b) = 3b - 2

Equating gives: 3a2=3b23a - 2 = 3b - 2

Solving this: 3a=3ba=b3a = 3b \Rightarrow a = b

Thus, f(x)f(x) is injective.

Step 4

Given that $f(x) = 3x - 2$, find a formula for $f^{-1}$, the inverse function of $f$. Show your work.

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Answer

To find the inverse function f1(x)f^{-1}(x):

  1. Start with y=f(x)=3x2y = f(x) = 3x - 2.
  2. Swap xx and yy: x=3y2x = 3y - 2
  3. Solve for yy:
    • Add 2 to both sides: x+2=3yx + 2 = 3y
    • Divide by 3: y=x+23y = \frac{x + 2}{3}

Thus, the inverse function is: f1(x)=x+23f^{-1}(x) = \frac{x + 2}{3}

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