Solving a Right-Angled Triangle Question Using Algebra
The question in the image involves a right-angled triangle, and it gives you two expressions for the lengths of the sides: \( x - 1 \) and \( 4x \), and for the hypotenuse, the expression is \( 5x - 9 \).
According to the Pythagorean theorem, in a right-angled triangle, the sum of the squares of the two shorter sides is equal to the square of the longest side (the hypotenuse). The equation is often expressed as \( a^2 + b^2 = c^2 \), where \( a \) and \( b \) are the lengths of the two shorter sides and \( c \) is the length of the hypotenuse.
Using the expressions provided for the lengths, we can create the following equation:
\[
(x - 1)^2 + (4x)^2 = (5x - 9)^2
\]
Now we'll expand and simplify this equation to solve for \( x \):
\[
x^2 - 2x + 1 + 16x^2 = 25x^2 - 90x + 81
\]
Combine like terms:
\[
17x^2 - 2x + 1 = 25x^2 - 90x + 81
\]
Bringing all terms to one side gives us:
\[
0 = 25x^2 - 17x^2 - 90x + 2x + 81 - 1
\]
Simplify and combine like terms:
\[
0 = 8x^2 - 88x + 80
\]
To make it easier to solve, divide the entire equation by 8:
\[
x^2 - 11x + 10 = 0
\]
Now we can factor this quadratic equation:
\[
(x - 10)(x - 1) = 0
\]
Setting each factor equal to zero gives us the possible values of \( x \):
\[
x - 10 = 0 \quad \Rightarrow \quad x = 10
\]
\[
x - 1 = 0 \quad \Rightarrow \quad x = 1
\]
Since \( x = 1 \) would result in one of the sides being zero length (since \( x - 1 = 0 \)), which is not possible for the side of a triangle, the only valid solution for \( x \) is:
\[
x = 10
\]
