Question

The sides of a right angled triangle are \[\left( {x - 1} \right)\] cm, \[3x\] cm, and \[\left( {3x + 1} \right)\] cm.
(a) Find the value of \[x\].
(b) Find the lengths of the sides of the triangle.
(c) Find the area of the triangle.

Answer 1

Hint:
Here, we need to find the value of \[x\]. We know that the longest side of a right-angled triangle is hypotenuse. We will identify which of the three given sides is the longest. Then, we will apply Pythagoras’s theorem and solve the equation to get the value of \[x\]. Using the value of \[x\], we will find the lengths of the sides of the triangle. Finally, using the length of the base and the height of the triangle, we will find the area of the triangle using the formula for area of a triangle.
Formula Used: The square of the sum of two numbers is given by the algebraic identity \[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\].
The square of the difference of two numbers is given by the algebraic identity \[{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab\].
The area of a triangle is given by the formula \[{\text{Area}} = \dfrac{1}{2} \times {\text{Base}} \times {\text{Height}}\].

Complete step by step solution:
(a)
We know that the hypotenuse of a right angled triangle is the longest side of a triangle.
Now, the sides of the triangle are given as \[\left( {x - 1} \right)\] cm, \[3x\] cm, and \[\left( {3x + 1} \right)\] cm.
If \[x\] is less than or equal to 1, then the length \[\left( {x - 1} \right)\] cm of the side becomes less than or equal to 0 cm.
Thus, \[x\] cannot be less than or equal to 1.
Therefore, we get that \[x\] is a positive number greater than 1.
Next, we can observe that if \[x\] is a positive number, then
\[x - 1 < 3x < 3x + 1\]
Therefore, the longest side is the side of length \[\left( {3x + 1} \right)\] cm.
This is the length of the hypotenuse of the right angled triangle.
Now, we will use Pythagoras’s theorem in the right angled triangle.
Pythagoras's theorem states that the square of the hypotenuse of a right angled triangle is equal to the sum of the squares of the other two sides.
Applying the Pythagoras’s theorem in the given triangle with hypotenuse of length \[\left( {3x + 1} \right)\] cm, we get
\[{\left( {3x + 1} \right)^2} = {\left( {x - 1} \right)^2} + {\left( {3x} \right)^2}\]
The square of the sum of two numbers is given by the algebraic identity \[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\].
The square of the difference of two numbers is given by the algebraic identity \[{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab\].
Using the algebraic identities to simplify the equation, we get
\[ \Rightarrow {\left( {3x} \right)^2} + {1^2} + 2\left( {3x} \right)\left( 1 \right) = {x^2} + {1^2} - 2\left( x \right)\left( 1 \right) + {\left( {3x} \right)^2}\]
Multiplying the terms in the equation, we get
\[ \Rightarrow 9{x^2} + 1 + 6x = {x^2} + 1 - 2x + 9{x^2}\]
Rewriting the equation, we get
\[ \Rightarrow {x^2} + 1 - 2x + 9{x^2} - 9{x^2} - 1 - 6x = 0\]
Adding and subtracting the like terms in the equation, we get
\[ \Rightarrow {x^2} - 8x = 0\]
Factoring the expression, we get
\[ \Rightarrow x\left( {x - 8} \right) = 0\]
Thus, we get either \[x = 0\] or \[x - 8 = 0 \Rightarrow x = 8\].
We know that \[x\] is a positive number greater than 1.
Thus, we get
\[x = 8\]
Therefore, the value of \[x\] is 8.
(b)
We will substitute the value of \[x\] on the sides of the triangle to get their lengths.
Thus, we get the lengths of the sides of the triangle as
\[x - 1 = 8 - 1 = 7{\text{ cm}}\]
\[3x = 3 \times 8 = 24{\text{ cm}}\]
\[3x + 1 = 3 \times 8 + 1 = 24 + 1 = 25{\text{ cm}}\]
Therefore, the lengths of the sides of the triangle are 7 cm, 24 cm and 25 cm respectively.
(c)
We know that the area of a triangle is given by the formula \[{\text{Area}} = \dfrac{1}{2} \times {\text{Base}} \times {\text{Height}}\].
Substituting the length of the base as 7 cm and the length of the height as 24 cm, we get
\[ \Rightarrow {\text{Area}} = \dfrac{1}{2} \times 7 \times 24\]
Multiplying the terms of the expression, we get

\[ \therefore\] The area of the triangle is \[84{\text{ c}}{{\text{m}}^2}\].

Note:
We have used Pythagoras’s theorem to find the sides. Pythagoras's theorem states that the square of the hypotenuse of a right angled triangle is equal to the sum of the squares of the other two sides. We need to remember that \[x = 0\] is not an acceptable value of \[x\] and thus we took \[x = 8\]. If \[x = 0\], then the sides of the triangle would be of length \[ - 1\] cm, 0 cm and \[ - 1\] cm. This is not possible.