
The sides of a right-angled triangle containing the right angle are $5x$ cm and $3x - 1$ cm. If its area is $60c{m^2}$, find its perimeter.
Answer
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Hint:
Use the given area and length of sides of the right triangle to form an equation. Solve the equation to find the value of $x$. Then, length of sides can be calculated by substituting the values of $x$ and applying Pythagoras theorem. And add all the sides to determine the perimeter of the triangle.
Complete step by step solution:
We will use the given length of sides of the triangle, $5x$ cm and $3x - 1$ cm to draw a corresponding diagram.
We know that the area of a triangle is given as $\dfrac{1}{2} \times {\text{base}} \times {\text{height}}$
The area of the given triangle is $60c{m^2}$
Then, $60 = \dfrac{1}{2} \times \left( {5x} \right) \times \left( {3x - 1} \right)$
Solve the above equation to find the value of $x$
$
60 = \dfrac{1}{2} \times \left( {5x} \right) \times \left( {3x - 1} \right) \\
\Rightarrow 120 = 15{x^2} - 5x \\
\Rightarrow 15{x^2} - 5x - 120 = 0 \\
\Rightarrow 3{x^2} - x - 24 = 0 \\
$
Factorise the above equation and then equate each factor to 0,
$
\Rightarrow 3{x^2} - 9x + 8x - 24 = 0 \\
\Rightarrow 3x\left( {x - 3} \right) + 8\left( {x - 3} \right) = 0 \\
\Rightarrow \left( {3x + 8} \right)\left( {x - 3} \right) = 0 \\
\Rightarrow x = - \dfrac{8}{3},3 \\
$
But, the length of the side cannot be negative, therefore, $x = 3$.
Therefore, the two sides of the triangle are $5\left( 3 \right) = 15$cm and $3\left( 3 \right) - 1 = 8cm$
We can find the remaining side using the Pythagoras theorem.
Pythagoras theorem states that the square of Hypotenuse is equal to the sum of squares of other two sides.
Therefore,
$
\Rightarrow A{C^2} = {\left( {15} \right)^2} + {\left( 8 \right)^2} \\
\Rightarrow A{C^2} = 225 + 64 \\
\Rightarrow A{C^2} = 289 \\
\Rightarrow AC = 17 \\
$
The perimeter is the sum of all sides of a triangle.
$15 + 8 + 17 = 40cm$
Note:
Area is the space covered by a shape and perimeter is the length of the boundary of the shape. Also, in a right triangle, the longest side is the hypotenuse and is opposite the right angle.
Use the given area and length of sides of the right triangle to form an equation. Solve the equation to find the value of $x$. Then, length of sides can be calculated by substituting the values of $x$ and applying Pythagoras theorem. And add all the sides to determine the perimeter of the triangle.
Complete step by step solution:
We will use the given length of sides of the triangle, $5x$ cm and $3x - 1$ cm to draw a corresponding diagram.
We know that the area of a triangle is given as $\dfrac{1}{2} \times {\text{base}} \times {\text{height}}$
The area of the given triangle is $60c{m^2}$
Then, $60 = \dfrac{1}{2} \times \left( {5x} \right) \times \left( {3x - 1} \right)$
Solve the above equation to find the value of $x$
$
60 = \dfrac{1}{2} \times \left( {5x} \right) \times \left( {3x - 1} \right) \\
\Rightarrow 120 = 15{x^2} - 5x \\
\Rightarrow 15{x^2} - 5x - 120 = 0 \\
\Rightarrow 3{x^2} - x - 24 = 0 \\
$
Factorise the above equation and then equate each factor to 0,
$
\Rightarrow 3{x^2} - 9x + 8x - 24 = 0 \\
\Rightarrow 3x\left( {x - 3} \right) + 8\left( {x - 3} \right) = 0 \\
\Rightarrow \left( {3x + 8} \right)\left( {x - 3} \right) = 0 \\
\Rightarrow x = - \dfrac{8}{3},3 \\
$
But, the length of the side cannot be negative, therefore, $x = 3$.
Therefore, the two sides of the triangle are $5\left( 3 \right) = 15$cm and $3\left( 3 \right) - 1 = 8cm$
We can find the remaining side using the Pythagoras theorem.
Pythagoras theorem states that the square of Hypotenuse is equal to the sum of squares of other two sides.
Therefore,
$
\Rightarrow A{C^2} = {\left( {15} \right)^2} + {\left( 8 \right)^2} \\
\Rightarrow A{C^2} = 225 + 64 \\
\Rightarrow A{C^2} = 289 \\
\Rightarrow AC = 17 \\
$
The perimeter is the sum of all sides of a triangle.
$15 + 8 + 17 = 40cm$
Note:
Area is the space covered by a shape and perimeter is the length of the boundary of the shape. Also, in a right triangle, the longest side is the hypotenuse and is opposite the right angle.
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