The hypotenuse of a right triangle is 25 cm. The difference between the lengths of the other two sides of the triangle is 5 cm. Find the perimeter of the triangle.
ANSWER
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Hint: In this question, let's assume the other two sides as some variables then by using the pythagoras theorem we can get a relation between the two sides. Now, by solving the two equations we can get the sides. Then, as we know all the sides we can get the perimeter.
Complete step-by-step answer: Let us assume the other two sides of the right triangle as a and b
By the pythagoras theorem, in a right triangle it follows that the hypotenuse of this triangle has the length of the square root of the sum of the squares of the other two sides.
\[c=\sqrt{{{a}^{2}}+{{b}^{2}}}\]
Where, a and b are the sides of the right triangle and c is the hypotenuse
Now, as given in the question that hypotenuse is 25 cm we get,
Now, we neglect the - case here as the side of a triangle cannot be negative.
\[\begin{align}
& \Rightarrow b=\dfrac{35-5}{2} \\
& \therefore b=15cm \\
\end{align}\]
Now, by substituting this value of b in the relation between a and b we get,
\[\begin{align}
& \Rightarrow a=\dfrac{300}{b} \\
& \Rightarrow a=\dfrac{300}{15} \\
& \therefore a=20cm \\
\end{align}\]
Now, let us assume the perimeter of this triangle as P.
\[\Rightarrow P=a+b+25\]
Now, on substituting the respective values of a and b we get,
\[\begin{align}
& \Rightarrow P=20+15+25 \\
& \therefore P=60cm \\
\end{align}\]
Note: Instead of writing a in terms of b we can also write b in terms of a and then substitute this value of b in any of the two equations and then solve for the value of a. Then by substituting this value of a in the relation between a and b we get the value of b. Both the methods give the same value of a and b. While getting the value of b from the quadratic equation formed instead of using the direct formula method we can also use factorisation method. But this may take a little more time as we need to check for the possible values. Here, from the direct formula we consider only the case which gives the positive value of b but there is possibility of another value of b as it is a quadratic equation. We can neglect that other value here as the side of a triangle cannot be negative.