Login/Sign Up
Home » The hypotenuse of a right-angled triangle is 1 meter less than twice the shortest side. If the third side 1 meter more than the shortest side, find the side, find the sides of the triangle.
The hypotenuse of a right-angled triangle is 1 meter less than twice the shortest side. If the third side 1 meter more than the shortest side, find the side, find the sides of the triangle.
CBSE | Class 10 | Exercise 10e | Maths | Quadratic Equations | RS Aggarwal
The hypotenuse of a right-angled triangle is 1 meter less than twice the shortest side. If the third side 1 meter more than the shortest side, find the side, find the sides of the triangle.

Let the shortest side be x \mathrm{~m}.

Therefore, according to the question:

Hypotenuse =(2 x-1) m

Third side =(x+1) m

On applying Pythagoras theorem, we get:

\begin{array}{l} (2 x-1)^{2}=(x+1)^{2}+x^{2} \\ \Rightarrow 4 x^{2}-4 x+1=x^{2}+2 x+1+x^{2} \\ \Rightarrow 2 x^{2}-6 x=0 \\ \Rightarrow 2 x(x-3)=0 \\ \Rightarrow x=0 \text { or } x=3 \end{array}

The length of the side cannot be 0 ; therefore, the shortest side is 3 \mathrm{~m}.

Therefore,

Hypotenuse =(2 \times 3-1)=5 m

Third side =(3+1)=4 m

i

More Material