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Home » The hypotenuse of a right-angled triangle is 20 meters. If the difference between the lengths of the ther sides be 4 meters, find the other sides.
The hypotenuse of a right-angled triangle is 20 meters. If the difference between the lengths of the ther sides be 4 meters, find the other sides.
CBSE | Class 10 | Exercise 10e | Maths | Quadratic Equations | RS Aggarwal
The hypotenuse of a right-angled triangle is 20 meters. If the difference between the lengths of the ther sides be 4 meters, find the other sides.

Let one side of the right-angled triangle be x \mathrm{~m} and the other side be (x+4) m.

On applying Pythagoras theorem, we have:

\begin{array}{l} 20^{2}=(x+4)^{2}+x^{2} \\ \Rightarrow 400=x^{2}+8 x+16+x^{2} \\ \Rightarrow 2 x^{2}+8 x-384=0 \\ \Rightarrow x^{2}+4 x-192=0 \\ \Rightarrow x^{2}+(16-12) x-192=0 \\ \Rightarrow x^{2}+16 x-12 x-192=0 \\ \Rightarrow x^{2}(x+16)-12(x+16)=0 \\ \Rightarrow(x+16)(x-12)=0 \\ \Rightarrow x=-16 \text { or } x=12 \end{array}

The value of x cannot be negative.

Therefore, the base is 12 \mathrm{~m} and the other side is \{(12+4)=16 \mathrm{~m} ).

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