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In a square root spiral, a student adds a perpendicular side of 1 unit at the end of the side \(\sqrt{5}\) and calls the new hypotenuse \(\sqrt{10}\). What is the correct correction?वर्गमूल सर्पिल में \(\sqrt{5}\) वाली भुजा के सिरे पर 1 इकाई की लंबवत भुजा जोड़ने के बाद एक विद्यार्थी नई कर्ण को \(\sqrt{10}\) कहता है। उसका सही सुधार क्या है?
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: \(\sqrt{6}\)\(\sqrt{6}\)
The new hypotenuse is \(\sqrt{6}\), since \(h^2=(\sqrt{5})^2+1^2=5+1=6\). It is not \(\sqrt{10}\), as the lengths are not added directly. Exam tip: apply Pythagoras’ theorem at every new step.नई कर्ण \(\sqrt{6}\) होगी, क्योंकि समकोण त्रिभुज में \(h^2=(\sqrt{5})^2+1^2=5+1=6\)। \(\sqrt{10}\) लंबाइयों को सीधे जोड़ने से नहीं मिलता। परीक्षा में पायथागोरस प्रमेय लगाएँ।
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