Try to extend this method for constructing line segments of lengths and using a ruler and a compass. Generalise this method to construct a line segment of any length of the form , where is a positive integer.

Question

KnowledgeBoat · Accepted Answer

Construction of : Step 1 : Construct OB = (as shown previously). Step 2 : At B, draw a perpendicular BD of length 1 unit. Step 3 : Join O to D. By the Pythagoras theorem : ⇒ OD2 = OB2 + BD2 ⇒ OD2 = ()2 + 12 ⇒ OD2 = 2 + 1 = 3 ⇒ OD = . Step 4 : With O as the centre and OD as the radius, draw an arc cutting the number line at point Q. Then OQ = . Construction of : Step 1: On the number line, measure a horizontal segment OA = 2 units. Step 2: At point A, draw a perpendicular line segment AC of length 1 unit. Step 3: Join O to C. The length OC is: ⇒ OC2 = OA2 + AC2 ⇒ OC2 = (2)2 + 12 ⇒ OC2 = 4 + 1 = 5 ⇒ OC = . Step 4 : With O as the centre and OC as the radius, draw an arc cutting the number line at point R. Then OR = . Generalisation to construct : Suppose we have a line segment of length . Then at one end of this segment, draw a perpendicular of length 1 unit. The hypotenuse of the resulting right triangle will have length : This method, when extended repeatedly starting from a unit segment, gives the famous "square root spiral". Hence, by repeatedly applying the Pythagoras theorem with a unit perpendicular to a known segment of length , we can construct a line segment of any length of the form for any positive integer n.

Mathematics

Try to extend this method for constructing line segments of lengths and using a ruler and a compass. Generalise this method to construct a line segment of any length of the form , where is a positive integer.

Whole Numbers

Related Questions

We have seen how to obtain a line whose length is a rational number. How do we obtain lines whose lengths are irrational?