The volume of a cone is V = 1/3*pi*r^2*h. Make r the subject of the formula.
Answering the question by acknowledging the student understands what the question is actually asking. The subject of the formula sits by itself on one side of the equation and doesn't appear at all on the other. Initially, V is the subject of the formula, the purpose of the question is to change to subject to r by rearranging. Recognising each term in the equation is multiplied together therefore division is necessary to get r 'by itself.' For a more detailed understanding, asking whether the student knows how r and h are related to get V ? (taught by drawing a cone and labelling r and h). Approaching the question by tackling fractions first. Writing out each step along the way, until it is easier for them to do it in fewer steps (but also acknowledging that a certain amount of steps are needed to be shown for method marks). Remarking to the student that what must be done to one side of the equation needs to be done to the other. Answer: 3V = pir2 h3V/pih = r2r = sqrt (3V/pih) This example can be used to demonstrate to students how rearranging the equation can be used for real world problems by substituting in numerical values to find the radius needed for a specific volume and height eg. what is the value of r when V = 47cm3 and h = 6cm
