Please solve the NCERT question No.4 Exercise-13.5 (p-258)

--------------------- +P -------------------- | /|\ | | / | \ | | / | \ S-s |H-h | / | \ | | / | \ | | /*****|**r**\ | | ** C+-----*+D-------------- H| S / *********** \ | | / | \ | | / | \ | | / | \ s | | / | \ |h | / | \ | | / | \ | | / **********|********** \ | | /***** | R *****\ | -----* A+----------------+B---- ****** ****** *********************We know R, r, and h, but not H, the total height of the cone from which the frustum was cut. If we can find it, then the volume of the frustum will be the volume of the whole cone, pi R^2 H/3, minus the volume of the cone we cut off the top, pi r^2 (H-h)/3.The triangles PAB and PCD are similar, so we can write the equation AB CD R r -- = -- or - = --- PA PC H H-hCross-multiplying [that is, multiplying both sides by H(H-h)], we get R(H-h) = rHWe can distribute the left side and collect H terms, then divide: RH - Rh = rH RH - rH = Rh (R-r)H = Rh Rh H = --- R-rNow let's write the volume formula and substitute this formula for H: pi pi V = -- R^2 H - -- r^2 (H-h) 3 3 pi = -- (R^2 H - r^2 H + r^2 h) 3 pi = -- [(R^2 - r^2) H + r^2 h] 3 pi Rh = -- [(R^2 - r^2) --- + r^2 h] 3 R-r pi R = -- [(R^2 - r^2) --- + r^2] h 3 R-rWe can write R^2 - r^2 as (R - r)(R + r) and cancel: pi = --- [(R + r) R + r^2] h 3 pi = --- [R^2 + Rr + r^2] h 3That's the formula.