Chapter 3.4, Problem 14PSC

To determine

To find:The sum of given two-line segments

Answer to Problem 14PSC

The sum of given two-line segments is 48 units

Explanation of Solution

Given information Two circles with centers O and P

Perimeter of triangle AOP = 80 units

The sum of OC and DP is 16 units

Line segment CD is 2 units longer than line segment OC.

Formula used Perimeter of a triangle is sum of its sides

Calculation$\begin{array}{l}\text{OA = OC + OD}\mathrm{.......}\left(a\right)\\ \text{AP = CD + DP}\mathrm{..........}\left(b\right)\end{array}$

On adding (a) and (b), we get,

  $\begin{array}{l}\text{OA = OC + OD}\mathrm{.......}\left(\text{a}\right)\\ \text{AP = CD + DP}\mathrm{..........}\left(\text{b}\right)\\ \text{OA + AP}\\ \text{= }\left(\text{OC + DP}\right)+2\text{CD}\\ =16+2\text{CD}\mathrm{......}\text{given: OC + DP =16}\mathrm{.....}\left(c\right)\\ =\text{P }-\text{ OP}\mathrm{......}\text{given P (perimeter) = OA + AP + OP}\\ \text{=80 }-\text{OP}\mathrm{........}\text{given: P = 80 units}\\ \therefore \text{16 + 2 CD = 80}-\text{16}-\text{CD}\\ \left(\because \text{from }\left(c\right),\text{ OA + AP = 16 + 2 CD and given P = 80 units}\right)\\ 3\text{CD= 80 }-\text{ 32 = 48}\\ \text{CD = }\frac{48}{3}=16\text{ units}\\ \text{Therefore, }\\ \text{OA + BP}\\ \text{= OA + AP}\\ \left(\text{since, AP and BP are radii of same circle}\right)\\ =16+(2\times 16)=16+32=48\text{ units}\end{array}$