Rectangle in Quadrant

There are several parameters we need to take into account. Let B equals base and H equals height. B*H = 48, B and H are greater than 0 and less than 10. Draw a line from O to Q , and call it OQ. Since RP is the same as OQ, therefore RP is the radius and equal to 10. So basically, the main equation is (10 - B) + 10 + (10 -H) = Ans. Substitute B = 48/H into the main equation, Then plug in values within the parameters until you get the right answer.

Sorry to confirm that rectangle mentioned in question is OPQR and not APQR. Solution:- For 48 area of rectangle ,length will be 8 cm and width 6 Cm Thus ,AP will be 2 Cm,Diagonal PR will be 10(Pythogorus theorem) and Rb will be 10-6 +4 Therefore, AP +PR + RB =16 ANS

K.K.GARG,India

The area of the rectangle is 48, so I assume that it's a six and eight rectangle by its height and base respectively. Then, the diagonal is 10 by using phytagoras triplets. Then, I summed up the rest of the sides, they are 2 and 4. The total three are 16.