Question

In the figure, AOB is a quarter circle of radius 10 and PQRO is a rectangle of perimeter 26. The perimeter of the shaded region is

A) $13 + 5\pi $
B) $17 + 5\pi $
C) $7 + 10\pi $
D) $7 + 5\pi $

Answer 1

Hint:
Firstly, find the lengths of line PA, arc AQB, line BR and line RP.
Then, to get the perimeter of the shaded region, add all the above values and choose the correct option.

Complete step by step solution:
It is given that AOB is a quarter circle and PQRO is a rectangle in the quarter circle

It is clear from the diagram that the radius of the quarter circle is the diagonal OQ of the rectangle OPQR. So, \[OQ = OB = OA = 10\] .
Also, the diagonals of a rectangle have the same length. So, \[OQ = PR = 10\] .
Now, as given in the question, the perimeter of the rectangle OPQR is 26.
Let, \[RO = QP = b\] and \[RQ = PO = l\] .
 $
  \therefore 26 = 2\left( {l + b} \right) \\
  \therefore \dfrac{{26}}{2} = l + b \\
  \therefore l + b = 13 \\
 $
Also, the radius of the quarter circle is given 10.
So, the perimeter of arc AQB .
Now, to find the perimeter of the shaded region, we add the perimeters of line PA, arc AQB, line BR and line RP.
$\therefore $ Perimeter of shaded region
Also, \[PA = OA-OP = r-l\] and \[BR = BO-RO = r-b\] .
Thus, perimeter of shaded region
    $
   = r - l + 5\pi + r - b + r \\
   = 3r - \left( {l + b} \right) + 5\pi \\
   = 3\left( {10} \right) - 13 + 5\pi \\
   = 30 - 13 + 5\pi \\
   = 17 + 5\pi \\
 $

So, option (B) is correct.

Note:
Here, students may go wrong by subtracting the perimeter of triangle OPR from the perimeter of the arc AOB which will be the wrong method to solve the given question.
Instead, we have to find the lengths of the line PA, arc AQB, line BR and line RP individually and add them to get the perimeter of the shaded region.