+0  
 
+2
45
4
avatar+285 

In the diagram, \(AOB\) is a quarter circle of radius 10 and \(PQRO\)is a rectangle of perimeter \(26\). What is the perimeter of the shaded region?

TheMathCoder  Apr 25, 2018
edited by TheMathCoder  Apr 25, 2018
Sort: 

4+0 Answers

 #1
avatar+86556 
+4

Here's my best effort  ....

 

If the perimeter of the rectangle is 26....then 

 

2(W + L )  = 26

W + L  = 13

Then  L  =  13 - W

 

Call point Q  =   ( W , L )  =  (W , 13 - W)

 

So.....using the equation of a circle with a center O located at  (0,0)  and a radius of 10, we have

 

W^2  + (13 - W)^2  =  100

W^2 + W^2 - 26W + 169  = 100

2W^2 - 26W + 69  =  0

 

Solving this for W  we get that W  =  [ 13 + √31] / 2  = OP

And L  =   [ 13  - √31 ] / 2  = PQ 

 

So  PR  =  (1/2) √ [  ( 13 + √31)^2 + (13 - √13)^2 ] =  (1/2) √[ 382 + 26√31 - 26√13 ]

 

So.... the perimeter  =

 

arc BQA  + AP  + PR  + BR  =

 

5pi + (10 - OP) + PR + (10 - PQ)  =

 

5pi  + (10 -  [ 13 + √31] / 2 )  + (1/2) √[ 382 + 26√31 - 26√13 ]  + (10 -  [ 13  - √31 ] / 2 )  ≈

 

33.1125  units

 

 

cool cool cool

CPhill  Apr 25, 2018
edited by CPhill  Apr 25, 2018
 #3
avatar+633 
+3

Oh wow, well done. 

 

I did not think of this method, writing an equation. 

 

I though we make it a whole circle and manipulate it somehow, but I just couldn't find the area of the rectangle. 

GYanggg  Apr 25, 2018
 #2
avatar
+3

CPhill: I will give it a go. Please check my reasoning! Thanks.

 

The diagonal OQ of the rectangle = radius of the circle = 10

Cos(45) =OP / 10 =~ 7.07 =10/sqrt(2)

OR = 26 / 2 - 7.07=~ 5.93 =13 - 5sqrt(2)

10 - OR = BR =~4.07=5sqrt(2) - 3

AP = 10 - 7.07 =~2.93=10 - 5sqrt(2)

Arc AQB =[10 x 2 x Pi] / 4 =~15.71 =5pi

So, the perimeter of the shaded area =10 + [5sqrt(2) - 3] + [10 - 5sqrt(2)] + 5pi =17 + 5pi

Guest Apr 25, 2018
edited by Guest  Apr 25, 2018
edited by Guest  Apr 25, 2018
edited by Guest  Apr 26, 2018
 #4
avatar+285 
+4

Noice guys! Thanks to CPhill and guest!

TheMathCoder  Apr 26, 2018

28 Online Users

avatar
avatar
New Privacy Policy (May 2018)
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  Privacy Policy