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sudhir18n
In the figure shown, ABCD is a square with side length 5 sqrt 2. Two quarter-circles BED and DFB are centered at A and C, respectively. What is the area of the shaded region?


A.50pi-25
B.25pi−50
C.50pi/3
D.25pi/3
E.25pi

I got the answer.. but whats the fastest way to do this?

Solved the same way Ian suggested:

\((5\sqrt{2})^2=\frac{\pi (5\sqrt{2})^2}{4}+\frac{\pi (5\sqrt{2})^2}{4}-Region\)

\(Region=25\pi\) − \(50\)

Ans:"B"
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sudhir18n
In the figure shown, ABCD is a square with side length 5 sqrt 2. Two quarter-circles BED and DFB are centered at A and C, respectively. What is the area of the shaded region?


A.50pi-25
B.25pi−50
C.50pi/3
D.25pi/3
E.25pi

I got the answer.. but whats the fastest way to do this?


The fastest way according to me is as below

Half of the required area is = the area of quarter of the circle centering at C- Area of the triangle BCD
= pi * \(Radius ^2\)* 1/2 - 1/2*Height of the triangle BCD * Base of the triangle BCD
=pi * \({5sqrt2} ^2\) -1/2 *\(5sqrt2\) *\(5sqrt2\)
= (25*pi )/2-25

so the required area is 25pi-50........ :) :)
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i solved the Q in one of the longer ways possible..i felt Ian's solution is the quickest way to reach the ans..thanks Ian
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for me easiest way was:

half Required area = sector area - area of triangle
=( (central angle*pi*r^2)/2*pi) - ((1/2 )* b*h)

Required area = 2 * half Required area

hence answer is B
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area of the square 50
area of 1/4 of circle: 25pi/2
now...from 50 we deduct 25pi/2 = and get value of 1 un-shaded portion.
(50-25pi/2)*2 = 100-25pi
from the whole area: 50, we deduct sum of the both un-shaded regions:
50-(100-25pi) = 25pi-50.
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sudhir18n
Attachment:
The attachment 166-1.gif is no longer available
In the figure shown, ABCD is a square with side length 5 sqrt 2. Two quarter-circles BED and DFB are centered at A and C, respectively. What is the area of the shaded region?

A. 50pi-25
B. 25pi−50
C. 50pi/3
D. 25pi/3
E. 25pi

This question appeared on an elementary school exam in Singapore, the leading country in mathematics as of now. When the teachers were asked how they were able to explain such a complex problem to children the answer was "using shapes and puzzle pieces."
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The diagram can be dissected as :
half of a circle and square

Thus , Required portion= Area of half of circle - area of square
25pi-50
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