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Perimeter of shaded region = AD + (AP+DP) = 10 + (more than 10)

Dear GMATGuruNY

Can you please elaborate how you deduce the highlighted part?

Thanks

The shortest distance between two points is a STRAIGHT LINE, implying that AD=10 is the shortest distance between A and D.
Since AP and PD link A to D but do not form a straight line, their combined length must be MORE THAN 10.
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=>

Attachment:
7.29ps.png
7.29ps.png [ 8.99 KiB | Viewed 3364 times ]

Since triangle \(PBC\) is equilateral, angle \(ABP\) has measure \(30°\) and the length of arc \([m]AP\) is 2π*10*(\frac{30}{360}) = (\frac{5}{3}) π.[/m]

Thus, the perimeter of the shaded area is \((\frac{5}{3}) π *2 + 10 = (\frac{10}{3}) π + 10.\)

Therefore, A is the answer.
Answer: A
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=>

Attachment:
7.29ps.png

Since @triangle \(PBC\) is equilateral, angle \(ABP\) has measure \(30°\) and the length of arc \([m]AP\) is 2π*10*(\frac{30}{360}) = (\frac{5}{3}) π.[/m]

Thus, the perimeter of the shaded area is \((\frac{5}{3}) π *2 + 10 = (\frac{10}{3}) π + 10.\)

Therefore, A is the answer.
Answer: A

MathRevolution

How the triangle is equilateral? Can you explain please to help us understand your solution?
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I am putting in a proper geometrical solution, which is lengthy.

The question could be answered by eliminating choices as only one option is a value that is 10 + 10 more.

Now since AB is a straight line AB will be shortest distance so APB has to greater than AB hence the logic that the answer choice should be 10 + more than 10.

Posted from my mobile device
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Part 2

Posted from my mobile device
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[quote="MathRevolution"]=>

[attachment=0]7.29ps.png[/attachment]

Since triangle [m]PBC[/m] is equilateral, angle [m]ABP[/m] has measure [m]30°[/m] and the length of arc [m][m]AP[/m] is 2π*10*([fraction]30/360[/fraction]) = ([fraction]5/3[/fraction]) π.[/m]

Thus, the perimeter of the shaded area is [m]([fraction]5/3[/fraction]) π *2 + 10 = ([fraction]10/3[/fraction]) π + 10.[/m]

Therefore, A is the answer.
Answer: A[/quote]

How can we deduce it to be equilateral ?

[size=80][b][i]Posted from my mobile device[/i][/b][/size]
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Given:
ABCD is a square with side 10.

We need to find: Arc (AxP) + Arc (PyD) + DA -> (a)

Since BAPCB and CDPBC are Quarter Circles, BP and PC are radii of the circles.
Thus, BP = PC = BC ->(b)

As shown in the figure: Triangle PBC can form an Equilateral Triangle -> From (b)
Thus, Angle (PBC) = Angle (PCB) = 60 degrees.

Hence, Angle (ABP) = Angle (DCP) = 90 - 60 = 30 degrees

Therefore, Arc (AxP) = Arc (PyD) = \(\frac{30}{360} * 2 * \pi * 10\) = \(\frac{5 * \pi}{3}\) -> (c)

The required perimeter is: \(\frac{5 * \pi}{3}\) + \(\frac{5 * \pi}{3}\) + 10 = \(\frac{10 * \pi}{3} + 10\) -> [From (a) and (c)]

Answer A
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