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01 Nov 2019, 10:42
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D
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Difficulty:
45%
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Question Stats:
67% (02:07) correct
33%
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wrong
based on 150
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In the figure above, ABCDEF is a regular hexagon with vertices A and C on the circle centered at point B. Is the perimeter of the figure greater than 40 ?
(1) The overlapping region between circle B and hexagon ABCDEF has an area of 12π.
(2) The area of the figure is 24\(\pi\) +54 \(\sqrt{3}\)
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01 Nov 2019, 11:27
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Before looking at the statements, there are few key points that we need to take care of..
First- this is a regular hexagon. Thus all the sides are equal in length and each internal angle is 120° { ((n-2)*180)/N, where N= 6}
Second- radius= side of hexagon..
Therefore to find whether perimeter is greater than 40, we just need to find the radius or length of each side of hexagon..
Statement 1- overlapping area is 12π.. we can get the radius and hence find the perimeter of entire figure. Sufficient
Statement 2- area of the figure is given..
=> Area of circle - area of sector + area of hexagon = 24π + 54√3..
We can find out the radius "r" and hence the perimeter. Sufficient.
Each statement is sufficient on its own.
D...
P.s - I didn't calculate the the values because we actually don't need to. Since we know that each statement will give a value of "r", a unique value, we will have a unique answer...
However for conceptual basis-
Statement 1 - 120/360*π*r^2
=> 1/3*π*r^2 = 12π
=> r^2 = 36
=> r = 6
Statement 2 - area of Figure = 24π + 54√3
=> π*r^2 - (1/3)*π*r^2 + (√3/4)*r^2*6
=> (2/3)π*r^2 + (3√3/2)*r^2
=> (2/3)π*r^2 = 24π
=> r^2 = 24*(3/2) => r^2 = 36
=> r = 6.
Hope it helps!
Posted from my mobile device
First- this is a regular hexagon. Thus all the sides are equal in length and each internal angle is 120° { ((n-2)*180)/N, where N= 6}
Second- radius= side of hexagon..
Therefore to find whether perimeter is greater than 40, we just need to find the radius or length of each side of hexagon..
Statement 1- overlapping area is 12π.. we can get the radius and hence find the perimeter of entire figure. Sufficient
Statement 2- area of the figure is given..
=> Area of circle - area of sector + area of hexagon = 24π + 54√3..
We can find out the radius "r" and hence the perimeter. Sufficient.
Each statement is sufficient on its own.
D...
P.s - I didn't calculate the the values because we actually don't need to. Since we know that each statement will give a value of "r", a unique value, we will have a unique answer...
However for conceptual basis-
Statement 1 - 120/360*π*r^2
=> 1/3*π*r^2 = 12π
=> r^2 = 36
=> r = 6
Statement 2 - area of Figure = 24π + 54√3
=> π*r^2 - (1/3)*π*r^2 + (√3/4)*r^2*6
=> (2/3)π*r^2 + (3√3/2)*r^2
=> (2/3)π*r^2 = 24π
=> r^2 = 24*(3/2) => r^2 = 36
=> r = 6.
Hope it helps!
Posted from my mobile device
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01 Nov 2019, 11:43
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The overlapping area is \(\frac{(πr^2)}{3}\). We will know the perimeter of the figure once we have the value of r (r=the radius of the circle and also the side of the regular hexagon)
St 1) We have the value of \(\frac{(πr^2)}{3}\). We can find r. SUFFICIENT
St 2) We have the total area of the figure, which is (\(\frac{2}{3}\)*\((πr^2)\)) +\(\frac{6\sqrt{3}}{4}*r^2\). We get the value of r from here too. SUFFICIENT
Hence D
Some formulas or value used in above calculation, that may not look so obvious to some
Area of a regular hexagon = \(\frac{6\sqrt{3}}{4}*r^2\)
Each angle of a hexagon is 120 degree, hence the sector area= \(\frac{120}{360} * πr^2\)= \(\frac{(πr^2)}{3}\)
Are of the the remaining sector = \((πr^2)\)- \(\frac{(πr^2)}{3}\)= \(\frac{2(πr^2)}{3}\)
St 1) We have the value of \(\frac{(πr^2)}{3}\). We can find r. SUFFICIENT
St 2) We have the total area of the figure, which is (\(\frac{2}{3}\)*\((πr^2)\)) +\(\frac{6\sqrt{3}}{4}*r^2\). We get the value of r from here too. SUFFICIENT
Hence D
Some formulas or value used in above calculation, that may not look so obvious to some
Area of a regular hexagon = \(\frac{6\sqrt{3}}{4}*r^2\)
Each angle of a hexagon is 120 degree, hence the sector area= \(\frac{120}{360} * πr^2\)= \(\frac{(πr^2)}{3}\)
Are of the the remaining sector = \((πr^2)\)- \(\frac{(πr^2)}{3}\)= \(\frac{2(πr^2)}{3}\)
