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GRE Math Challenge #31-area of the hexagon region
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18 Sep 2014, 12:18
18 Sep 2014, 12:18
Expert Reply
Attachment:
figure 20.jpg [ 8.43 KiB | Viewed 2755 times ]
What is the area of the hexagon region shown in the figure above?
a) 54√3
b) 108
c) 108√3
d) 216
e) It cannot be determined from the information given
Re: GRE Math Challenge #31-area of the hexagon region
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09 Aug 2018, 09:51
09 Aug 2018, 09:51
1
So my explanation here is:
From the figure we have 4 right triangles and 2 rectangles.
Since it is a 6 side figure using (N-2)*(180º), we have (6-2)*(180º) = 720 must be the sum of the 6 angles, xº, which means each xº = 120
Now, with that in mind:
a) The right triangles: the hypotenuse = 6 and using the properties of a right triangle since we are cutting in half the 120º we have a 60º in one corner so the last angle must be a 30º. With this we can see the property 2x: x: x*sqrt(3) -> 6: 3: 3*sqrt(3), so the area of the triangle is: (bh/2) = (3*3sqrt(3))/2 = (9/2)*sqrt(3)
b) Now we have the rectangles, which we know now from the triangles that the width is 6 and the length is 3* sqrt(3), the Area = 3 *sqrt(3) * 6 = 18 * sqrt (3)
c) we have 4 triangles and 2 rectangles to add which will give us:
(4 * (9/2) * sqrt(3)) + (2 * 18 * sqrt(3)) = 54 * sqrt(3)
So answer is A
Sorry for not being able to put an image and hopefully it helps.
Cheers
From the figure we have 4 right triangles and 2 rectangles.
Since it is a 6 side figure using (N-2)*(180º), we have (6-2)*(180º) = 720 must be the sum of the 6 angles, xº, which means each xº = 120
Now, with that in mind:
a) The right triangles: the hypotenuse = 6 and using the properties of a right triangle since we are cutting in half the 120º we have a 60º in one corner so the last angle must be a 30º. With this we can see the property 2x: x: x*sqrt(3) -> 6: 3: 3*sqrt(3), so the area of the triangle is: (bh/2) = (3*3sqrt(3))/2 = (9/2)*sqrt(3)
b) Now we have the rectangles, which we know now from the triangles that the width is 6 and the length is 3* sqrt(3), the Area = 3 *sqrt(3) * 6 = 18 * sqrt (3)
c) we have 4 triangles and 2 rectangles to add which will give us:
(4 * (9/2) * sqrt(3)) + (2 * 18 * sqrt(3)) = 54 * sqrt(3)
So answer is A
Sorry for not being able to put an image and hopefully it helps.
Cheers
Re: GRE Math Challenge #31-area of the hexagon region
[#permalink]
18 Nov 2018, 17:32
18 Nov 2018, 17:32
Can any body explain with diagram?
