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Test Smart Circle question

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Test Smart Circle question

by jawad » Sat Aug 30, 2008 9:29 am
Hi ,

Below question appears in PR manual as TEST SMART question. Can anyone help understand how to calculate the area.

A circle is inscribed in equilateral triangle ABC such that point D lies on circle and on line segment AC , point E lies on circle and line segment AB , and point F lies on circle and on line segment BC. If line segment AB=6 , what is the area of figure created by line segments AD , AE and minor arc DE.

regards.jawad
Jawad Shah
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by ELYAC Realty » Sat Aug 30, 2008 10:18 am
I believe there are 4 steps to this questions:

1) find the area of the triangle. If I am not mistaken this triangle should be an equilateral triangle where all sides are 6 (so (6 squared x square root of 3)/4) or the a squared + 3 squared = 6 squared (https://www.mathwords.com/a/area_equilat ... iangle.htm)
2) find the area of the circle
3) subtract the area of the circle from the area of the triangle
4) multiple by one third

I would be curious to know if this is right.
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Test Smart Circle question

by jawad » Sun Aug 31, 2008 8:01 am
My issue is how can we calculate area of circle.

Please anyone can help in this regard.

regards/jawad
Jawad Shah
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by ELYAC Realty » Sun Aug 31, 2008 12:51 pm
Radius is equal to:
area of triangle/(1/2(triangle side 1 + triangle side 2 + triangle side 3)
(https://www.efunda.com/math/areas/Circle ... gleGen.cfm)

Area of equilateral triangle from the formula I posted earlier is: 15.59

So, Radius = 15.59/9 = 1.732

Area of circle = 9.42

Subtract area of circle from area of triangle to get: 6.170

Multiply by one third to get the area the question is asking for = 2.036
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It is easier than it looks

by Alex_mp » Mon Sep 08, 2008 3:12 pm
My approach:

1- We are looking for the area of the small triangle ADE.
2- In order to obtain the area's triangle we need the base and height.

As we know from the squares inscribed in circles, this figure is inscribed in a perfect equilateral triangle ABC. The information that we derive out of this is that the intersection of the circle over every side is exactly in the middle point of each side of the triangle. Therefore, the side AD = 3 = base of the triangle. Now we only need the height.

If we draw the subtriangle DEF is clear to notice that it is an equilateral triangle as well, but of sides equal to 3. This triangle DEF shares the same height of the triangle ADE. From the Pythagorean equation we can get that (3)square(2) - (3/2)square(2) = 27/4; this is the squared value of the height. From this value :
height = square root of (27/4)
height = 3(SR(3))/2

Well, from here is just multiply 3 (side AD) by the height and divide it by 2.

The problem is proved if you get the sine of 60 degrees (SR(3)/2) multiply it by 3 (side AE) and this will get you the same height.

I Hope it works.

Alex.
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by mmtang1 » Tue Sep 23, 2008 6:38 pm
My try: Draw a line from each point (D, E, and F) to the opposite corner to create 6 non-overlapping 30-60-90 degree right triangles. The base for each triangle is 3. Since we know for 30-60-90 triangles that shortest side is X and second longest side (base, in this case) is X*sqrt(3), then X = radius = 3/sqrt(3).
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