Grade 6 Singapore Mathematics Circle Area Question

[senseimichael]

Grade 6 Singapore Mathematics question. The area of the equilateral triangle is 63, the radius of each circle is 7. I need to find the area of the dark portion in the middle. I'm quite sure it has to do with how the triangles interact with the circles. I tried so many angles, but I am truly stumped!

 

[lev888]

View attachment 29070
Grade 6 Singapore Mathematics question. The area of the equilateral triangle is 63, the radius of each circle is 7. I need to find the area of the dark portion in the middle. I'm quite sure it has to do with how the triangles interact with the circles. I tried so many angles, but I am truly stumped!

Wondering why we are given both values. If the triangle is inscribed, I think we can calculate the radius given triangle area and vice versa.
Anyway, please post what you tried.

 

[jonah2.0]

Beer induced observation follows.

View attachment 29070
Grade 6 Singapore Mathematics question. The area of the equilateral triangle is 63, the radius of each circle is 7. I need to find the area of the dark portion in the middle. I'm quite sure it has to do with how the triangles interact with the circles. I tried so many angles, but I am truly stumped!

Maybe it's just the beer talking, but given the equilateral triangle area of 63, the distance from the center of the circle in the middle (the one the triangle is inscribed in) to any corner of the triangle is definitely not 7.

 

[Dr.Peterson]

given the equilateral triangle area of 63, the distance from the center of the circle in the middle (the one the triangle is inscribed in) to any corner of the triangle is definitely not 7.

It's close. Area 63 implies radius 6.964; radius 7 implies area 63.65.

View attachment 29070
Grade 6 Singapore Mathematics question. The area of the equilateral triangle is 63, the radius of each circle is 7. I need to find the area of the dark portion in the middle. I'm quite sure it has to do with how the triangles interact with the circles. I tried so many angles, but I am truly stumped!

How much does a grade 6 student know of geometry and trigonometry there? Apparently they aren't expected to know enough to do what I did above! So this must require only ... what?

 

[jonah2.0]

Beer induced comment follows

How much does a grade 6 student know of geometry and trigonometry there? Apparently they aren't expected to know enough to do what I did above! So this must require only ... what?

Wondering much about that too.

 

[Dr.Peterson]

You can find the area of the red region, which is the same as the yellow region:


Then, you can use that to find the required region, at least if you can convince yourself that what looks like the center of the circle really is. That may or may not require some deeper knowledge of geometry. The main tool is addition and subtraction of areas.

 

[senseimichael]

The model answer is this, which I went WTH to...

I don't even understand how on earth 35=9.33. I think the teacher who set this question must have been drinking...

 

[jonah2.0]

Beer induced ramblings follow.

The model answer is this, which I went WTH to...
View attachment 29092
I don't even understand how on earth 35=9.33. I think the teacher who set this question must have been drinking...

It would be interesting indeed to see how that teacher, a fellow drinker, would justify those figures. I speak drunk fluently and I have trouble figuring out how them figures ended up as the model answer.

 

[Dr.Peterson]

The model answer is this, which I went WTH to...
View attachment 29092
I don't even understand how on earth 35=9.33. I think the teacher who set this question must have been drinking...

First, it troubles me that they are taking pi to be exactly 22/7; it should at least be stated that you are to use that approximation (though perhaps that is understood within this course).

Second, when I realized that, I supposed that that could explain the inconsistent values for area and radius, but those depend on the square root of 3, not on pi. Is it possible that they are taught to approximate $\sqrt{3}$ as 12/7? More likely, they were just given approximate values for both numbers because they are not expected to be able to relate them, or to try to do so.

Taking pi as 22/7, my answer does become 9 1/3. But where this has 49, I have 2*63. So it looks like they wrongly used the 49 (and got 35 as their answer), but then tacked on the correct answer as if they had obtained it.

Yes, some drinking was probably involved. But someone in the vicinity did get the right answer!