Year 7 geometry help

Hello, and welcome to FMH! :)

What are your thoughts on the choices given?
 
Joshua76 said:
Just need help on this one.
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Please read our submission guidelines, which tell you that, in order to help you, we need to work with you. We don't just do the work for you, which wouldn't help you learn.

So, tell us what you have thought, so we can guide you in the right direction (or just affirm your ideas).
 
Come on, you have to think.
Choice A: If AB is a line, then how can you even find the midpoint?
Choice B: If line segment CD bisects AB is is because line segment CD crossed AB at the midpoint of AB. Line CD would cross AB at the same point as line segment CD
Choice C: Maybe
Choice D: Your figure shows an example where the radii are different. Does it look like CD bisects AB?
Choice E: Maybe
 
Dr.Peterson said:
Please read our submission guidelines, which tell you that, in order to help you, we need to work with you. We don't just do the work for you, which wouldn't help you learn.

So, tell us what you have thought, so we can guide you in the right direction (or just affirm your ideas).
Click to expand...
Did i say that no, i need a explaination. My friend was right i should of went to another math network
 
Joshua76 said:
Did i say that no, i need a explaination. My friend was right i should of went to another math network
Click to expand...

I would guess, based on notions of left-right symmetry, that the circles must have the same radius. But, I wouldn't be satisfied with that. I would likely construct a coordinate geometry based proof. Let the centers of the two circles lie along the \(x\)-axis, separated by a distance \(2a\) where \(0<a<1\). The circle on the left is given by:

[MATH](x+a)^2+y^2=r_1^2[/MATH]
And the circle on the right is given by:

[MATH](x-a)^2+y^2=r_2^2[/MATH]
To find the \(x\)-coordinate where the circles meet, we must have:

[MATH]r_1^2-(x+a)^2=r_2^2-(x-a)^2[/MATH]
Now, in order for the condition we want to be met, the two circles must meet along the \(y\)-axis, where \(x=0\), and so we must have:

[MATH]r_1^2-a^2=r_2^2-a^2[/MATH]
And this implies:

[MATH]r_1=r_2[/MATH]
And so we see, that the two radii being equal is a necessary condition. :)
 
Joshua76 said:
Did i say that no, i need a explaination. My friend was right i should of went to another math network
Click to expand...
I agree with your friend that you should have gone to another network---an English network
By, see you later.
 
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