Geometry word problem

amathproblemthatneedsolve

amathproblemthatneedsolve

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The distance from the centre of the hole to the edge AC where the triangle ABC joins the square needs to be at least 2 centimetres. Assume that the center of the hole is on the median from B to AC. How much closer towards the edge AC could the hole be drilled? Justify your answer.

Use the following information in your investigation. The vertices of the triangle are A (4, −2), B (1, 9) and C (10, 2). The points D (14, −4) and E (8, –8) are the other vertices of the quadrilateral ACDE. The line 2y = −3x + 21 is an axis of symmetry of the total shape ABCDE. The scale of the diagram is 1 unit = 1 cm.

centre of the hole is at (5,3) I've found the distance from Centre hole (5,3) to (7,0)Ac to be √13
 
You appear to have dropped us into the middle of a problem, since you refer to "the hole", "the square", etc. without introduction. Please state the problem as given to you, so we can be sure what you are given, what you are asked for, and what parts you have worked out yourself.

But it appears that you have found how far the center of the hole is from AC; just subtract 2 from that to find how far the center can be moved and still be 2 cm away.
 
A circular hole with radius 1 centimetre is drilled with the centre at the centroid of the
triangle (the centroid is also known as the centre of gravity). The centroid is the point where
the medians of the triangle intersect. A median is a line from a vertex to the midpoint of the
opposite side. Find the centre of the circle.


Triangle cooridnates

A=(4,-2)

B=(1,9)

C=(10,2)

I've worked the midpoint of AB to be (2.5,3.5)

midpoint of CB=(5.5,5.5)

gradient of c=-0.2
y-2==-0.2(x-10)

gradient of A = 5

y--2=5(x-4) y=5x-22 y=3 x=5


and for the inital question can the hole be moved aprox 1.6 cm closer?
 
Thanks.

It's not clear what lines you are calling "c" and "A"; that is not typical notation. If c means the median through C and A means the median through A, you have the correct equations, and the correct centroid. So I've figured out what you mean ...

The distance to AC is as you said, about 3.61, so the answer to the part of the problem you initially quoted is indeed about 1.61 cm. Good work.

I imagine there is more to come, as nothing has yet made use of the quadrilateral you mentioned originally. If you ask more, please be sure to quote the entire problem from beginning to end (or at least through the last part you ask about).
 
Dr.Peterson said:
Thanks.

It's not clear what lines you are calling "c" and "A"; that is not typical notation. If c means the median through C and A means the median through A, you have the correct equations, and the correct centroid. So I've figured out what you mean ...

The distance to AC is as you said, about 3.61, so the answer to the part of the problem you initially quoted is indeed about 1.61 cm. Good work.

I imagine there is more to come, as nothing has yet made use of the quadrilateral you mentioned originally. If you ask more, please be sure to quote the entire problem from beginning to end (or at least through the last part you ask about).
Click to expand...

Thank you.
the quadrilateral is there because I was asked to prove it's a square at the start. And for C and A you are correct. What does the question mean by justify your answer? thanks
 
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