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1. Medians $\overline{AX}$ and $\overline{BY}$ of $\triangle ABC$ intersect perpendicularly at point $O$. We know the lengths $AX = 12$ and $BC = 4\sqrt {13}$. Find the length of the third median, $CZ$. (Is not: 200 or 4sqrt22)

2. As shown in the diagram, $\overline {AD}, \overline {BE},$ and $\overline {CF}$ are concurrent. We know that $CD=CE=BF=4$ and $BD=AF=6$. What is $AE?

$ (Is not: 16/9)

 Nov 27, 2019
 #1
avatar+2551 
+2

So much easier to read with this post:

 

https://web2.0calc.com/questions/instant-latex-conversion-to-handle-those-lazy-guests

 

it has blessed and healed my eyeballs from un-rendered LaTeX.

 

What I see on MY screen:

 

much better compared to what you see on YOUR screen, right?

 Nov 27, 2019
edited by CalculatorUser  Nov 27, 2019
 #2
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Thank you for blessing my eyes with this holy water. 🙏 Anyway, this guest is really rich-- there are a lot of dollar signs lol

artsyleilani  Nov 27, 2019
 #3
avatar+2551 
+1
 

Also, for number 1. This is how far I got.

 

Using the properties of a median through a centroid, I found OB through pythagorean theorem

 

I also found AO through median properties

 

Notice that AOB is a right angle through supplements.

 

I you used pythagorean theorem to find AB?

 

Notice triangles AOB and ZOB are similar.

 

Since it is a median, we know that the triangles ZOB and AOB are in a ratio of 1:2, respectively. I also know the area of AOB because base and height are known.

 

So, how would I find OZ?

 Nov 27, 2019
 #4
avatar+2551 
+1
 

For number 2.

 

You have to use similar triangles.

 

Sorry I ran out of time gotta go angry

 

also, why is "help me please" in the top of Sticky Topics?

 Nov 27, 2019
 #5
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1. CZ = sqrt(16^2 - 8^2) = 8 sqrt(3).

 

2. AE = (6 - 4*4)*3/2 = 15/2.

 Nov 27, 2019

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