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Let triangle ABC be a right triangle such that B is at the right angle. A circle with a diameter of BC meets side AC at D.  If the area of triangle ABC  is 150 and AC=25 then what is BD?

(no visual given)

 Apr 10, 2019
 #1
avatar+107348 
+1

Kinda' hard to answer without a pic, SS.....

 

 

cool cool cool

 Apr 10, 2019
 #2
avatar+256 
+1

this was all that was given and that's what I thought as well but I figured anyone on here had more experience and might be able to figure it out differently.

SydSu22  Apr 10, 2019
 #3
avatar+107348 
+1

 

 

OK....I think I have it.....letting BC = 15  and  and BA    = 20

Then AC  = 25

And the product of the legs BC * BA  / 2  =  the area  = 150

 

The circle is centered at  (0, 7.5)  and has a radius of 7.5

 

Notice in the diagram....angle ADB  = 90°.....so  BD  is an altitude of triangle ABC

So...letting AC  be a base...we have that

 

Area of ABC  = (1/2) base * altitude

 

150 = (1/2)AC * ( BD)

 

150  = (1/2)AC * BD

 

150  = (1/2)(25) (BD)

 

300 = 25 (BD)

 

300 /25  = BD  =   12

 

 

cool cool cool

 Apr 10, 2019
 #4
avatar+256 
+1

OMGosh that is amazing and this is one of the reasons I come here when I need help!

SydSu22  Apr 10, 2019
 #5
avatar+107348 
+1

Well...I'm not great at Geometry.....but....I can do a little   !!!!

 

 

cool coolcool

CPhill  Apr 10, 2019

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