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Triangles \(ABC\) and \(BDC\), shown here, are \(30\)-\(60\)-\(90\) right triangles, and \(AC=12\) cm. The area of \(\triangle{BDC}\) can be written in simplest radical form \(\frac{a\sqrt{c}}{b}\) cm\(^2\), where \(a\)\(b\), and \(c\) are positive integers. What is the value of \(a+b+c\)?

 

 Aug 22, 2017
 #1
avatar+558 
+1

We know that AC =12cm. And triangle ABC is a 30-60-90 triangle. So Angle A must be 30 degrees.

So we can do sin (30)=AC/BC, which is sin(30)=12/BC. 

Multiply both sides by BC and we get sin(30)*BC =12. 

Divide both sides by sin(30) and we get BC=sin(30)*12.

Plug sin(30)*12 into a calculator and you get 6. So BC is 6.

 

Now, we also know that triangle BDC is a 30-60-90 triangle.

The ratio of side DC to side DB to side BC is 1:sqrt(3):2

Using that ratio, we can find that DC is 3 and DB is 3*sqrt(3) because we already found the length of BC, which is 6.

And the triangle's area formula is base *height*1/2. So 3*(3*sqrt(3))*1/2= 9*sqrt(3)/2

Then 'a' must be 9, 'b' must be 2, and 'c' must be 3.

 

So a+b+c= 9+2+3= 14.

The answer is 14.

 

Correct me if I'm wrong. I might've made some errors in calculation and making equations. smiley

 Aug 22, 2017
 #2
avatar+7350 
+2

It looks like    m∠A  =  30°   , but I don't know how to tell that for sure....

 

If   m∠A  =  60°   , then   m∠C  =  30°   and   BC =  6√3   .

 

That means   BD  =  3√3   and   CD  =  3√3 * √3  =  9

 

So    the area of triangle BDC  =  \(\frac{(3\sqrt3)(9)}{2} \,=\, \frac{27\sqrt3}{2}\)

 

And.....  27 + 3 + 2  =  32

 

 

If  m∠A  =  30°  , then I get  14  too. smiley

 Aug 22, 2017

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