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# MATHCOUNTS State

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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 $$a+b+c$$?

benjamingu22  Aug 22, 2017
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#1
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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.

Gh0sty15  Aug 22, 2017
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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.

hectictar  Aug 22, 2017

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