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# Circles? PLS HELP ASAP

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In triangle ABC, AB=AC=5 and BC=6. Let O be the circumcenter of triangle ABC. Find the area of triangle OBC

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MIRB15  Jul 8, 2017
edited by MIRB15  Jul 8, 2017
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Here's my approach.....however.....there may be better methods of attack

Draw  altitude AD (through O ) of triangle ABC to base BC

This altitude will bisect BC  so that DC  = 3

The triangle ADC is right  and since AC = 5  and DC = 3, then this is a "Pythagorean" Triple" 3 - 4 - 5  right triangle......so.....AD  = 4

And  the triangle ODC formed is also right.... and  we have  that

OD = √[OC^2 - DC^2]  =  √[OC^2 - 3^2] = √[ OC^2 - 9  ]  =  height of  ODC

And since OC = OA....we have that

OA + OD  =   AD =  4      ....so...substituting

OC + √[ OC^2 - 9  ]  = 4     subtract OC from each side

√[ OC^2 - 9  ] = 4 - OC      square both sides

OC^2 - 9 =  16 - 8OC  + OC^2     subtract OC^2  from each side

- 9 = 16 -  8OC     subtract 16 from both sides

-25 =  -8 OC    divide both sides by -8

25/8 = OC

So the height of triangle OBC = OD  = √[ OC^2 - 9  ] = √[ (25/8)^2 - 9  ]

And its area  =  (1/2) *BC * OD  =

(1/2) * 6 * √[ (25/8)^2 - 9  ]   =

3 * √ (625/ 64 - 576/64 )  =

3 * √ (49/64)  =

3 * 7/8  =

21 / 8  units^2

CPhill  Jul 8, 2017
edited by CPhill  Jul 9, 2017

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