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1)An equilateral triangle has an area of 64sqrt3 cm^2 . If each side of the triangle is decreased by 4 cm, by how many square centimeters is the area decreased?

 

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

 

 May 4, 2018
 #1
avatar+98061 
+1

1)

The  area of an equilateral triangle   is  given by

 

√3/4 side^2

 

So

 

64√3  =  √3 /4  side^2

256 = side^2

16 cm  = side  of original equilateral triangle

 

Decreasing each side by 4 cm  produces an area of  

 

√3/ 4 * 12^2   =  

 

36√3  cm^2

 

So  the area decrease  is    64√3  - 36√3   =  28√3 cm^2

 

 

cool cool cool

 May 4, 2018
 #2
avatar+98061 
+1

2.

 

We can find angle BAC as follows

 

6^2  = 2(5)^2  - 2(5^2) cos(BAC)

 

[36 - 50]  / (-50)  = cos(BAC)

 

-14/-50  = cos(BAC)

 

7/25 = cos (BAC)

 

arccos ( 7/25) = BAC

 

And since BAC is an inscribed angle, the BOC has twice its measure =  2arccos(7/25)

 

And   cos [ 2 arccos (7/25) ]  =    - 527/ 625

 

And OC  = OB

 

So......using the Law of Cosines

 

6^2  = 2(OC)^2 - 2(OC)^2 cos [2arccos(7/25) ]

 

36 = 2(OC)^2 - 2(OC)^2 [ -527/ 625]

 

36 = 2(OC)^2 + 2(OC)^2 [ 527/625]

 

18 = OC^2 [ 1 + 527/625]

 

18 = OC^2 [ 1152/ 625]

 

18 * 625 / 1152  = OC^2  =  11250/ 1152  =  625/ 64

 

So.....the area of triangle OBC  =

 

(1/2) OC^2 * sin (BOC)

 

(1/2) (625/64) * sin [ 2 arccos(7/25) ]  = 

 

(625/ 128) * sin [ 2 arccos (7/25) ]  =

 

(625 / 128) * (336/625) =

 

336/128    =

 

( 21 /  8  )   units^2

 

 

cool cool cool

 May 4, 2018
 #3
avatar
+1

2) Area of the isosceles triangle ABC =12 [By bisecting BC, we have, 3, 4, 5 right triangle]

Since O is the circumcenter of triangle ABC, then:

The circumradius OC=OB=3.125. The circumradius R = a.b.c / 4*Area =5.5.6 / 4*12=3.125.

Then the area of the isosceles triangle OBC with sides OC=OB=3.125 and BC=6 and h=0.875

=2 5/8 units^2

 May 4, 2018
edited by Guest  May 4, 2018
edited by Guest  May 4, 2018
edited by Guest  May 5, 2018

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