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Can someone help with this...

 

 Feb 13, 2019
 #1
avatar+101871 
+2

The width is no problem.....it is just 4 times the radius of each circle = 4r = 4(2mm) = 8mm    (1)

 

Note that joining the 3 radii will form an equilateral triangle  with sides = 2r  = 2 * 2mm = 4mm

 

The height of this triangle, h, can be found as

 

tan 60 = h/2

sqrt (3) = h / 2

2sqrt(3) = h

 

So....the height of the rectangle =  [ 2sqrt(3)  +  2r ]  mm   = [ 2sqt(3) + 2(2)] mm  = [ 2qrt(3) + 4 ] mm  (2)

 

So....the area of the rectangle =  (1) * (2)  =

 

[ 8 ] *  [ 2sqrt (3) + 4 ]  mm^2 =

 

 [ 16sqrt (3) + 32 ]  mm ^2

 

 

cool cool cool

 Feb 13, 2019
 #2
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could you also help with this?

Guest Feb 13, 2019
 #3
avatar+101871 
+1

OK....

 

Note that   3sqrt (2)  =  = sqrt (9 * 2) =  sqrt (18)

And 2sqrt (5)  = sqrt (4 * 5) =  sqrt (20)

 

So....we can find the height o the triangle by the Pythagorean Theorem...

h ^2  = [ sqrt (20) ] ^2 - [sqrt (18) ] ^2 

h^2  = 20 - 18

h^2 = 2

h = sqrt (2)

 

And the base = 3sqrt (2) + sqrt (2)  = 4sqrt (2)

 

So....the area  =  (1/2)base * height =  (1/2) [ 4 sqrt (2) ] [ sqrt (2) ]  =  (1/2) (4) [sqrt (2) ]^2  =

 

(1/2) (4) (2)  =

 

4 cm ^2

 

 

cool cool cool

 Feb 13, 2019

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