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Given two equilateral triangles whose sides are of length a and b respectively, draw a diagram to show how to construct a third equilateral triangle whose area is the sum of the areas of the other two triangles. (By "construct" we mean straightedge and compass construction.)

jonathanxu999  Feb 10, 2018
edited by jonathanxu999  Feb 10, 2018
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Note that the area of the first equilateral triangle  is just (1/2)a^2 sin (60°)

 

And the area of the second equilateral triangle is just  (1/2)b^2sin(60°)

 

So....the sum of their areas is just

 

(1/2) sin (60°)  (a^2 + b^2)

 

So....we actually  need   the side of our constructed  triangle to be  √[a^2 + b^2]

 

Here's a way to do this  ....a pic will definitely help !!!!

 

 

1. Draw any segment AC  and then constuct any segment AB  perpendicular to this....Connect BC...call AB, "a"  and AC, "b" and BC  =  sqrt (a^2 + b^2)

2. Put your compass at A  and draw a circle with radius AC.....then put it at C and draw a circle with radius CA.....let these circles intersect at "D"  and connect AD and CD....this will create equilateral  triangle  ADC

3. Similarly......put your compass at A and draw a circle with radius AB....then put your compass at B and draw a circle with radius BA...let these intersect at E....this will create equilateral triangle BEA

4. Finally.....put your compass at C and  and draw a circle with radius CB......then put your compass at B  and draw a circle with radius BC.....let these circles intersect at F....this will create equilateral triangle CFB

 

So.....area of ADC  =  (1/2) a^2 sin (60°)   and area of BEA =  (1/2) b^2 sin (60°)

 

So  their combined area  =   (1/2) sin (60°) (a^2 + b^2) 

 

And the area of  CFB  =  (1/2)sin (60°)√[a^2 + b^2] * √ [a^2 + b^2]  =

 

(1/2) sin(60°) (a^2 + b^2)..... the combined areas of ADC  and BEA  !!!

 

Here's a pic without the circles:

 

 

cool cool cool

CPhill  Feb 10, 2018

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