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An equilateral triangle has sides $8$ units long. An equilateral triangle with sides $2$ units long is cut off at the top, leaving an isosceles trapezoid. What is the ratio of the area of the smaller triangle to the area of the trapezoid? Express your answer as a common fraction.

 Jan 2, 2024
 #1
avatar+129771 
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The height of the trapezoid is 3 times the height of the  equilateral triangle

 

Area of equilateral triangle = (1/2)base * height  = (1/2) (2) (H) =  H

 

Bases  of trapezoid = 2 , 8......height =  3H

 

Area of trapezoid=   (1/2) (3H)( 2 + 8)  =   15H

 

Ratio of areas =   H / ( 15H)  =   1 / 15

 

 

cool cool cool

 Jan 3, 2024

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