Equilateral Triangle

Set 6.1
David Cohen

 

49.
show that area of an equaterial triangle of side s is given by

(sqrt(3)/4)s^2




Area of Triangle = 1/2* base * height

Here, base = s, and height = h

Now, apply Pythagoras Theorem in the triangle.

s^2 = h^2 + (s/2)^2

h^2 = s^2 – (s^2/4)

h^2 = (3s^2)/4

Or, h = (1/2)(√3*s)

Now, put the value of “h” in the area of the triangle equation.

A =1/2* s * (1/2)(√3*s)

Or, Area of Equilateral Triangle = (√3/4)*s^2


50.
show that the area of the shaded segment is given by

s^2((2pi-3sqrt(3))/12)

the area of the shaded segment= the area of the circle sector ABC- the area of the triangle ABC

the area of the circle sector ABC=(theta/360)*pi*r^2

central angle theta=60 degrees, r=s

the area of the circle sector ABC=(60/360)*pi*s^2=(1/6)pi*s^2

use the area of the triangle ABC from 49 which is = (√3/4)*s^2

then

the area of the shaded segment=(1/6)pi*s^2- (√3/4)*s^2
the area of the shaded segment=(2pi/12)*s^2- (3√3)/12*s^2
the area of the shaded segment=(2pi- 3√3)/12*s^2

 

I hope you don't mind the occasional David Cohen textbook questions. David's textbook is a lot more involved than the Ron Larson book. I just love the Cohen textbook questions which are above my current level of mathematics.

 

I don't mind , both textbooks are good

 

Ok. Cool. I will continue using Ron Larson's book to review and complete precalculus before calculus 1 but occasionally post challenging questions from the Cohen book. The next set of questions will be right triangle trigonometry applications from the Larson book Section 4.3.