A B C D E F is a regular hexagon with each side of 2 cm . Then find the area of the △ A B G to four decimal places.
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I just calculated the apothem and used the traditional triangle area formula.
Look at the figure below: A B C D E F and in the second figure it is also divided into 6 equilateral triangles.The height of the equilateral triangles divides them into 2 right triangles with base 1 cm and hypotenuse 2 cm.Therefore by the Pythagorean theorem we get: ( H e i g h t ) 2 + 1 2 = 2 2 ( H e i g h t ) 2 = 4 − 1 = 3 H e i g h t = 3
In the figure is the hexagonThe height of the equilateral triangles is half the height of A B G .So the height of A B G is 2 × 3 = 2 3 .The base of A B G is 2cm so: A r e a = 2 1 × b r e a d t h × h e i g h t = 2 1 × 2 × 2 3 = 2 3 = 3 . 4 6 4 T o 4 d e c i m a l p l a c e s
Isn't the upload image feature great
Very easy simply adding the points A,E and B,D then we get ∆ ABG=(1/2)* rectangle ABDE . Again it is very simple that the height of the triangle is a√3 so the area of the triangle is (1/2) a a√3 sq unit so ar(∆ ABG)=3.4641 sq cm. Can't be it done like it!!!!!!!!!!
△ B C D −
InBy sine rule,
sin 1 2 0 B D = sin 3 0 B C
B D = 4 × sin 1 2 0
Area of Shaded △ = 2 1 × D E × B D
= 2 1 × 2 × 4 × sin 1 2 0
= 4 × 0 . 8 6 6
= 3 . 4 6 4
Did the same thing.
C D = A E = 1 , A B = 2 A D = 2 A C 2 − A D 2 = 2 3 ∆ A E F = 2 1 × A B × B F = 2 1 × 2 3 × 2 = 2 3 = 3 . 4 6 4
The apothem is calculated with the formula sqrt(side^2 - (side/2)^2)
sqrt(2^2 - (2/2)^2) = sqrt(3)
The height of the triangule is 2 * apothem => 2 sqrt(3)
The area of a triangle is calculated with the formula (base * height)/2
=> 2*2sqrt(3)/2 = 2 sqrt(3)
Ans: 2 sqrt(3)
Area of ABG = 1/2(AB)(h) join E and A and h=EA such that angle F meets EA at O, F is divided in two angles F= EFO+AFO=120 (Hexagon's angle) EFO=30 and AFO=90 EO=FE Cos30 , OA= FA Cos30 EO=2(0.866), OA=2(0.866) EO=1.732, OA=1.732 EA=EO+OA EA=3.464 A=1/2(h)(AB) A=1/2(3.464)(2) A=3.464 sq cm
A=1/2 bh; a=1cot 30=sqrt3 and h=2a Substituting, A=1/2 x 2 x 2sqrt3 =2x sqrt3=3.464sqcm.
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Area of △ A B G = 2 1 × A B × A E = 2 1 × 2 × 2 × A F × cos 3 0 ∘
= 2 1 × 2 × 2 × 2 × 2 3 = 2 3 = 3 . 4 6 4