Asked by Kate
The area of a regular pentagon with a perimeter of 35 units.
Join the centre to each vertex to form 5 eq
Join the centre to each vertex to form 5 equal triangles. Consider one of those triangles
the central angle is 72º and each of the other two equal angles is 54º, and the base is 7 units.
Can you find the height of that triangle using simple trig?
then the area of the triangle is 1/2 base*height
Multiply that answer by 5.
I am not sure what the formula is for the height in my Trig book it says H=a sin 0... but i don't know what a is equal to.
O never mind sorry i was making the problem to hard then what it was
I would use tangent
let the height be h
then tan 54=h/3.5
h=3.5*tan54
=4.817
the formula your book uses requires the hypotenuse of the right-angled triangle.
O i did sin 36/3.5 =sin54/c
then a= .5(7)(4.8)= 16.8 units^2
then times 16.8*5=84ft squared
good, I had the same answer (84.3)
Join the centre to each vertex to form 5 eq
Join the centre to each vertex to form 5 equal triangles. Consider one of those triangles
the central angle is 72º and each of the other two equal angles is 54º, and the base is 7 units.
Can you find the height of that triangle using simple trig?
then the area of the triangle is 1/2 base*height
Multiply that answer by 5.
I am not sure what the formula is for the height in my Trig book it says H=a sin 0... but i don't know what a is equal to.
O never mind sorry i was making the problem to hard then what it was
I would use tangent
let the height be h
then tan 54=h/3.5
h=3.5*tan54
=4.817
the formula your book uses requires the hypotenuse of the right-angled triangle.
O i did sin 36/3.5 =sin54/c
then a= .5(7)(4.8)= 16.8 units^2
then times 16.8*5=84ft squared
good, I had the same answer (84.3)
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