~F.2 Maths X2

1. 先將個五邊形係正中間cut開做5個相等嘅三角形 (△AOB)
Let y 做三角形係中間隻角 (∠AOB)
5y=360°(∠s at a pt.)
y=72°

將個三角形垂直切開2均等份 (△AOP)
Let z 做三角形係中間隻角 (∠AOP)
z=72°/2
z=36°

AP=AB/2
AP=6/2
AP=3 (cm)

AP/OP=tan36°
3/OP=tan36°
OP=3/tan36° [留返呢個數黎計會準確d]

Area of the pentagon:
(AB*OP/2)*5
=6*(3/tan36°)/2*5
=61.9(cm2) (corr. to 3 sig. fig.)

無圖好難表達
希望你明啦

1.
if you cut the regular pentagon into a half, there will be two trapeziums and it can be cut into 3 equilateral triangles.
the upper base of the trapezium will be 6cm and the lower base is 3*6cm
the x be the high of the triangle.
x(2)+3(2)=6(2) *number in ( )is the power of the number
x=√27
x=5.20(corr. to 3 sig. fig.)
the area=(6+3*6)*5.2
=125cm2(corr. to 3 sig. fig.)

no.2's is not clear enough to answer.

1)
let ABCDE be vertex of pentagon, O be centre
angle OAB = 108 / 2 = 54 deg. (五邊形內角為 108 度,一半為 54 度)
OA = 3 / cos54 (AB既中心點劃條直線至點 O)
area of triangle OAB = (1/2)(6)(OA)sin54
area of pentagon
= (5)(area of triangle OAB)
= (5)(1/2)(6)(3 / cos54)sin54
= 61.9 cm^2

2)
係咪打少錯,問得唔清唔楚,唔明

1)a regular pentagon of side 6 cm can be divided to 6 identical triangle
area of each triangle
=6x6√3/4
=9√3cm^2
the area of regular pentagon =6x9√3=54√3cm^2
2)it has not enough infomation to tell thre radius of OCD