麻煩各位師兄幫幫小弟 ~
諗咗好耐都諗唔到點計 ..
唔好意思幅圖好蒙 ..


In the figure, a regular pentagon with side 12 cm is inscribed in a circle with centre O.
(a) Find the radius of the circle.
(b) Find the area of the pentagon.
(c) Hence find the area of the shaded region.

a part 我唔知揾完一隻角幾多度之後應該點計
btw b part 應該要用到 " 1/2 ab sin C = Area of triangle " 呢條 formula
thz ~

JN568@80K :a part 我唔知揾完一隻角幾多度之後應該點計

(a)

∠AOB = 360°/5 = 72°
∵ BO = AO
∴ ∠BAO = ∠ABO
∠ABO + ∠BAO + ∠AOB = 180°
2∠ABO + 72° = 180°
∠ABO = 54°

AO / sin 54° = 12 cm / sin 72°
AO = 10.2 cm (corr. to 3 sig. fig.)

∴ The radius of the circle is 10.2 cm.

---

(b)

　Area of △ABO
= 1/2(12cm)(BO)sin54°
= 49.5497 … cm²

∴ Area of the pentagon
= 49.5497 … cm² * 5
= 247.75 cm² (corr. to 2 d.p.)

(P.S. 如計錯請指)

JY4066~HT1391~JD7408~LOVEMUSIC :

JN568@80K :a part 我唔知揾完一隻角幾多度之後應該點計

(a)

∠AOB = 360°/5 = 72°
∵ BO = AO
∴ ∠BAO = ∠ABO
∠ABO + ∠BAO + ∠AOB = 180°
2∠ABO + 72° = 180°
∠ABO = 54°

AO / sin 54° = 12 cm / sin 72°
AO = 10.2 cm (corr. to 3 sig. fig.)

∴ The radius of the circle is 10.2 cm.

---

(b)

　Area of △ABO
= 1/2(12cm)(BO)sin54°
= 49.5497 … cm²

∴ Area of the pentagon
= 49.5497 … * 5
= 247.75 cm² (corr. to 2 d.p.)

(P.S. 如計錯請指)

我有兩題問題想問下 ~
1. 點解 ∠AOB = 360°/5 ? 係咪因為 ∠ at a point ? 洗唔洗係前面補充番 ?
2. AO / sin 54° = 12 cm / sin 72° 係咪用咗 sine formula ?
thz a lot

JN568@80K :

JY4066~HT1391~JD7408~LOVEMUSIC :(a)

∠AOB = 360°/5 = 72°
∵ BO = AO
∴ ∠BAO = ∠ABO
∠ABO + ∠BAO + ∠AOB = 180°
2∠ABO + 72° = 180°
∠ABO = 54°

AO / sin 54° = 12 cm / sin 72°
AO = 10.2 cm (corr. to 3 sig. fig.)

∴ The radius of the circle is 10.2 cm.

我有兩題問題想問下 ~
1. 點解 ∠AOB = 360°/5 ? 係咪因為 ∠ at a point ? 洗唔洗係前面補充番 ?
2. AO / sin 54° = 12 cm / sin 72° 係咪用咗 sine formula ?
thz a lot

1. (我慣咗做公開試係skip steps... 補番先)
∵ AB = BC = CD = DE = EA (given)
∴ ∠AOB = ∠BOC = ∠COD = ∠DOE = ∠EOA (eq. chords, eq. ∠s)
∠AOB * 5 = 360° (∠s at a pt.)
∠AOB = 72°

---

2. 係

唔想用sine formula就咁樣計：

：隱藏內容，按一下這裏顯示

(加番晒reasons先..)

Let F be the mid-point of A and B such that AB⊥FO.

BF = 12cm / 2 = 6cm

∠AOB = 72°
∵ BO = AO (radii)
∴ ∠BAO = ∠ABO (base ∠s, isos. △)
∠ABO + ∠BAO + ∠AOB = 180° (∠ sum of △)
2∠ABO + 72° = 180°
∠ABO = 54°

In △BFO,
cos 54° = 6 / BO
BO = 10.2 cm (corr. to 3 sig. fig.)

∴ The radius of the circle is 10.2 cm.