急!急! F4 Properties Circles q13

17.
(a)
arc BCA = arc CAD (given)
arc BC + arc CA = arc CA + arc AD
Then, arc BC = arc AD
∠DBE = ∠BDE (equal arcs subtendequal ∠s at circum.)

In ΔEDB : ∠DBE + ∠BDE + ∠BED = 180° (∠ sum of Δ)
∠DBE + ∠DBE + 74° = 180°
∠DBE = 53°

(b)
Join AC.

It is proven that : arc BC = arc AD
∠ECA = ∠EAC (equal arcs subtend equal ∠s at circum.)
In ΔEAC: AE = EC (equal ∠s to equal sides)
AE = 5 cm


18.
(a)
Join OC and OB.

arc AB = arc BC = arc CD (given)
∠AOB = ∠BOC = ∠COD (equal arcs subtendequal ∠s at centre)
But ∠AOB + ∠BOC + ∠COD = 90° (given)
Then ∠AOB = ∠BOC = ∠COD = 30°
∠AOC = ∠AOB + ∠BOC = 60°

OA = OC (radii of the same circle)
In ΔAOC: ∠OAE = ∠OCA (equal sides to equal ∠)
In ΔAOC: ∠OAE + ∠OCA + ∠AOC = 180° (∠ sum of Δ)
∠OAE + ∠OAE + 60° = 180°
∠OAE = 60°

(b)
∠OAE = 60°
Similarly ∠ODE = 60°
In quad. OAED: ∠AED + ∠OAE + ∠AOD + ∠ODE = 360°
∠AED + 60° + 90° + 60° = 360°
∠AED = 150°

(c)
arc AB = arc CD (given)
∠AOB = ∠COD (equal arcs subtendequal ∠s at centre)
∠AOB + ∠BOC = ∠COD + ∠BOC
Then, ∠AOC = ∠BOD
OA = OB (radii of same circle)
OC = OD (radii of same circle)
ΔAOC ≡ ΔBOD (SAS)
AC = BD (corr. sides of cong. Δ)
AC = 7 cm


34.
Denote AB as the horizontal ceiling, and O as the centre.
Join XO, and XO cuts AB at Y.
Let r m be the radius of the circle.

In ΔOYA:
OA² = AY² + OY² (Pythagorean theorem)
r² = (7/2)² + (r - 1.6)²
r² = 12.25 + r² - 3.2r + 2.56
3.2r = 14.81
r = 14.81/3.2
2r = 2 * (14.81/3.2)
2r = 9.3
Hence, diameter = 9.3 m