i. Line BD = 6√2 cm
ii. Perimeter = 3 (π + 4) cm
iii. Area = 18 ( π - 2) cm
How to determine the parameters
Let the length of BD = x
Since B is the midpoint of the arc AC, then ∠BOC = 15°
In ΔBDB
Let's use the trigonometric ration, tan
[tex]Tan 45 = \frac{BD}{CD}[/tex]
CD = x
Then, using the Pythagorean theorem
[tex]x^2 + x^2 = 12^2[/tex]
[tex]2x^2 = 144[/tex]
[tex]x^2 = \frac{144}{2}[/tex]
[tex]x = \sqrt{72}[/tex]
[tex]x = \sqrt{36 *2}[/tex]
[tex]x = 6\sqrt{2}[/tex]
Thus, line BD is [tex]6\sqrt{2}[/tex] cm
ii. Perimeter = BC + BD + CD
= 2πr/8 + x + r - x
= π(12)/4 + r
= 3π + 12
= 3 ( π + 4) cm
iii. Area of shaded area = Area of BOC + Area of BOD
⇒ πr^2/8 × 1/2x^2
= π144/8 × 1/2(72)
= 18π - 36
= 18 ( π - 2) cm²
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