Therefore, m∠BD is twice the m ∠x, which we know equals 28^(∘).
m∠BD=2(28) ⇔ m∠BD = 56
Finding m BD
Let's pay close attention to the arcs whose AD and BD.
An arc whose endpoints are the endpoints of a diameter has a measure of 180^(∘). Therefore, by the Arc Addition Postulate the sum of a and 56^(∘) is equal to 180.
mAD+56^(∘)=180^(∘) ⇔ mAD=124^(∘)
b
The area of a circle with radius r is calculated using the formula below.
A=Ï€ r^2
Therefore, we want to find the radius of C, which we can do by first finding its diameter.
Finding The Diameter
One of the corollaries of the Inscribed Angle Theorem says that an angle inscribed in a semicircle is a right angle. Therefore, m ∠ADB = 90^(∘). Let's also add the given lengths to the diagram.
Notice that â–³ ADB is a right triangle. Therefore, we can find the length AB using the Pythagorean Theorem
AD^2 + BD^2 = AB^2
Let's substitute the known lengths to find the length AB.
c We know that the radius of C is 8 units and mBD = 100^(∘). We want to find the value of BD from the following diagram.
To do so, let's add the line segment CD and the measure of BD to the diagram. Also, since the radius of C is 8 units, both BC and CD are 8 units long.
Since mBD = 100^(∘), we have that m∠BCD = 100^(∘). We can use the Law of Cosines to write the equation for BD in terms of BC, CD, and m ∠BCD.
BD^2 = BC^2 + CD^2 - 2 BC * CD * cos m ∠BCD
Let's substitute 8 for BC and CD and 100^(∘) for m∠BCD in the above equation. Then, we will solve the resulting equation for BD.
