The perimeter of the polygon is 53.6 units.
Solution:
The reference image to the answer is attached below.
AP = 8, BQ = 7.8 and CR = 11
AP and AR are tangents to a circle from an external point A.
BP and BA are tangents to a circle from an external point B.
CQ and CR are tangents to a circle from an external point C.
<em>Tangents drawn from an external point to a circle are equal in length.</em>
⇒ AP = AR, BP = BQ and CQ = CR
AR = 8
BP = BQ
⇒ BP = 7.8
CQ = CR
⇒ CQ = 11
Perimeter of the polygon = AP + BP + BQ + CQ + CR + AR
= 8 + 7.8 + 7.8 + 11 + 11 + 8
= 53.6
The perimeter of the polygon is 53.6 units.
Hi! Let me help you!
This problem is a Venn Diagram question. It is much easier to answer this question with the visual it usually comes with. However, we can employ logic to give this one a smart and logical answer.
What we have so far:
P(A) = 25. This means that the shaded area of A is equal to 25.
P(B/A) = 920. This means that area A subtracted 25 to the value of area B which resulted to 920.
P(B) = 945. How did I arrive with 945? Simple, I just added 25 to 920. Go and be a smart lad and find why I did that.
Solution:
Solving P(A∩B) is quite challenging because like I said, a visual is needed for us to give it a pefect answer. Employing logic and using what we have so far:
If P(A∩B) means A and B intersection, we can assume that it is simply as A + B.
If that is the case, then we can simply pull out our P(A) and P(B), 25 and 945 respectively and add them:
P(A) + P(B) = P(A∩B)
P(A∩B) = 25 + 945
P(A∩B) = 970 <----- What we are looking for
Therefore, P(A∩B) = 970!
Given that the length's of the secant's are equal, it implies that the distance of a perpendicular bisector from the secant to the center of the circle will also be the same. This implies that the value of x will be given by:
x=16
Answer:
I think answer is B Altitude
