calculate the area of minor segment of circle of radius 5cm if the deggre measure of the arc of the segment 80degree and the lenght of the chord is 8cm
Answers
Answer:
Answer: 34.364761 sq.cm (to 6 dp)
Step-by-step explanation:
We want to find the area of the segment, which is the shaded part.
D is the mid point of the line AB.
From the question:
The radius is 5 cm, thus OA=OB=5 cm
The chord length is 8 cm, thus AB=8 cm and AD=BD=4 cm
The angle ODA is a right angle, thus, from Pythagoras, we have:
OA2=OD2+AD2⇒OD2=OA2−AD2
=52 cm2−42 cm2=25 cm2−16 cm2=9 cm2
⇒OD=3 cm
The area of the triangle OAB is half the length (AB) multiplied by the height (OD), thus:
Area=0.5×8 cm×3 cm=12 cm2
From the question, the angle AOB is 80 degrees, which means that the angle DOB is 40 degrees and thus the angle OBD is 50 degrees. As the angle DOB is less than the angle OBD, the sine of angle DOB is less than the sine of the angle OBD.
sin(∠DOB)=DBOB=45=0.8
sin(∠OBD)=DOOB=35=0.6
So, your question is telling us that 0.8<0.6
This is utter rubbish. You’ve obviously got something wrong in the question. I’m going to assume it’s the angle (which saves me having to redo my earlier calculations).
Angle AOB is twice the angle DOB
=2arctan(BDDO)=2arctan(43)
≈106.260205∘≈1.85459044 radians
Area of the Sector OAB is the angle AOB (in radians) multiplied by the square of the radius
≈1.85459044×(5 cm)2≈46.3647609 cm2
Area of Segment is simply the Area of the Sector OAB less the Area of the Triangle OAB
≈46.3647609 cm2−12 cm2≈=34.3647609 cm2
Answer: 34.364761 sq.cm (to 6 dp)
MARK BRANIEST