The area of a quadrant of a circle with radius 28 cm is ____ cm2
Answers
Answer:
In △ BAC, by pythagoras theorem,
BC
2
=AB
2
+AC
2
BC
2
=784+784
BC=28
2
cm
2
BC
=14
2
cm
Now,
A=ar(BDCEB)
A=ar(BCEB)−ar(BCDB)
A=ar(BCEB)−[ar(BACDB)−ar(BAC)]
A=[
2
1
(
7
22
×(14
2
)
2
)−(
4
1
×
7
22
×28
2
−
2
1
×28×28)]
A=[
2
1
×
7
22
×196×2−
4
1
×
7
22
×28×28+
2
1
×28×28]
A=616−616+392=392 cm
2
Answer:
The area of a quadrant of a circle having a radius of 28 cm is 616 cm².
Step-by-step explanation:
A circle is a particular type of ellipse when the eccentricity is zero and both foci are present. The locus of points drawn equally apart from the centre is also referred to as a circle. The radius of a circle is the separation between its centre and its periphery. The circle's diameter, which is equal to twice its radius, is the line that splits the circle into two equal pieces.
The quadrant of a circle is one-fourth of the circle.
Given,
r = 28 cm
A = πr²
The area of the quadrant of a circle is one-fourth i.e., AQ = A/4
AQ = πr²/4
AQ = π×28²/4
AQ = 616 cm²
Hence, the area of a quadrant of a circle having a radius of 28 cm is 616 cm².
To learn more about the quadrant, click on the link below:
https://brainly.in/question/85794
To know more about the circle, click on the link below:
https://brainly.in/question/252726
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