Question

ABC is an equilateral triangle inscribed in a circle of radius 4 cm find the area between the circle and the triangle

Answer 1

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an equilateral triangle is inscribed in a circle of radius 4 cm. find the area of the part of the circle other than the part covered by the triangle.

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EXPERT ANSWERS

HALA718 | CERTIFIED EDUCATOR

The area between the circle and the triangle = area of the circle - area of the triangle.

Area of the circle = r^2*pi = 4^2*22/7 = 50.29

Area of the triangle:

Let ABC be a triangle, O is the center of the circle.

Let us connect between OA, OB, and OC

OA= OB= OC = r = 4

Now we have divided the triangle into three equl triangles.

Then the area of the triangle ABC = 3*area of the treiangle AOB.

The tiangle AOB is an isoscele. and the angle AOB = 120 degree, then angle OAB = angle OBA = 30 degree.

Let OD be perpindicular on AB ,

==> OA^2 = OD^2 +(AB/2)^2

==> 4^2 = OD^2 + AB^2/4

But sinA = OD/ OA

==> sin30 = OD / 4 = 1/2

==> OD = 2

==> 4^2 = 2^2 +AB^2/4

==> 16 = 4 +AB^2/4

==> 12 = AB^2/4

==> AB^2 = 48

==> AB = sqrt48 = 4sqrt3

Then area of the small triangle = (1/2)*4sqrt3*2 = 4sqrt3

The area of the big triangle = 3*4sqrt3 = 12sqrt3= 20.78

Answer 2

Heya!!!

Here's your answer friend,

The area of circle = πr²
= 22/7 x 4 x 4 = 352/7
= 50.48 cm²
Now area of equilateral triangle = √3a²/4
Draw perpendicular as shown in the figure..
from A to O on BC
therefore, the angle is bisected at angle A
and it divides the line BC into equal half
let BO = x = OC

Now, again
In triangle O'OB

tan 30° = opp/ adj
1/√3 = O'O/BO
1/√3 = 4/x

this implies x = 4 √3 units

therefore, BC = 2 x 4√3 = 8√3 units

Now we got one side of equilateral triangle
therefore now area of the triangle is √3a²/4
√ 3 x 8√3 x 8 √3 = 192√3

= 192 x 1.732
= 332.544

Now area remaining = area of triangle - area of circle = 332.544 - 50.48= 282.062 units...

Hope it helps you that O
which is centre is O' ok