We can discover the area using trigonometry. Given that the radii of the circles are the exact same, we can deduce that the shaded location in the circles are the same, so to solve the difficulty, we will find the shaded location in among the circles and also double our answer.First, we know that the radius can be drawn from the facility to any suggest on the edge of the circle. Because of this we deserve to produce an isosceles triangle by drawing the radius from the center to the various other finish of the chord of size 4 (view attached photo for triangle). Next off, we deserve to discover the angle created at the center of the circle using a trigonometry attribute. We understand that the chord opposite of the angle is 4 lengthy and also that both nearby sides are 4 lengthy. If we draw the amplitude of the triangle which bisects the center angle, we will certainly have 2 appropriate triangles. The side opposite to the new center angle is 2 long. Now we have the right to usage the sine feature to uncover the angle:
If half of the facility angle is 30 degrees, then we understand that the angle between the radii is 60 degrees. Due to the fact that a full circle is 360 degrees, we recognize that we are taking care of one-6th of the circle.Now we should find the location of one-sixth of the circle and also subtract the location of the triangle to discover the shaded area.To find the area of the triangle, we can use one-fifty percent base times elevation, yet first we have to find the height. We deserve to usage the tangent attribute to discover it:
So the location of the shaded area is:
So multiply that by two to acquire the area for both circles and also the last answer is:A = 2.8988 

5 & 1/3 pi - 8 sqrt 3

Step-by-step explanation:

1) Find the location of one circle: A = pi * r^2 > A = 16 pi.

You are watching: Find the area of the shaded portion intersecting between the two circles.

2) Draw or imagine one more radius drawn with its endsuggest on the endpoint of the chord. This creates an equilateral triangle. This additionally means that each angle of the triangle has actually a meacertain of 60 levels. So, considering that the triangle is part of the sector, the meacertain of the sector"s arc is 60 degrees.

3) With this information, you have the right to discover out the area of the sector: carry out 60/360 = 1/6. Then multiply by the location of the circle:

1/6 * 16 pi = 2 & 2/3 pi. So the area of the sector is 2 & 2/3 pi.

4) Then discover the area of the equilateral triangle: A = 1/4 * s^2 * sqrt 3 > A = 4 sqrt 3. So the area of the triangle is 4 sqrt 3.

5) Finally, discover the location of the segment or one of the shaded components. You would certainly carry out this by doing the location of the sector minus the location of the triangle > 2 & 2/3 pi - 4 sqrt 3. You also must multiply this by 2 since tright here are 2 segments that are component of the shaded percent. So your last exact answer would certainly be: 5 & 1/3 pi - 8 sqrt 3. The approximate answer would be 2.90.

Hope this helps! :)

Answer from: DIGlBICK9402

A=16/3 pi - 8Г3

Step-by-step explanation:

Answer from: rhettnyah

Area shaded percent = 16/3 π - 8√3

Step-by-action explanation:

The shaded percentage consists of 2 equal segment

∵ Two circles have the very same radii = 4

∵ The the size of the prevalent chord of the 2 circles = 4

∴ The central angle of each segment = π/3 (60° equilateral Δ)

∵ Area segment = area sector - area Δ

∵ Area sector = 1/2 r²Ф = 1/2 × (4)² × π/3 = 8/3 π

∵ Area Δ = 1/4 s² √3 = 1/4 × (4)² × √3 = 4√3

∴ Area segment = 8/3 π - 4√3

∴ Area shaded percentage = 2(8/3 π - 4√3) = 16/3 π - 8√3


Join other finish of chord from the center of 2 circles.

As these two circles are concentric circles .

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And the triangles are equilateral triangle.

Area of sector



Area of equilateral triangle having actually side 4 cm=


Area of area I = Area of sector - Area of equilateral triangle having actually side 4 cm



Area of colored area = 2 × Area of area I

= 2 × 1.461

= 2.922 cm² (Approx)


Answer from: kierraware04
Are you going to include a picture for us to see?
Answer from: TH3L0N3W0LF
Look at the photo.ΔABC and ΔBDC are equilateral triangles.Area of sector of a circle:
Area of the triangle ABC:
Area of the shaded portion:

So let"s try to discover the part of it in one circle then multiply it by 2.if the vertical line of shaded reason is 4, then 2 would certainly form a rt triangle with radius. thus the cos € = 2/4 = 1/2, so € = arccos (1/2)€ = 60°, and also 2× that = 120°so now one component of the shaded region is the entirety sector (pi×r^2) - the triangle within (base×ht): - <2(4cos30)> = <16pi×1/3> - <2×3.46> = 16.76 - 6.93 = 9.83currently double that for both parts of the shaded region: 9.83×2 = 19.65 sq. units