The figures below shows two intersecting circles with centre A and B. <CBD =900
and BC = BD =10cm
<CAD = 720
and line CD is a common chord.
Calculate
i.
The length of common chord CD.
ii.
The radius AC to the nearest whole number.
iii.
The area of intersection of the two circles that is shaded.
To solve this problem, we can start by labeling the given information on the diagram. Let's call the point of intersection of the two circles point C and the intersection of the common chord CD with circle A point E.
i. The length of common chord CD:
Since BC = BD = 10 cm, triangle BCD is isosceles. Therefore, angle CBD = angle BCD = 90/2 = 45 degrees.
Since triangle CAD is isosceles with angle CAD = angle CDA = 72 degrees, we can find angle ECD by subtracting angle CAD from angle BCD:
angle ECD = angle BCD - angle CAD = 45 - 72 = -27 degrees.
Since triangle CDE is a triangle, the sum of its angles must be 180 degrees. Therefore, we can find angle CED:
angle CED = 180 - (-27) = 207 degrees.
Now, we can use the Law of Cosines to find the length of CD.
In triangle CDE, CD^2 = CE^2 + DE^2 - 2(CE)(DE)cos(angleCED).
Since CE = DE (as they are both radii), we have:
CD^2 = 2(CE^2) - 2(CE^2)cos(angleCED).
Since angle CED = 207 degrees, we can substitute the values into the equation:
CD^2 = 2(10^2) - 2(10^2)cos(207).
Using a calculator, we find:
CD^2 ≈ 69.611
Therefore, the length of common chord CD is approximately sqrt(69.611) = 8.34 cm.
ii. The radius AC to the nearest whole number:
Since triangle CAD is isosceles, we can use the Law of Cosines to find AC.
In triangle CAD, AC^2 = CD^2 + AD^2 - 2(CD)(AD)cos(angleCAD).
Since CD = 8.34 cm and angle CAD = 72 degrees, we can substitute the values into the equation:
AC^2 = (8.34)^2 + 10^2 - 2(8.34)(10)cos(72).
Using a calculator, we find:
AC^2 ≈ 153.919
Therefore, the radius AC to the nearest whole number is approximately sqrt(153.919) ≈ 12 cm.
iii. The area of the intersection of the two circles that is shaded:
To find the area of the shaded region, we need to find the area of the sector of circle A formed by angle CAD and subtract the area of triangle CAD.
The formula for the area of a sector is given by: A = r^2(angle/360).
Using the given information, the radius of circle A is AC ≈ 12 cm and the angle of sector CAD is 72 degrees. Therefore, the area of the sector CAD is:
A_sector = (12)^2(72/360) = 8.64 cm^2.
To find the area of triangle CAD, we can use Heron's formula since we know all three side lengths:
s = (CD + AD + AC)/2 = (8.34 + 10 + 12)/2 = 30.34/2 = 15.17 cm.
Area_triangle = sqrt(s(s-CD)(s-AD)(s-AC)) ≈ sqrt(15.17(15.17-8.34)(15.17-10)(15.17-12)) ≈ √(244.697815) ≈ 15.66 cm^2.
Therefore, the area of the shaded region is approximately 8.64 - 15.66 = -7.02 cm^2. However, since the area cannot be negative, we can conclude that there is no shaded area in the intersection of the two circles.