This problem actually has two interpretations.
I. After the circle is cut, the isosceles triangle fills up the gap. In this scenario you know the radius of the circle.
II. The circle is cut, a piece of length 3pi is taken, probably stretched in such a way that a forms a semicircle of another circle (actually it can be contorted in a variety of ways to form part of a new circle.) And an Isoceles triangle of height 5ft is fitted to it.
The phrase “ ¾ of a circle of radius two feet on top of a triangle of height 2 feet " Is it the same circle with the angle it subtended before being cut or are we allowed to change that angular form and superimpose it on the base of triangle? You can put ¾ of any circle of radius 2 feet on top of an infinite array of isosceles triangles of height 5 feet. If I is intended, then I believe the proper wording should have been “¾ of a circle of radius 2 ft is cut and the base of an isosceles triangle of height 5ft is used in its place”
These two solutions will actually differ. Under scenario I, B is the answer, but not under scenario two, when we don’t know the radius of the new circle.
The solution under I can be found by drawing radii on both ends of the base of the triangle from the center of the circle and vertically upwards. Then draw lines connecting the end points of the radii. You have another triangle. Divide 3pi by 2 and you find the length of one of the arcs. This length subtends a central angle. Whenever given the length of an arc in pi and you know the radius then use:
Arc Length = Central Angle x Radius ( The angle measure here is in radians)
3pi/2 = Central Angle x 2
So Central Angle = 3pi/4. If you want it in degrees multiply by 180/pi and you get 135 degrees. So you know the two arcs subtend 270 degrees and since there are only 360 degrees in any circle the other central angle is 90. The two sides of making this 90 degree angle are the two radii of length 2 and the base of the triangle is the hypotenuse. So the base is 2(2^.5 ). When you divide this by 2 you get one right triangle whose height is 5 and base is 2^.5. By Pythagoras this gives a hypotenuse of 27^.5. Since there are two such triangles, we have 2(27^.5) and simplifying this gives 6(3)^.5. Add this to the 3pi and you have your answer.
Under II , you don’t know the radius of the new circle.
