Quote:
Originally Posted by 80199 3 Marks: --/1
Two identical circles intersect so that their centers, and the points at which they intersect, form a square of side 1 cm. The area in sq. cm of the portion that is common to the two circles is:
Choose one answer.
a. p/4 b. p/2 -1 c. p/5 d. p/4 e. √2 -1
plz help me solving this |
Let O and O' be the centers of the two identical triangles and A and B be the intersecting points. Now OO'AB is a unit square.
Now the area common to the circles is obtained by subtracting the area of the triangle OAB with the area of the sector OAB in the circles..
Area of the sector OAB = pi/4
area of the triangle OAB = 1/2
so common area is = [pi/4 -1/2]*2 ... ( it is mutiplied by 2 since we get only one of the common area after subtracting)
so area is [pi/2 - 1] .. option b