Maths Puzzles

 A rectangle is inscribed in a quadrant of a circle with the distances as shown in the diagram. Find out the length of diagonal AC.  Hint : the solution is much simpler than you think

Answer : 4 units  Area of a hexagon which is inscribed in a circle of radius r is 3 √3 / 2 * r 2  This hexagon has its side = radius of circle.  But the circumscribed hexagon has radius of the circle as its perpendicular to a side from center of circle.      As shown in the diagram, if the side of outer hexagon is s, then the radius from center of circle perpendicular to the side, cuts the side in half. So the radius forms a right angled triangle with three sides as r, s/2 and hypotenuse as s. Using Pythagoras theorem s 2  = r 2   +s 2  /4 3s 2  /4 = r 2  s 2   =r 2  * 4/3 Now replacing this value for the area of outer hexagon, we get area =  3 √3 / 2 *s 2   =  3 √3 / 2* r 2  * 4/3 But area of inner hexagon is 3 =>    3 √3 / 2* r 2     is 3 Area of circumscribed hexagon = 3 * 4/3 = 4 units. ...

 2 Regular hexagons are inscribed and circumscribed in a circle. If the area of inscribed hexagon is 3 units, what is the area of circumscribed hexagon?  

Answer : 96 cm   The original width of 9cm. After 3 invocations of magic spell the width of the belt becomes 9/3 , 3/3, 1/3  = 1/3cm The final area is 4 Therefore final length is 4/width = 4/1/3 = 12 cm. The length before 3rd invocation is 12*2 = 24 cm Length of the belt before 2nd invocation is 24*2 = 48 cm. Length of the belt before 1st invocation is 48*2 = 96 cm.     

There is a magic belt. When ever the wizard utters a spell its length reduces to half its original length and width reduces to 1/3 of its original width.  The original width was 9 cm. After the wizard invoked the spell 3 times, the area of the belt now is 4 cm 2  . What was the original length of the belt?