Page 245 - ICSE Math 7
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10.  ABCD is a diameter of a circle of radius 6 cm. The lengths AB,
                        BC and CD are equal. Semicircles are drawn on AB and BD as           A         B      C       D
                        diameters as shown in the given figure. Find the area of the shaded
                        region. (Take p = 3.14)







                     11.  Two  circles  are  drawn  inside  a  big  circle  of  diameter  24  cm. The
                                                         1      2
                        diameter of the two circles are   and   of the diameter of the big
                                                         3      3
                        circle as shown in the adjoining figure. Find the ratio of the areas of
                        the shaded part to the unshaded part of the circle.


                     12.  The adjoining figure represents a rectangular lawn with a
                        circular flowerbed in the middle. Find:
                                                                                                     2 m
                         (a)  the area of the whole land.                                                            5 m
                         (b)  the area of the flowerbed.

                         (c)    the area of the lawn excluding the area of the flowerbed.
                         (d)  the circumference of the flowerbed.                                  10 m


                                                              AT A GLANCE

                    ¾   Perimeter is the length of the boundary of any closed plane figure.
                    ¾   Area is the surface or region enclosed inside a closed boundary.
                    ¾  (a)  Perimeter of a rectangle = 2 × (length + breadth)
                        (b)  Perimeter of a square = 4 × side

                        (c)  Area of a rectangle = length × breadth
                        (d)  Area of a square = side × side
                                                  1
                    ¾   (a)  Area of a triangle =   × base × height
                                                  2
                        (b)  Area of a parallelogram = base × altitude

                    ¾   (a)  Circumference of a circle 2pr = p × d
                        (b)  Area of a circle = pr 2

                    ¾   (a)  Area between two rectangles                                      h                       b
                             = Area of outer rectangle – area of inner rectangle
                             =     ab – (a – 2h)(b – 2h)                                                  a
                        (b)  Area of a circular ring                               r

                             = Area of outer circle – area of inner circle        O    R
                                  2
                             =     pR  – pr 2


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