Let the radius of the park be r, then?r + 2r = 288(? + 2)r = 288? [(22/7) + 2]r = 288? r = (288 x 7)/36 = 56 ? Area of the park = (1/2)?r2= (1/2) x (22/7) x 56 x 56= 4928
Let the side of the square(ABCD) be x metres. Then, AB + BC = 2x metres. AC = √2x = (1.41x) m. Saving on 2x metres = (0.59x) m. Saving % = ❨ 0.59x x 100 ❩% = 30%
According to the question, Area of semi- circle = 77 m(1/2) x ? x r2 = 77? r2 = (77 x 2 x 7)/22? r = 7m ?Circumference of semi- circle =?r + 2r= ( ? + 2)r = [(22/7) + 2] x 7 = 36 m
Area of the room=(544 * 374) size of largest square tile= H.C.F of 544 & 374 = 34 cm Area of 1 tile = (34 x 34) c m 2 Number of tiles required== [(544 x 374) / (34 x 34)] = 176
EXAMIANS