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BACK Complex Probems: 1.If the area of a square with side a s equal to the area of a triangle with base a, then the altitude of the triangle is sol: area of a square with side a = a ² sq unts area of a triangle with base a = ½ * a*h sq unts a ² =1/2 *a *h => h = 2a altitude of the triangle is 2a 2.An equilateral triangle is described on the diagonal of a square. What is the ratio of the area of the triangle to that of the square? Sol: area of a square = a ² sq cm length of the diagonal = √2a cm area of equilateral triangle with side √2a = √3/4 * (√2a) ² required ratio = √3a² : a ² = √3 : 2 3.The ratio of bases of two triangles is x:y and that of their areas is a:b. Then the ratio of their corresponding altitudes wll be sol: a/b =(½ * x*H) /(1/2 * y * h) bxH = ayh =>H/h =ay/bx Hence H:h = ay:bx 4 .A parallelogram has sides 30m and 14m and one of its diagonals is 40m long. Then its area is sol: let ABCD be the given parallelogram area of parallelogram ABCD = 2* (area of triangle ABC) now a = 30m, b = 14m and c = 40m s = ½(30+14+40) = 42m Area of triangle ABC = √[ s(s-a)(s-b)(s-c) = √(42*12*28*2 = 168sq m area of parallelogram ABCD = 2 *168 =336 sq m 5.If a parallelogram with area p, a triangle with area R and a triangle with area T are all constructed on the same base and all have the same altitude, then which of the following statements is false? Sol: let each have base = b and height = h then p = b*h, R = b*h and T = ½ * b*h so P = R, P = 2T and T = ½ R are all correct statements 6.If the diagonals of a rhombus are 24cm and 10cm the area and the perimeter of the rhombus are respectively. Sol: area = ½*diagonal 1 *diagonal 2= ½ * 24 * 10= 120 sq cm ½ * diagonal 1 = ½ * 24 = 12cm ½ * diagonal 2 = ½ *10 =5 cm side of a rhombus = (12) ² + (5) ² = 169 => AB = 13cm 7.If a square and a rhombus stand on the same base, then the ratio of the areas of the square and the rhombus is: sol: A square and a rhombus on the same base are equal in area 8.The area of a field in the shape of a trapezium measures 1440sq m. The perpendicular distance between its parallel sides is 24cm. If the ratio of the sides is 5:3, the length of the longer parallel side is: sol: area of field =1/2 *(5x+3x) *24 = 96x sq m 96x = 1440 => x = 1440 /96 = 15 hence, the length of longer parallel side = 5x = 75m 9.The area of a circle of radius 5 is numerically what percent its circumference? Sol: required percentage = (5)²/(2*5) *100 = 250% 10.A man runs round a circular field of radius 50m at the speed of 12m/hr. What is the time taken by the man to take twenty rounds of the field? Sol: speed = 12 k/h = 12 * 5/18 = 10/3 m/s distance covered = 20 * 2*22/7*50 = 44000/7m time taken = distance /speed = 44000/7 * 3/10 = 220/7min 11.A cow s tethered in the middle of a field with a 14feet long rope.If the cow grazes 100 sq feet per day, then approximately what time will be taken by the cow to graze the whole field? Sol: area of the field grazed = 22/7 * 14 * 14 = 616 sq feet 12.A wire can be bent in the form of a circle of radius 56cm. If it is bent in the form of a square, then its area will be sol: length of wire = 2 r = 2 *22/7 *56 = 352 cm side of the square = 352/4 = 88cm area of the square = 88*88 = 7744sq cm 13.The no of revolutions a wheel of diameter 40cm makes in traveling a distance of 176m is sol: distance covered in 1 revolution = 2 r = 2 *22/7 *20 = 880/7 cm required no of revolutions = 17600 *7/880 = 140 14.The wheel of a motorcycle 70cm in diameter makes 40 revolutions in every 10sec. What is the speed of motorcycle n km/hr? Sol: distance covered in 10sec = 2 *22/7 *35/100 *40 =88m distance covered in 1 sec =88/10m = 8.8m speed =8.8m/s = 8.8 * 18/5 *k/h = 31.68 k/h 15.Wheels of diameters 7cm and 14cm start rolling simultaneously from x & y which are 1980 cm apart towards each other in opposite directions. Both of them make the same number of revolutions per second. If both of them meet after 10seconds. The speed of the smaller wheel is sol: let each wheel make x revolutions per sec. Then (2 *7/2 *x)+(2 * 7*x)*10 = 1980 (22/7 *7 * x) + (2 * 22/7 *7 *x) = 198 66x = 198 => x = 3 distance moved by smaller wheel in 3 revolutions = 2 *22/7 *7/2 *3 = 66cm speed of smaller wheel = 66/3 m/s = 22m/s 16.A circular swimming pool is surrounded by a concrete wall 4ft wide. If the area of the concrete wall surrounding the pool is 11/25 that of the pool, then the radius of the pool is? Sol: let the radius of the pool be R ft radius of the pool including the wall = (R+4)ft area of the concrete wall = [(R+4)2 - R2 ] => = [R+4+R][R+4-R] = 8(R+2) sq feet 8(R+2) = 11/25 R2 => 11 R2 = 200 (R+2) Radius of the pool R = 20ft 17.A semicircular shaped window has diameter of 63cm. Its perimeter equals sol: perimeter of window = r +2r = [22/7 * 63/2 +63] = 99+63 = 162 cm 63.The area of the largest triangle that can be inscribed in a semicircle of radius is sol: required area = ½ * base * height = ½ * 2r * r = r 2 18.Three circles of radius 3.5cm are placed in such a way that each circle touches the other two. The area of the portion enclosed by the circles is sol: required area = (area of an equilateral triangle of side 7 cm) - (3 * area of sector with Ø = 6o degrees and r = 3.5cm) = ( √ ¾ * 7 * 7) – (3* 22/7 *3.5 *3.5*60/360 ) sq cm = 49√3/4 – 11*0.5*3.5 sq cm = 1.967 sq cm 19. Four circular cardboard pieces, each of radius 7cm are placed in such a way that each piece touches two other pieces. The area of the space encosed by the four pieces is sol: required area = 14*14 – (4 * ¼ * 22/7 * 7 *7) sq cm = 196 – 154 = 42 sq cm |