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BACK Medium Problems: 11.Find the ratio of the areas of the incircle and circumcircle of a square. Sol: let the side of the square be x, then its diagonal = √2 x radius of incircle = x/2 and radius of circmcircle =√2 x /2 = x/√2 required ratio = x²/4 : x²/2 = ¼ : ½ = 1:2 12.If the radius of a circle is decreased by 50% , find the percentage decrease in its area. Sol: let original radius = r and new radius = 50/100 r = r/2 original area = r² and new area = (r/2)² decrease in area = 3 r²/4 * 1/ r² * 100 = 75% 13.Two concentric circles form a ring. The inner and outer circumference of the ring are 352/7 m and 528/7m respectively. Find the width of the ring. sol: let the inner and outer radii be r and R meters then, 2r = 352/7 => r = 352/7 * 7/22 * ½ = 8m 2R = 528/7 => R= 528/7 * 7/22 * ½ = 12m width of the ring = R-r = 12-8 = 4m 14.If the diagonal of a rectangle is 17cm long and its perimeter is 46 cm. Find the area of the rectangle. sol: let length = x and breadth = y then 2(x+y) = 46 => x+y = 23 x²+y² = 17² = 289 now (x+y)² = 23² =>x²+y²+2xy= 529 289+ 2xy = 529 => xy = 120 area =xy=120 sq. cm 15.A rectangular grassy plot 110m by 65cm has a gravel path .5cm wide all round it on the inside. Find the cost of gravelling the path at 80 paise per sq.mt sol: area of theplot = 110 * 65 = 7150 sq m area of the plot excluding the path = (110-5)* (65-5) = 6300 sq m area of the path = 7150- 6300 =850 sq m cost of gravelling the path = 850 * 80/100 = 680 Rs 16. The perimeters of ttwo squares are 40cm and 32 cm. Find the perimeter of a third square whose area is equal to the difference of the areas of the two squares. sol: side of first square = 40/4 =10cm side of second square = 32/4 = 8cm area of third squre = 10² – 8² = 36 sq cm side of third square = √36 = 6 cm required perimeter = 6*4 = 24cm 17. A room 5m 44cm long and 3m 74cm broad is to be paved with squre tiles. Find the least number of squre tiles required to cover the floor. sol: area of the room = 544 * 374 sq cm size of largest square tile = H.C.F of 544cm and 374cm= 34cm area of 1 tile = 34*34 sq cm no. of tiles required = (544*374) / (34 * 34) = 176 18. The diagonals of two squares are in the ratio of 2:5. Find the ratio of their areas. sol: let the diagonals of the squares be 2x and 5x respectively ratio of their areas = ½ * (2x)² : ½*(5x)² = 4:25 19.If each side of a square is increased by 25%. Find the percentage change in its area. sol: let each side of the square be a then area = a ² new side = 125a/100 = 5a/4 new area = (5a/4)² = 25/16 a² increase in area = (25/16)a² - a² = (9/16)a² increase % = (9/16)a² * (1/a²) * 100 = 56.25% 20.The base of triangular field os three times its altitude. If the cost of cultivating the field at Rs. 24.68 per hectare be Rs. 333.18. Find its base and height. sol: area of the field = total cost/ rate = 333.18 /24.68 = 13.5 hectares => = 13.5 * 10000 = 135000 sq m let the altitude = x mt and base = 3x mt then ½ *3x * x = 135000 => x² = 90000 => x = 300 base= 900m and altitude = 300m 21.In two triangles the ratio of the areas is 4:3 and the ratio of their heights is 3:4. Find the ratio of their bases? Sol: let the bases of the two triangles be x &y and their heights be 3h and 4h respectively. (1/2*x*3h)/(1/2*y*4h) =4/3 => x/y = 4/3 *4/3 = 16/9 22.Find the length of a rope by which a cow must be tethered in order that it may be able to graze an area of 9856 sq meters. Sol: clearly the cow will graze a circular field of area 9856 sq m and radius equal to the length of the rope. Let the length of the rope be r mts then r²=9856 => r²=9856*7/22 = 3136 => r=56m 23.The diameter of the driving wheel of a bus is 140cm. How many revolutions per minute must the wheel make inorder to keep a speed of 66 kmph? Sol: Distance to be covered in 1min = (66*1000)/60 m =1100m diameter = 140cm => radius = r =0.7m circumference of the wheel = 2*22/7*0.7 = 4.4m no of revolutions per minute = 1100/4.4 = 250 24.The inner circumference of a circular race track, 14m wide is 440m. Find the radius of the outer circle. Sol: let inner radius be r meters. Then 2r =440 => r=440*7/22*1/2 = 70m radius of outer circle = 70+4 =84m 25.A sector of 120 degrees, cut out from a circle, has an area of 66/7 sq cm. Find the radius of the circle. Sol: let the radius of the circle be r cm. Then r²ø/360 =66/7=> 22/7*r²*120/360 = 66/7 =>r² = 66/7 *7/22*3 =9 radius = 3cm 26.The length of the room is 5.5m and width is 3.75m. Find the cost of paving the floor by slabs at the rate of Rs.800 per sq meter. Sol: l=5.5m w=3.75m area of the floor = 5.5 * 3.75 = 20.625 sq m cost of paving = 800 *20.625 =Rs. 16500 27.A rectangular plot measuring 90 meters by 50 meters is to be enclosed by wire fencing. If the poles of the fence are kept 5 meters apart. How many poles will be needed? Sol: perimeter of the plot = 2(90+50) = 280m no of poles =280/5 =56m 28.The length of a rectangular plot is 20 meters more than its breadth. If the cost of fencing the plot @ 26.50 per meter is Rs. 5300. What is the length of the plot in meter? Sol: let breadth =x then length = x+20 perimeter = 5300/26.50 =200m 2(x+20+x) =200 => 4x+40 =200 x = 40 and length = 40+20 = 60m 29.A rectangular field is to be fenced on three sides leaving a side of 20 feet uncovered. If the area of the field is 680 sq feet, how many feet of fencing will be required? Sol: l=20feet and l*b=680 => b= 680/20 = 34feet length of fencing = l+2b = 20+68 =88 feet 30.A rectangular paper when folded into two congruent parts had a perimeter of 34cm foer each part folded along one set of sides and the same is 38cm. When folded along the other set of sides. What is the area of the paper? Sol: when folded along the breadth we have 2(l/2 +b) = 34 or l+2b = 34...........(1) when folded along the length, we have 2(l+b/2)=38 or 2l+b =38............(2) from 1 &2 we get l=14 and b=10 Area of the paper = 14*10 = 140 sq cm 31.A took 15 seconds to cross a rectangular field diagonally walking at the rate of 52 m/min and B took the same time to cross the same field along its sides walking at the rate of 68m/min. The area of the field is? Sol: length of the diagonal = 52*15/60 =13m sum of length and breadth = 68*15/60 = 17m √(l²+b²)=13 or l+b = 17 area =lb = ½ (2lb) = ½[(l+b)² – (l²+b²)] = ½[17² -169] =1/2*120 = 60 sq meter 32 . A rectangular lawn 55m by 35m has two roads each 4m wide running in the middle of it. One parallel to the length and the other parallel to breadth. The cost of graveling the roads at 75 paise per sq meter is sol: area of cross roads = 55*4 +35*4-4*4 = 344sq m cost of graveling = 344 *75/100 =Rs. 258 33.The cost of fencing a square field @ Rs. 20 per metre is Rs.10.080.How much will it cost to lay a three meter wide pavement along the fencing inside the field @ Rs. 50 per sq m sol: perimeter = total cost / cost per m = 10080 /20 = 504m side of the square = 504/4 = 126m breadth of the pavement = 3m side of inner square = 126-6 = 120m area of the pavement = (126*126)-(120*120)=246”*6 sq m cost of pavement = 246*6*50 = Rs. 73800 34.Amanwalked diagonally across a square plot. Approximately what was the percent saved by not walking along the edges? Sol: let the side of the square be x meters length of two sides = 2x meters diagonal = √2 x = 1.414x m saving on 2x meters = .59x m saving % = 0.59x /2x *100% = 30% (approx) 36.A man walking at the speed of 4 kmph crosses a square field diagonally in 3 meters. The area of the field is sol: speed of the man = 4*5/18 m/sec = 10/9 m/sec time taken = 3*60 sec = 180 sec length of diagonal = speed * time = 10/9 * 180 = 200m Area of the field = ½ *(dioagonal)² = ½ * 200*200 sq m = 20000sq m 37.A square and rectangle have equal areas. If their perimeters are p and q respectively. Then sol: A square and a rectangle with equal areas will satisfy the relation p < q 38.If the perimeters of a square and a rectangle are the same, then the area a & b enclosed by them would satisfy the condition: sol: Take a square of side 4cm and a rectangle having l=6cm and b=2cm then perimeter of square = perimeter of rectangle area of square = 16 sq cm area of rectangle = 12 sq cm Hence a >b 39.An error of 2% in excess is made while measuring the side of a square. The percentage of error in the calculated area of the square is sol: 100cm is read as 102 cm a = 100*100 sq cm and b = 102 *102 sq cm then a-b = 404 sq cm percentage error = 404/(100*100) = 4.04% 40.A tank is 25m long 12m wide and 6m deep. The cost of plastering its walls and bottom at 75 paise per sq m is sol: area to be plastered = [2(l+b)*h]+(l*b) = 2(25+12)*6 + (25*12)= 744 sq m cost of plastering = Rs . 744*75/100 = Rs. 5581 41.The dimensions of a room are 10m*7m*5m. There are 2 doors and 3 windows in the room. The dimensions of the doors are 1m*3m. One window is of size 2m*1.5m and the other 2 windows are of size 1m*1.5m. The cost of painting the walls at Rs. 3 per sq m is sol: Area of 4 walls = 2(l+b)*h =2(10+7)*5 = 170 sq m Area of 2 doors and 3 windows = 2(1*3)+(2*1.5)+2(1*1.5) = 12 sq m area to be planted = 170 -12 = 158 sq m cost of painting = Rs. 158 *3 = Rs. 474 42.The base of a triangle of 15cm and height is 12cm. The height of another triangle of double the area having the base 20cm is sol: a = ½ *15*12 = 90 sq cm b = 2a = 2 * 90 = ½ * 20 *h => h= 18cm 43.The sides of a triangle are in the ratio of ½:1/3:1/4. If the perimeter is 52cm, then the length of the smallest side is sol: ratio of sides = ½ :1/3 :1/4 = 6:4:3 perimeter = 52 cm, so sides are 52*6/13 =24cm 52*4/13 = 16cm 52 *3/13 = 12cm length of smallest side = 12cm 44.The height of an equilateral triangle is 10cm. Its area is sol: a² = (a/2)² +(10)² a² – a²/4 = 100 =>3a² = 100*4 area = √3/4 *a² = √3/4*400/3 = 100/√3 sq cm 45.From a point in the interior of an equilateral triangle, the perpendicular distance of the sides are √3 cm, 2√3cm and 5√3cm. The perimeter of the triangle is sol: let each side of the triangle be ‘a’ cm then area(AOB) +area(BOC)+area(AOC) = area(ABC) ½ * a *√3 +1/2 *a *2√3 +1/2 * a*5√3 = √3/4 a ² a/2√3(1+2+5) = √3/4 a ² => a=16 perimeter = 3*16 = 48cm |