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 = πx²/4 : πx²/2 = 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 = πr² and new area = π(r/2)²
decrease in area = 3 πr²/4 * 1/ π² * 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, 2πr = 352/7 => r = 352/7 * 7/22 * ½ = 8m
2πR = 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 17 cm 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 graveling the path at 80 paise per sq.mt
sol: area of the plot = 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 graveling the path = 850 * 80/100 = 680 Rs
16. The perimeters of two squares are 40 cm 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 πr²=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 rev
