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PROBLEMS
1.One side of a rectangular field is 15m and one of its diagonal is 17m.
Find the area of field?
Sol:Other side = √[(17*17) – (15*15)] = √(289-225) = 8m
Area = 15 * 8 =120 sq. m
2.A lawn is in the form of a rectangle having its sides in the ratio 2:3
The area of the lawn is 1/6 hectares. Find the length and breadth of the lawn.
Sol:
let length = 2x meters and breadth = 3x mt
Now area = (1/6 * 1000)sq m = 5000/3 sq m
2x * 3x = 5000/3 =>x * x =2500 / 9
x = 50/3
length = 2x = 100/3 m and breadth = 3x = 3*(50/3) = 50m
3.Find the cost of carpeting a room 13m long and 9m broad with a carpet
75cm wide at the rate of Rs 12.40 per sq meter
Sol:
Area of the carpet = Area of the room = 13* 9 =117 sq m
length of the carpet = (Area/width) = 117 * (4/3) = 156 m
Cost of carpeting = Rs (156 * 12.40) = Rs 1934.40
4.The length of a rectangle is twice its breadth if its length is decreased
by 5cm and breadth is increased by 5cm, the area of the rectangle is
increased by 75 sq cm. Find the length of the rectangle.
Sol:
let length = 2x and breadth = x then
(2x-5) (x+5) – (2x*x)=75
5x-25 = 75 => x=20
length of the rectangle = 40 cm
5.In measuring the sides of a rectangle, one side is taken 5% in excess
and the other 4% in deficit. Find the error percent in the area, calculate
from the those measurements.
Sol:
let x and y be the sides of the rectangle then
correct area = (105/100 * x) * (96 / 100 *y)
=(504/500 xy) – xy = 4/500 xy
Error% = 4/500 xy*(1/xy)*100 % = 4/5% = 0.8%
6.A room is half as long again as it is broad. The cost of carpeting the room
at Rs 5 per sq m is Rs 2.70 and the cost of papering the four walls at Rs 10
per sq m is Rs 1720.If a door and 2 windows occupy 8 sq cm. Find the
dimensions of the room?
Sol:
let breadth=x mt ,length= 3x/2 mt and height=h mt
Area of the floor = (total cost of carpeting /rate) = 270/5 sq m = 54 sq m
x * 3x/2=54 => x*x= 54*(2/3)=36 => x = 6m
so breadth = 6m and length=3/2*6 = 9m
now papered area = 1720 /10 = 172 sq m
Area of one door and 2 windows =8 sq m
total area of 4 walls = 172+8 = 180 sq m
2(9+6)*h = 180 => h=180/30 = 6m
7.The altitude drawn to the base of an isosceles triangle is 8cm and the perimeter is 32cm. Find the area of the triangle?
Sol:
let ABC be the isosceles triangle, the AD be the altitude
let AB = AC=x then BC= 32-2x
since in an isoceles triange the altitude bisects the base so BD=DC=16-x
in ∆ADC,(AC) 2 = (AD) 2 + (DC) 2
x*x=(8*8) + (16-x)*(16-x)
32x =320 => x = 10
BC = 32-2x = 32-20 = 12 cm
Hence, required area = ½ * BC * AD
= ½ * 12 * 10 = 60 sq cm
8.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 * a
New side = 125a / 100 = 5a / 4
New area =(5a * 5a)/(4*4) = (25a²/16) – a²
= 9a²/16
Increase %= 9a²/16 * 1/a² * 100%
= 56.25%
9.Find the area of a Rhombus one side of which measures 20cm and one diagonal 24cm.
Sol:
Let other diagonal = 2x cm
since diagonals of a rhombus bisect each other at right angles, we have
20² = 12² + x² => x = √[20² -12²]= √256 = 16cm
so the diagonal = 32 cm
Area of rhombus = ½ * product of diagonals
= ½ * 24 * 32
= 384 sq cm
10. The area of a circular field is 13.86 hectares. Find the cost of
fencing it at the rate of Rs. 4.40 per meter.
Sol:
Area = 13.86 * 10000 sq m = 138600 sq m
r²= 138600 => r² = 138600 * 7/22 => 210 m
circumference = 2r = 2 * 22/7 * 210m = 1320 m
cost of fencing = Rs 1320 * 4.40 = Rs. 5808
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
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