business mathematics

Areas problems

Areas problems


Simple 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 = 2r = 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, 2r = 352/7 => r = 352/7 * 7/22 * ½ = 8m
2R = 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 2r =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

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

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