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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
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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)=246â€*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
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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
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<b>Complex Probems:</b>
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
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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
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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
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