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<body bgcolor="aqua"><center><h2>Areas</h2></center>
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<strong>Important Facts and Formulae:</strong>
<b>Results On Triangle</b>
1.Sum of the angles of a triangle is 180 degrees.
2.The sum of any two sides of a triangle is greater
than third side.
3.Pythagoras Theorem:
In a right angled triangle (Hypotenuse)2 = (Base)2 +(Height)2
4.The line joining the mid point of a side of a triangle
to the opposite vertex is called the MEDIAN.
5.The point where the three medians of a triangle meet,
is called CENTROID. The centroid divides each of the
medians in the ratio 2:1
6.In an isosceles triangle, the altitude from the
vertex bisects the base
7.The median of a triangle divides it into two triangles
of the same area.
8.The area of the triangle formed by joining the mid points
of the sides of a given triangle is one-fourth of the area
of the given triangle.
<b>Results On Quadrilaterals</b>
1.The diagonals of a Parallelogram bisect each other.
2.Each diagonal of a Parallelogram divides it into two
triangles of the same area.
3.The diagonals of a Rectangle are equal and bisect
each other
4.The diagonals of a Square are equal and bisect each
other at right angles.
5.The diagonals of a Rhombus are unequal and bisect
each other at right angles.
6.A Parallelogram and a Rectangle on the same base
and between the same parallels are equal in area.
7.Of all he parallelogram of given sides the parallelogram
which is a rectangle has the greatest area.
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<b>Formulae</b>
1.Area of a RECTANGLE = length * breadth
Length = (Area/Breadth) and Breadth = (Area/Length)
2.Perimeter of a RECTANGLE = 2(Length + Breadth)
3.Area of a SQUARE = (side)2 = ½ ( Diagonal)2
4.Area of four walls of a room = 2(length + breadth) * height
5.Area of a TRIANGLE = ½ * base * height
6.Area of a TRIANGLE = √[s * (s-a) * (s-b) * (s-c)],
where a,b,c are the sides of the triangle and s = 1/2(a+b+c)
7.Area of EQUILATERAL TRIANGLE = √(3/4)* (side)2
8.Radius of in circle of an EQUILATERAL TRIANGLE of
side a = r / 2√3
9.Radius of circumcircle of an EQUILATERAL TRIANGLE
of side a = r / √3
10.Radius of incircle of a triangle of area ∆ and
semi perimeter S = ∆ / s
11.Area of a PARALLELOGRAM = (base * height)
12.Area of RHOMBUS = 1/2 (product of diagonals)
13.Area of TRAPEZIUM =
=1/2 * (sum of parallel sides)* (distance between them)
14.Area of a CIRCLE =  r2 where r is the radius
15.Circumference of a CIRCLE = 2r
16.Length of an arc = 2 rø / 360, where ø is central angle
17.Area of a SECTOR = ½ (arc * r) = r2ø / 360
18.Area of a SEMICIRCLE = r2 / 2
19.Circumference of a SEMICIRCLE = r
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<b>Simple Problems</b>
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
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<b>Medium Problems:</b>
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
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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.
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