Archive for January 2014
SAMVID 14 (SSTC/SSGI Bhilai)
SAMVID
SAMVID 14 (SSTC/SSGI Bhilai)
Nexus of SSTC campus is to embark on one of the biggest and most happening event of Central India SAMVID 2014, the 3 days of the event worth a life, get ready to experience the Central India's elite Fest. SAMVID 2014, only 20 days to kill.
SAMVID is TECHNO CULTURAL fest held every year @ SSTC(SHRI SHANKARACHARYA TECHNICAL CAMPUS)
Last year it is done by student of SHRI SHANKARACHARYA GROUP OF INSTITUTIONS
BUT THIS YEAR IT IS TO BE DONE BY
SAMVID 14 (SSTC/SSGI Bhilai)
SAMVID is TECHNO CULTURAL fest held every year @ SSTC(SHRI SHANKARACHARYA TECHNICAL CAMPUS)
Last year it is done by student of SHRI SHANKARACHARYA GROUP OF INSTITUTIONS
BUT THIS YEAR IT IS TO BE DONE BY
- SHRI SHANKARACHARYA GROUP OF INSTITUTIONS (SSGI)(includes MCA , MBA , PHARMACY)
- SHRI SHANKARACHARYA ENGINEERING COLLEGE(SSEC)
- SHRI SHANKARACHARYA INSTITUTE OF TECHNOLOGY & MANAGEMENT (SSITM)
- Shri Shankaracharya Institute of Engineering & Technology(SSIET)
BOOM OF SAMVID 2014
SAMVID, an exalted national level Inter-Collegiate fest, came into being due to a small number of dedicated students who dared to step over the line and decided to bring about a change for a reason they believed in.
Samvid is fortunate of being a protégé of Shri Shankaracharya Technical campus without which its present stature couldn't have been at par with excellence.
In its debutant year 2013, the fest was planned and executed with the coordination of 500+ students along with the never-tiring assistance of the management team of the college. The vibes of healthy competition amongst the competitors and the never-say-die spirit within the coordinators and volunteers was indeed exhilarating. The annual fest was receptive of a staggering response from across the nation. Participants were magnetized to register from Bangalore, Bhopal, Nagpur, covering entire Chhattisgarh and from many other states across the length and the breadth of the country. An unprecedented entry of more than 15000 college students emboldened the success of Samvid on its very debut.
The upcoming SAMVID 2014 will commence with the promise of new innovation, new thrill, new breathtaking events & a platform that'll catapult the participants towards bullseye!
With expectations to surpass the lofty benchmarks set earlier in SAMVID 2013, Team Samvid is poised to sweep off everyone of their feet this time. Its once again Come Compete and Conquer!
Good Luck!
This time, its something new approaches to all the people out here. Through Samvid, we are giving you all a chance to donate what you could to a well versed NGO.
Samvid, since it has been certified as a Tech-Green Fest, we are inviting all of you to donate whatever possible on the following topics to the same NGO.
Topic included:
• Donate Blood • Donate a Book • Donate a Coin • Donate a Cloth
LIST OF MANAGEMENT EVENT !!
Samvid 2014 releases its first poster for the MANAGEMENT EVENT. This event has 3 categories in it.
1) BUSINESS PLAN
2) SALE KA KHEL
3) MANAGEMENT QUIZ
1) BUSINESS PLAN
2) SALE KA KHEL
3) MANAGEMENT QUIZ
LIST OF TECHNICAL EVENT !!
- QUIZ
- APTITUDE
- MY INNOVATION
- BLIND CODING
- CAD DESIGNING
- TECHNICAL MODEL
- STRUCTURE MAKING
LIST OF CULTURAL EVENT !!
- DANCE
- DRAMA
- RAMP WALK
- TALENT HUNT
- PHOTOGRAPHY
- MOVIE MAKING
- WAR OF BAND
- TREASURE HUNT
NOTE DETAILS OF THIS EVENT WILL WE UPLOADED SOON
VISIT OFFICIAL SITE samvid '14
INCOMING SEARCHES
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college in bhilai
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AREA
AREA
FUNDEMENTAL CONCEPTS
I.RESULTS ON TRIANGLES:
1.Sum
of the angles of a triangle is 180
degrees.
2.Sum of
any two sides of a triangle is greater than the third side.
3.Pythagoras
theorem:
In a right
angle triangle,
(Hypotenuse)^2
= (base)^2 + (Height)^2
4.The line
joining the midpoint 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.
Centroid
divides each of the medians in the ratio 2:1.
6.In an
isosceles triangle, the altitude from the vertex bi-sects the base
7.The
median of a triangle divides it into two triangles of the same area.
8.Area of a
triangle formed by joining the midpoints of the sides of a given triangle is
one-fourth of the area of the given triangle.
II.RESULTS
ON QUADRILATERALS:
1. The diagonals of a parallelogram bisects 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 the
parallelograms of a given sides , the parallelogram which is a rectangle has the greatest area.
IMPORTANT FORMULAE
I.1.Area
of a rectangle=(length*breadth)
Therefore
length = (area/breadth) and breadth=(area/length)
2.Perimeter
of a rectangle = 2*(length+breadth)
II.Area
of a square = (side)^2 =1/2(diagonal)^2
III Area
of four walls of a room = 2*(length + breadth)*(height)
IV
1.Area of the triangle=1/2(base*height)
2. Area of a triangle =
(s*(s-a)(s-b)(s-c))^(1/2), where a,b,c are the sides of a triangle and s= ½(a+b+c)
3.Area of the equilateral triangle
=((3^1/2)/4)*(side)^2
4.Radius of incircle of an equilateral
triangle of side a=a/2(3^1/2)
5.Radius of circumcircle of an equilateral
triangle of side a=a/(3^1/2)
6.Radius of incircle of a triangle of area
del and semiperimeter S=del/S
V.1.Area
of the parellogram =(base *height)
2.Area of the rhombus=1/2(product of the
diagonals)
3.Area of the trapezium=1/2(size of
parallel sides)*distance between them
VI 1.Area
of a circle =pi*r^2,where r is the radius
2. Circumference of a circle = 2∏R.
3. Length of an arc = 2∏Rθ/(360) where θ is
the central angle
4. Area of
a sector = (1/2) (arc x R) = pi*R^2*θ/360.
VII. 1. Area of a semi-circle = (pi)*R^2.
2.
Circumference of a semi-circle = (pi)*R.
SOLVED EXAMPLES
Ex.1. One side of a rectangular field is 15 m and
one of its diagonals is 17 m. Find the area of the field.
Sol. Other side = ((17) 2-
(15)2)(1/2) = (289- 225)(1/2) = (64)(1/2) = 8
m.
Area =
(15 x 8) m2 = 120 m2.
Ex. 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
metres and breadth = 3x metre.
Now, area = (1/6 )x 1000 m2
= 5000/3m2
So, 2x * 3x = 5000/3
<=> x2 = 2500/9 <=> x = 50/3
therefore Length = 2x = (100/3) m = 33(1/3) m and Breadth
= 3x = 3(50/3) m = 50m.
Ex. 3. Find the cost of carpeting a room 13 m long
and 9 m broad with a carpet
75 cm wide at the rate of Rs. 12.40 per square metre.
Sol. Area of the carpet = Area of the room =
(13 * 9) m2 = 117 m2.
Length of the carpet = (area/width) = 117 *(4/3)
m = 156 m.
Therefore Cost
of carpeting = Rs. (156 * 12.40) = Rs. 1934.40.
Ex. 4. 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 or
x + y = 23 and x2 + y2 = (17) 2 = 289.
Now, (x
+ y) 2 = (23) 2 <=> (x2 +
y2) + 2xy = 529 <=> 289 + 2xy = 529 óxy=120
Area = xy = 120 cm2.
Ex. 5. The length of a rectangle is twice its breadth. If its length
is decreased by 5 cm and breadth is increased by 5 cm, the area of the
rectangle is increased by 75 sq. cm. Find the length of the rectangle.
Sol. Let breadth = x. Then, length = 2x. Then,
(2x - 5) (x + 5) - 2x * x = 75 <=> 5x -
25 = 75 <=> x = 20.
:. Length of the rectangle = 20 cm.
Ex. 6. 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 calculated from these measurements. (M.B.A.
2003)
Sol. Let x and y be the sides of the
rectangle. Then, Correct area = xy.
Calculated area = (105/100)*x * (96/100)*y = (504/500
)(xy)
Error In measurement = (504/500)xy-
xy = (4/500)xy
Error % = [(4/500)xy *(1/xy) *100]
% = (4/5) % = 0.8%.
Ex. 7. A rectangular grassy
plot 110 m. by 65 m has a gravel path 2.5 m wide all round it on the inside.
Find the cost of gravelling the path at 80 paise per sq. metre.
Sol. Area of the plot = (110 x 65) m2
= 7150 m2
Area of the plot excluding the path = [(110 - 5) *
(65 - 5)] m2 = 6300 m2.
Area of the path = (7150 - 6300) m2 = 850
m2.
Cost of gravelling the path = Rs.850 * (80/100)= Rs.
680
Ex. 8. 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. (S.S.C. 2003)
Sol. Side of first square = (40/4) = 10 cm;
Side of second square = (32/4)cm = 8 cm.
Area of third square = [(10) 2 - (8)
2] cm2 = (100 - 64) cm2 = 36 cm2.
Side of third square = (36)(1/2) cm = 6
cm.
Required perimeter = (6 x 4) cm =
24 cm.
Ex. 9. A room 5m 55cm long and 3m 74 cm broad is to be paved with
square tiles. Find the least number of square tiles required to cover the
floor.
Sol. Area of the room = (544 x 374) cm2.
Size of largest square tile =
H.C.F. of 544 cm and 374 cm = 34 cm.
Area of 1 tile = (34 x 34) cm2.
Number of tiles required
=(544*374)/(34*34)=176
Ex. 10. Find the area of a
square, one of whose diagonals is 3.8 m long.
Sol. Area of the square
= (1/2)* (diagonal) 2 = [(1/2)*3.8*3.8 ]m2 = 7.22 m2.
Ex. 11. The diagonals of two
squares are in the ratio of 2 : 5. Find the ratio of their areas. (Section
Officers', 2003)
Sol. Let the diagonals of the squares be 2x and
5x respectively.
Ratio of their areas = (1/2)*(2x)
2 :(1/2)*(5x) 2 = 4x2
: 25x2 = 4 : 25.
Ex.12. 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 = a2.
New side =(125a/100) =(5a/4). New area = (5a/4)
2 =(25a2)/16.
Increase in area = ((25 a2)/16)-a2
=(9a2)/16.
Increase% = [((9a2)/16)*(1/a2)*100]
% = 56.25%.
Ex. 13. If the length of a certain rectangle is
decreased by 4 cm and the width is increased by 3 cm, a square with the same
area as the original rectangle would result. Find the perimeter of the original
rectangle.
Sol. Let x and y be the
length and breadth of the rectangle respectively.
Then, x - 4 = y + 3 or x - y = 7
----(i)
Area of the rectangle =xy;
Area of the square = (x - 4) (y + 3)
(x - 4) (y + 3) =xy <=>
3x - 4y = 12 ----(ii)
Solving (i) and (ii), we get x = 16 and
y = 9.
Perimeter of the rectangle = 2 (x
+ y) = [2 (16 + 9)] cm = 50 cm.
Ex. 14. A room is half as long again as it is broad. The cost of
carpeting the at Rs. 5 per sq. m is Rs. 270 and the cost of papering the four
walls at Rs. 10 per m2 is Rs. 1720. If a door and 2 windows
occupy 8 sq. m, find the dimensions of the room.
Sol. Let breadth = x metres, length = 3x metres,
height = H metres.
Area of the floor=(Total cost of carpeting)/(Rate/m2)=(270/5)m2=54m2.
x* (3x/2) = 54 <=> x^2 = (54*2/3) = 36
<=> x = 6.
So, breadth = 6 m and length =(3/2)*6 = 9 m.
Now, papered area = (1720/10)m2 = 172 m2.
Area of 1 door and 2 windows = 8 m2.
Total area of 4 walls = (172 + 8) m2 = 180 m2
2*(9+ 6)* H = 180 <=> H = 180/30 = 6 m.
Ex. 15. Find the area of a triangle whose sides measure 13 cm, 14 cm
and 15 cm.
Sol. Let a = 13, b = 14
and c = 15. Then, S =
(1/2)(a + b + c) = 21.
(s- a) = 8, (s - b) = 7 and (s - c)
= 6.
Area = (s(s- a) (s - b)(s
- c))(1/2) = (21 *8 * 7*6)(1/2) = 84 cm2.
Ex. 16. Find the area of a right-angled triangle
whose base is 12 cm and hypotenuse is 13cm.
Sol. Height of the triangle = [(13)
2 - (12) 2](1/2) cm = (25)(1/2)
cm = 5 cm.
Its area =
(1/2)* Base * Height = ((1/2)*12 * 5) cm2 = 30 cm2.
Ex. 17. The base of a
triangular field is 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 = (333.18/25.6)hectares = 13.5 hectares
ó (13.5 x 10000) m2 = 135000 m2.
Let altitude = x metres and base = 3x metres.
Then, (1/2)* 3x *
x = 135000 <=>x2= 90000 <=>x = 300.
Base = 900 m and Altitude = 300 m.
Ex. 18. The altitude drawn to the base of an isosceles triangle is 8
cm and the perimeter is 32 cm. Find the area of the triangle.
 |
B D C
Sol.
Let ABC be the isosceles triangle and AD be the altitude.
Let AB = AC = x. Then, BC = (32 - 2x).
Since, in an isosceles triangle, the altitude bisects the base,
so BD = DC = (16 - x).
In triangle ADC, AC2= AD + DC2=>x2=(82)+(16-x)
2
=>32x = 320 =>x= 10.
BC = (32- 2x) = (32 - 20) cm = 12 cm.
Hence, required area = ((1/2)x*BC * AD) = ((1/2)*12
*10)cm2 = 60 cm2.
Ex. 19. Find the length of the altitude of an equilateral triangle of
side 3Ö3
cm.
Sol. Area of the triangle = (Ö3/4) x
(3Ö3)2
= 27Ö3. Let the height be h.
Then, (1/2) x 3Ö3 x h
= (27Ö3/4)
X(2/Ö3) = 4.5 cm.
.
Ex. 20. 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
and y and their heights be 3h and 4h respectively.
Then,
((1/2) X x X 3h)/(1/2) X
y X 4h) =4/3 ó x/y =(4/3 X 4/3)=16/9
Required ratio = 16 : 9.
Ex.21. The base of
a parallelogram is twice its height.
If the area of the parallelogram
is 72 sq. cm, find its height.
Sol. Let the height of
the parallelogram be x. cm. Then, base = (2x) cm.
2x X x =72 ó
2x^2 = 72 ó
X ^2=36 ó
x=6
Hence, height of the
parallelogram = 6 cm.
Ex. 22. Find the area of a rhombus one side of
which measures 20 cm and 01 diagonal 24 cm.
Sol.
Let other diagonal = 2x cm.
Since diagonals of a rhombus
bisect each other at right angles, we have:
(20)2 = (12)2 + (x)2 _ x = Ö(20)2
– (12)2= Ö256=
16 cm. _I
So, other diagonal = 32 cm.
Area of rhombus = (1/2) x (Product of diagonals) = ((1/2)x 24 x
32) cm^2 = 384 cm^2
Ex. 23. The difference between two parallel sides
of a trapezium is 4 cm. perpendicular distance between them is 19 cm. If the
area of the trapezium is 475 find the lengths of the parallel sides. (R.R.B. 2002)
Sol. Let the two parallel sides of the trapezium be a em
and b em.
Then, a - b = 4
And, (1/2) x (a + b) x
19 = 475 ó (a + b) =((475 x 2)/19) ó
a + b = 50
Solving (i) and (ii), we get: a = 27,
b = 23.
So, the two parallel sides are 27 cm and 23 cm.
Ex. 24. Find the length of a rope by which a cow
must be tethered in order tbat it
may be able to graze an area
of 9856 sq. metres. (M.A.T.
2003)
Sol. Clearly, the cow will graze a circular
field of area 9856 sq. metres and radius
equal to the length of the rope.
Let the length of the rope be R metres.
Then, Õ(R)^2 = (9856 X (7/22)) = 3136ó R = 56.
Length of the rope = 56 m.
Ex. 25. The area of a circular field is 13.86
hectares. Find the cost of fencing it at
the rate of Rs. 4.40 per metre.
Sol. Area = (13.86 x 10000) m2=
138600 m2.
Õ(R2=
138600 ó(R)2 = (138600 x
(7/22)) ó
R = 210 m.
Circumference = 2ÕR = (2 x (22/7) x
210) m = 1320 m.
Cost of fencing = Rs. (1320 x 4.40) = Rs. 5808.
Ex. 26. The diameter of the driving wheel of a bus is
140 em. How many revolution, per minute must the wheel make in order to keep a
speed of 66 kmph ?
Sol. Distance to be covered in 1 min. = (66 X_1000)/(60) m =
1100 m.
Circumference of the wheel = (2 x (22/7) x 0.70) m = 4.4 m.
Number of revolutions per min. =(1100/4.4) = 250.
Ex, 27. A wheel makes 1000 revolutions in covering a
distance of 88 km. Find the radius of the wheel.
Sol. Distance covered in
one revolution =((88 X 1000)/1000)= 88m.
2ÕR = 88 ó 2 x (22/7) x R = 88 ó R = 88 x (7/44) = 14 m.
Ex, 28. The inner circumference of a circular race
track, 14 m wide, is 440 m. Find
radius of the outer circle.
Sol . Let inner radius be r metres. Then, 2Õr =
440 ó
r = (440 x (7/44))= 70 m.
Radius of outer circle =
(70 + 14) m = 84 m.
Ex, 29. Two concentric circles form a ring. The
inner and outer circumferences of ring are (352/7) m and (518/7) m
respectively. Find the width of the ring.
Sol.. Let the inner and outer radii be r and R
metres.
Then 2Õr =
(352/7) ó
r =((352/7) X (7/22) X (1/2))=8m.
2ÕR=(528/7)
ó
R=((528/7) X (7/22) X (1/2))= 12m.
, ', Width of the ring = (R - r)
= (12 - 8) m = 4 m.
Ex, 30. A sector of 120', 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,
( Õ( r )2 q)
/360=(66/7) ó
(22/7) X ( r ) 2 X(120/360)= (66/7)
ó ( r )2=((66/7) X
(7/22) X 3) ó
r=3.
Hence, radius = 3 cm.
Ex, 31. 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 = Ö2 x.
Radius of incircle = (x/2)
Radius of circum circle= (Ö2x/2) =(x/Ö2)
Required ratio = ((Õ ( r
)2 )/4 : (Õ(
r )2) /2) = (1/4) : 1/2) = 1 : 2.
Ex. 32. If the radius of a
circle is decreased by 50%, find the percentage decrease
in its area.
Sol. Let original radius
= R. New radius =(50/100) R = (R/2)
Original area=Õ( R )2= and new area= Õ ((R/2))2= (Õ ( R
)2)/4
Decrease in area =((3Õ (R )2 )/4 X (1/Õ(
R)2) X 100) % = 75%
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