Download To File - KENDRIYA VIDYALAYA No. 3 Amritsar

MINIMUM PACKAGE FOR SA-II
CHAPTERS
EXPECTED MARKS
1. QUADRATIC EQUATION
a. Nature of roots
b. Finding roots by factorization and by using quadratic formula.
08
c. Problems related to day today activities.
2. ARITHMETIC PROGREESION
a. nth term of A.P.
b. Sum of n terms, sum of series which is multiple of a given number.
12
c. Practice given to differentiate between an and n
3. COORDINATE GEOMETRY
a. Distance formula and its applications.
b. Section formula and its applications.
08
c. Area of triangle, finding area and checking collinearity.
4. CIRCLES
a. Enough practice for theorems i.e.,
i.
The tangent at any point of the circle is perpendicular to the radius through the
point of contact.
ii.
The lengths of tangents drawn from an external point to circle are equal.
06
iii.
Simple problems based on theorems.
5. CONSTRUCTIONS
a) Division of line segment in a given ratio.
b) Tangent to a circle from a point outside it.
04
c) Construction of a triangle similar to a given triangle.
6. AREA RELATED TO CIRCLE
a) Formulas related to sector and segment of a circle
b) Simple applications.
06
7. SURFACE AREAS AND VOLUMES
a) Problems on surface areas and volumes.
10
b) Direct application of formula related to frustum of a cone.
8. PROBABILITY
06
TOTALMARKS
60
Q
CHAPTER : 4
QUADRATIC EQUATIONS
SECTION A











A polynomial of the form p(x)=ax2+bx+c, where a≠0 ; a, b, c are real numbers and x is a
real variable, is called a quadratic polynomial.
An equation p(x) = 0 , where p(x) is a quadratic polynomial is called a quadratic
equation. i.e. ax2+bx+c = 0, a≠0
Standard form of quadratic equation is ax2+bx+c = 0, where a≠0 ; a, b, c are real
numbers and x is a real variable.
Those values of x for which ax2+bx+c = 0 is satisfied are called solutions of quadratic
equation.
There are three methods in this chapter to find the solution of quadratic equation
namely : method of factorization , method of completing the square and quadratic
formula.
We can find the nature of root without finding the roots of equations.
If x2 = m, then x= ± √ .
The fundamental theorem of algebra guarantees that a quadratic equation has two
solutions. These solutions may be either real, or both complex (non real).
If α and β are the roots of p(x) = 0, then P (α) =0 & P (β) =0.
The zeroes of the quadratic polynomial ax2+bx+c and the root of the quadratic equation
ax2+bx+c =0 are the same.
b2 - 4ac is called discriminant of the quadratic equation ax2+bx+c =0
SECTION B
• Finding the roots of a quadratic equation by the method of factorisation : If we can
factorise the quadratic polynomial ax2 + bx + c, then the roots of the quadratic
equation ax2 + bx + c = 0 can be found by equating to zero the linear factors of ax2 + bx
+ c.
• Finding the roots of a quadratic equation by the method of completing the square : By
adding and subtracting a suitable constant, we club the x2 and x terms in the quadratic
equation so that they become a complete square, and solve for x.
 Quadratic formula: The quadratic formula to find the roots of ax2+bx+c = 0 is

Existence of roots of a quadratic equation : A quadratic equation ax2+bx+c =0 has
1. Two distinct real roots , if b2-4ac>0,
Q
2. Two equal roots, if b2-4ac = 0,
3. No real roots, if b2-4ac < 0.
SECTION C
TYPE –I
Sample Question 1 : Find the roots of 2x – 5x +3 = 0 by the factorisation of the corresponding
quadratic polynomial.
Solution : 2x2– 5x +3 = 0
 2x2 –3x -2x + 3 =0
 x (2x– 3 ) - 1 (2x – 3 ) = 0
 (2x -3 ) (x – 1 ) = 0
 2x-3 = 0 or x-1 = 0
 x= or 1
2
So, the roots are
and 1.
Sample Question 2 : Find the roots of x2+ 5x +6 = 0 by the factorisation of the
corresponding quadratic polynomial.
Solution : x2 + 5x +6 = 0
 x2 +3x + 2x + 6 =0
 x (x+ 3 ) + 2 (x + 3 ) = 0
 (x +3 ) (x + 2 ) = 0
X+3 = 0 or x + 1 = 0
i.e. x= -3 or -1
So, the roots are
and 1.
EXERCISE
Find the roots of the following quadratic equations by the factorisation method:
(i) x2 – 3x –10 = 0
(ii) 2x2 + x - 6 = 0
(iii) 100x2 -20x + 1 = 0
(iv) x2 -11x + 10 = 0
ANSWERS: (i) -2 & 5 (ii) -2 &
(iii)
&
(iv) 1 & 10
TYPE –II
Sample Question 1 : Find the roots of the quadratic equation x2 – 3x –10 = 0 using the
quadratic formula.
Solution :
b2-4ac = (-3)2-4x1x(-10) = 9 + 40 = 49
√
Therefore , the roots are

√
&
√
Q

&

&
 5 & -2
Sample Question 2 : Find the roots of the quadratic equation 2x2 – √ x – 2 = 0 using
the quadratic formula.
Solution :
b2-4ac = ( √ )2-4x2x(-2) = 5 + 16 = 21
Therefore , the roots are
i.e.
√
√
√
√
√
√
EXERCISE
1. Find the roots of the quadratic equations by using the quadratic formula in each of
the following:
(i) 2x2 – 3x – 5 = 0
(ii) 5x2 + 13x + 8 = 0
(iii) –3x2 + 5x + 12 = 0
(iv) –x2 + 7x – 10 = 0
(v) x2 + 2√ x – 6 = 0
(vi) x2 – 3 √ x + 10 = 0
Answers : (i)5/2 & -1 (ii) -1 &
(iii) 3 & -4/3 (iv) 5 & 2 (v) √ & -3√
(vi) √ & 2√
TYPE- III
Sample Question 1: Determine the nature of roots of the quadratic equation
4x2 – 4x +1 = 0.
Solution : The given equation is 4x2 – 4x +1 = 0.
Here a = 4, b = -4, c= 1
D = b2-4ac = (-4)2-4X4X1 = 16 -16 = 0
Since D =0. Therefore the roots of given equations are real and equal.
Sample Question 2: Determine the nature of roots of the quadratic equation
2x2 +x -1 = 0.
Solution : The given equation is 2x2 +x -1 = 0.
Here a = 2, b = 1, c= -1
D = b2-4ac = (1)2-4 X 2X (-1) = 1 +8 = 9
Since D > 0. Therefore the roots of given equations are real and different.
Sample Question 3 : Determine the nature of roots of the quadratic equation
x2 +x +1 = 0.
Solution : The given equation is x2 +x +1 = 0.
Here a = 1, b = 1, c= 1
D = b2-4ac = (1)2-4 X 1 X 1 = 1 -4 = -3
Since D < 0. Therefore the roots of given equations are not real
Q
EXERCISE
Q1: Determine the nature of roots of the quadratic equation 2x2 -3x +4 = 0.
Q2: Determine the nature of roots of the quadratic equation 2x2 -7x +5 = 0.
Q3: Determine the nature of roots of the quadratic equation 3x2 -2√ x +2 = 0.
Q4: Determine the nature of roots of the quadratic equation x2 +2√ x -1 = 0.
Answers : (i) not real (ii) real and distinct (iii) real and equal (iv) real and distinct.
TYPE- IV
Sample Question 1 : Find the value of K for which the equation 2x2 -Kx +1 = 0 has real and
equal roots.
Solution: The given equation is 2x2 -Kx +1 = 0
For real & equal roots, D = 0
 b2-4ac =0
 i.e. (-K)2 -4X2X1=0
 i.e. k2-8 = 0
 i.e. k2 = 8
 i.e. K = ±2√
Sample Question 2: Find the value of K for which the equation 3x2 -5x +2K = 0 has real roots.
Solution: The given equation is 3x2 -5x +2K = 0
For real & equal roots, D ≥ 0
 b2-4ac ≥ 0
 (-5)2-4X3X2K ≥ 0
 25-24K ≥ 0
 -24K ≥ -25
 24K ≤ 25
 K≤
Sample Question 3 : Find the value of K for which the equation Kx2 +6x +1 = 0 has real and
unequal roots.
Solution: The given equation is Kx2 +6x +1 = 0
For real & unequal roots, D> 0
 b2-4ac>0
 (6)2-4XKX1 > 0
 36 - 4K > 0
 -4K > -36
 4K < 36
 K< 9

Q
EXERCISE
Q1: Find the value of K for which the equation 2Kx2 -40x +25 = 0 has real and equal roots.
Q2: Find the value of K for which the equation x2 -kx +9 = 0 has real and distinct roots.
Q3: if the roots of the equation (b-c)x2 +(c-a)x +(a-b) = 0 are equal , then prove that
2b= a+c.
Q4: Find the value of K for which the equation Kx2+6x 1 = 0 has real roots.
Answers: (i) k =8
(ii) K< -6 or k > 6
(iv) K ≤ 9.
TYPE - V
Sample Question 1 : Check whether the equation 6x2 – 7x + 2 = 0 has real roots, and
if it has, find them by the method of completing the squares.
Solution : The discriminant = b2 – 4ac = 49 – 4 × 6 × 2 = 1 > 0
So, the given equation has two distinct real roots.
Now, 6x2 – 7x + 2 = 0

X2 - X + = 0
 X2 - X +( )2 - ( )2+ = 0
 (X
2
=
 (X
= √
 x=
√
 x=
√
-
 x=
 x=
,
 x= ,
Sample Question 2 : Had Ajita scored 10 more marks in her mathematics test out of 30 marks,
9 times these marks would have been the square of her actual marks. How many marks did she
get in the test?
Solution : Let her actual marks be x
Q
Therefore, 9 (x +10) = x2
 x2 – 9x – 90 = 0
 x2 – 15x + 6x – 90 = 0
 x(x – 15) + 6(x –15) = 0
 (x + 6) (x –15) = 0
Therefore, x = – 6 or x =15
Since x is the marks obtained, x ≠ – 6. Therefore, x = 15.
So, Ajita got 15 marks in her mathematics test.
Sample Question 3 :: A motor boat whose speed is 18 km/h in still water takes 1 hour more to
go 24 km upstream than to return downstream to the same spot. Find the speed of the stream.
Solution : Let the speed of the stream be x km/h.
Therefore, the speed of the boat upstream = (18 – x) km/h and the speed of the boat
downstream = (18 + x) km/h.
The time taken to go upstream =
=
hours.
Similarly, the time taken to go downstream =
hours.
According to the question,
-
=1
 24(18 + x) – 24(18 – x) = (18 – x) (18 + x)
 324 + 24x -324 + 24x = 324 – x2
 x2 + 48x – 324 = 0
Using the quadratic formula, we get
Since x is the speed of stream it cannot be negative. So, we ignore the root x = – 54. Therefore,
x = 6 gives the speed of the stream as 6 km/h.
Sample Question 4 : A train travels at a certain average speed for a distance of 63 km and then
travels a distance of 72 km at an average speed of 6 km/h more than its original speed. If it
takes 3 hours to complete the total journey, what is its original average speed?
Solution : Let its original average speed be x km/h. Therefore,
+
Q
= 3

63(X+6)+72X=3(X+6)X

63x + 378 + 72x = 3x2 + 18x

3x2 + 18x-72x -63x -378 =0

3x2 -117x – 378 = 0

x2 – 39x – 126 = 0

x2 – 42x +3x – 126 = 0

(x + 3) (x – 42) = 0

x = – 3 or x = 42
Since x is the average speed of the train, x cannot be negative. Therefore, x = 42.
So, the original average speed of the train is 42 km/h.
1.
2.
3.
4.
5.
EXERCISE
Find a natural number whose square diminished by 84 is equal to thrice of 8 more than
the given number.
A natural number, when increased by 12, equals 160 times its reciprocal. Find the
number.
A train, travelling at a uniform speed for 360 km, would have taken 48 minutes less to
travel the same distance if its speed were 5 km/h more. Find the original speed of the
train.
If Zeba were younger by 5 years than what she really is, then the square of her age (in
years) would have been 11 more than five times her actual age. What is her age now?
A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would
have taken 1 hour less for the same journey. Find the speed of the train.
Answers: (1) 12
(2) 8
(3) 45km/hr
Q
(4) 14 yrs
(5) 40km/hr
Chapter 5
Arithmetic Progressions
Key Contents
•
Concept of an A.P ( its definitions criteria, basic terms, initial terms & common
difference).
•
General form of A.P.
•
terms of an A.P.
•
Sum of first terms of A.P.
•
Simple statements Sum on A.P.
Basic formulae
•
General form of an A.P. is
a, a+d, a+2d, a+3d,……….
Where a is the first term and d is the common difference.
d= (a+d)-a=d
•
i) nth term of an A.P.
= a+(n-1)×d
term from the last = l-(n-1) ×d , where l= last term & d= common difference.
•
Sum of first n term of A.P
=
•
[2a+(n-1)d]
= [a+l], where l is the last term
Question Bank For Slow Bloomers
Type 1
Q1) For the A.P 3,1,-1,-3…., Write the first term and common difference?
Sol. Here, first term ‘a’=3
Q
& common difference ‘d=1-3
=-2
Q2) For the A.P 3,5,7,9…. Write the first term and common difference?
Sol. Consider given A.P
3,5,7,9….
first term ‘a’= 3
& common difference ‘d’=5-3
=2.
Exercise
Q1) Write the first term and common difference of the A.P 2,7,12….?
Ans) 2 & 5
Q2) Write the first term and common difference the A.P 1,6,11,16…..?
Ans) 1 & 5
Q3) Write the first term of and common difference the A.P 1/3,5/3,9/3,13/3?
Ans) 1/3 & 4/3
Q4) Write the first term and common difference of the A.P -5,-1,3,7?
Ans) 5 & 4
Q5) Write the first term and common difference of the A.P 15,22,29,36?
Ans) 15 & 7
Type 2
Q1) Find the 10th Term of the A.P 2,7,12,….?
Sol. Here, A.P=2,7,12,…
first term ‘a’=2
Q
& common difference ‘d’=7 - 2
=5.
Now,
n= 10
=a+(n-1)×d
Or,
=2+ (10-1)×5
=2+45
=47
Q2) Find the
term from the end of the A.P. 7,10,13,…..184?
Sol. Given A.P=7,10,13,…..184
a = 7 ,d = 10 – 7 =3
Also ,last term ‘l’= 184
Therefore,the term from the end = l-(n-1) ×d
= 184-(8-1)
= 184-21
=163
Exercise
Q1) Find the
Term of the A.P -6,-3,0,3…?
Ans) 36.
Q2) Find the
Term of the A.P 10,7,4,…?
Ans) -20
Q3) Find the
Term of the A.P 117,104,91,78…?
Ans) 26
Q4) Find the
Term of the A.P from the last Term of the A.P : 10,7,4,….,-62?
Q
Ans) -32.
Q5) Find the
Term of the A.P from the end of the A.P : 8,10,12,…,126.
Ans) 106
Type 3
Q1) Which term of the A.P : 21,18,15,….. is -81?
Sol) Here a = 21 , d= 18-21= -3
Let
term = -81
= -81
a+(n-1)×d = -81
or, 21+(n-1)(-3)= -81
or, (n-1)(-3)= -81-21
or, (n-1)=-102/-3
or, (n-1)=34
Or, n=35
Hence
term is -81.
Q2) Which term of the A.P : 3,8,13,18………….. is 78?
Sol) Consider given A.P=3,8,13,18………
Here a = 3 , d=8-3=5
Let an=78
a+(n-1)×d =78
or, 3+(n-1)×5 = 78
or,
(n-1)×5 = 78-3
or,
(n-1) = 75/5
or,
(n-1) = 15
Q
Or,
n = 16
Hence,
term is 78.
Exercise Questions
Q1) Which term of the A.P : 7,13,19,………….. is 205?
Ans) -34
Q2) Which term of the A.P : 72,68,64,60,,………….. is 0?
Ans) 19
Q3) Which term of the A.P : 5,2,-1,-4,………….. is -55?
Ans) 21
Q4) Which term of the A.P : 3,8,13,18,………….. is 58?
Ans) 14
Q5) Which term of the A.P : 1,7,13,18,………….. is 85?
Ans) 15
Type 4
Q1) Determine the A.P whose
term is 5 and
Sol) Let first term of A.P =a & common difference =d
since,
=5
or, a+(3-1)d=5
or, a+2d =5 …………..(1)
&
=9
Or, a+(7-1)d=9
Or, a+6d=9 …………..(2)
Subtracting (1) & (2),we get
a+6d-a-2d = 9-5
Q
term is 9?
or, 4d =4
Or, d=1
Putting d=1 in eq. (1), we get
a+2×1 = 5
a+2 =5
a= 5-2
or, a=3
Required A.P is a, a+d, a+2d, a+3d
i.e. 3, (3+1),3+2×1,3+3×1
i.e. 3,4,5,6….
Q2) Find the sum of 16th term of the A.P 2,5,8,11.......?
Sol) first term = 2
& common difference =5-2
=3
Also, n =16
S16 = ?
= [ 2a + (n-1) ×d ]
=
[ 2×2 + (16-1)×3 ]
=8 [ 4+15×3 ]
Or,
= 8 [4+45]
Or,
=8×49
= 392
Exercise
Q1) If the first term of an A.P is 5 & its 100th term is -292.Find 50th term?
Q
Ans) -142
Q2) Find the sum of first 13 terms of the A.P : 13,8,3,-2,…..?
Ans) -221
Q3) If the 6th term of an A.P is -10 & its 10th term is -26.Find 15th term of A.P?
Ans) -46
Q4) Find the sum of first 10 terms of an A.P 2,7,12,17,….?
Ans) 245
Q5) Ram started work in 1995 at an annual salary of Rs.5000 & received an increment of Rs.
200 each year.In which year did his income reach Rs.7000.
Ans) 11th year.
HOTS QUESTIONS
Q1)
Find the sum of all natural number’s between 100 & 1000 which are multiples of 7?
Ans) 70336
Q2)
200 logs are stacked in the following manner : 20 logs in the bottom row ,19 logs in the
next row ,18 in the row next to it & so on. In how many rows the 200 logs are placed
and how many logs are in the top row?
Ans)
16 rows: 5 logs on the top row.
Q3)
The sum of n terms of a progression is 3
.P & the sum of its rth term.
Ans)
7,13,19,25,…. : (3 +4r).
Q4)
A theif runs away from a Police Station with a uniform speed of 100 m/min.After a
minute a Policeman runs behind the theif to catch him. He goes at a speed of 100
m/min in the first minute and increases his speed 10m each succeeding minute . After
how many minutes,the Policeman will catch the thief?
Ans)
5 minutes
Q
+4n. Is this progression in A.P. If so find the A
MINIMUM PACKAGE OF LEARNING
CHAPTER-7
CO-ORDINATE GEOMETRY
SECTION –A
Contents of the Chapter:








Review the concept of Co-ordinate Geometry.
Concept of X-axis ,Y-axis and Origin.
Coordinates of a point.
Distance between two points.
Section formula(internal).
Coordinates of centroid of a triangle.
Area of a triangle.
Condition for collinear points.
SECTION -B
 “Co-ordinate Geometry is that branch of geometry in which two numbers called
coordinates, are used to locate the position of a point in a plane.”
The distance of a point from the y-axis is called its x-co-ordinates or abscissa. The
distance of a point from the x-axis is called its y-co-ordinates or ordinate.
The co-ordinates (abscissa, ordinate) of a point on the x-axis,are of the form (x,0) and a
point on y-axis will have the co-ordinates of the form (0,y).
 Distance Formula:
If we take the coordinate of point P and Q as (x1,y1) and (x2,y2) resp. then
P(x1,y1)
Q(x2,y2)
PQ = √
 Section Formula :
Let A(x1,y1) and B(x2,y2) be the two points such that
=
i.e.,P divides AB internally in the ratio m1 : m2 .
A(x1,y1)
m1
P(x,y)
Then co-ordinates of P = [
m2
B(x2,y2)
]
 Co-ordinates of middle point.:
Let A(x1,y1) and B(x2,y2) ,and P be the middle point of line segment AB i.e.,
Then co-ordinates of middle point P
= [
]
Q
=
 Area of Triangle :
If we are given three vertices of a triangle ABC, then area of triangle will be
A(
B(
Area
C(
[
]
 Condition for collinear points.
The three points A,B and C are collinear if and only if the area of
SECTION –C
Type-1
Example: Find the distance between two points A(2,3) and B(4,1).
Solution : Here x1=2,y1=3 and x2=4,y2=1
Therefore from distance formula
AB = √
√
 √
 √
 √
 √
Q1. Find the distance of the point (3,-4) from the origin.
Q2. Find the distance between (a,b) and (-a,-b).
Q3. Find the distance between (2,3) and (4,1).
Type-2
Example: Find a relation between x and y such that the point (x , y) is equidistant
from the points (7, 1) and (3, 5).
Solution : Let P(x, y) be equidistant from the points A(7, 1) and B(3, 5).
We are given that AP = BP. So, AP2= BP2
i.e., (x – 7)2 + (y – 1)2 = (x – 3)2 + (y – 5)2
i.e., x2– 14x + 49 + y2– 2y + 1 = x2 – 6x + 9 + y2 – 10y + 25
i.e., x – y = 2
Q
which is the required relation.
Q1. What point on Y- axis is equidistant from (-5,-2) and (3,2).
Q2. Find the co-ordinates of a point P on y-axis, equidistant from two points
A(-3,4) and B(3,6) on the same plane.
Q3. Find the point on the x-axis which is eqyidistant from (2,-5) and (-2,9).
Type-4
Example: The line joining the points (4,-5) and (4,5) is divided by the point P
such that
Solution : Given
Subtracting 1 from both sides, we get
Or
Here (x1,y1)= (4,-5), (x2,y2)= (4,5) and m1:m2= 2:3
(
)
(
(
)
) = (4,-1)
Q1. Find the coordinates of the point which divide the line segment
joining the points (3,5) and (7,9) internally in the ratio of 2:3.
Q2.Find the coordinates of a point A, where AB is the diameter of a circle whose centre is
(2,-3) and B is (1,4).
Q3. Find the coordinates of the points of trisection of the line segment joining (4,-1) and
(-2,-3)
Type-5
Example : In what ratio does the point (– 4, 6) divide the line segment joining
the points A(– 6, 10) and B(3, – 8)?
Solution : The ratio m1 : m2 can also be written as
, Let (– 4, 6)
divide AB internally in the ratio k : 1. Using the section formula,
we get
(– 4, 6) = (
)
So, -4 =
-7k = -2
k = or k :1 = 2 :7
You can check it for y coordinate also.
So, the point (– 4, 6) divides the line segment joining the points A(– 6, 10)
Q
and B(3, – 8) in the ratio 2 : 7.
Q1. Find the ratio in which the point (-3,10) divides the line segment joining the points
(6,-8) and (-1,6) .
Q2. Find the ratio in which the y-axis divides the line segment joining the points (5,-6)
and (-1,-4).
Q3.Find the ratio in which the x-axis divides the line segment joining the points (-8,-3)
and (-2,4).
Type-6
Example : Find the area of a triangle formed by the points A(5, 2), B(4, 7)
and C (7, – 4).
Solution : The area of the triangle formed by the vertices A(5, 2), B(4, 7) and
C (7, – 4) is given by
[

[

[

[
]
]
]
]
 [ ]
Since area is a measure, which cannot be negative, we will take the numerical value of – 2, i.e.,
2. Therefore, the area of the triangle = 2 square units.
Q1. Find the area of the triangle whose vertices are (2, 3), (–1, 0), and (2, – 4) .
Q2. What will be the area of a triangle whose vertices are (–5, –1), (3, –5),and (5, 2).
Q3. Find the area of the triangle PQR whose vertices are P(-1.5,3),Q(6,-2) and R(-3,4).
Type-7
Example : Find the value of k if the points A(2, 3), B(4, k) and C(6, –3) are
collinear.
Solution : Since the given points are collinear, the area of the triangle formed
by them must be 0, i.e.,
[
]
[

]
[
]
Q

[

[
]
]
therefore

Q1. In each of the following find the value of ‘k’, for which the points are
collinear.
(i) (7, –2), (5, 1), (3, k)
(ii) (8, 1), (k, – 4), (2, –5)
Q2. Find the value of k, if the points (
) (
Q
) (
) are collinear.
Applications of Trigonometry (Height and distance)
Chapter 9
Content :






Angle of elevation
Angle of depression
Line of sight
Horizontal level
Point of observation
Height and distance
Section B :-
Formulae used.
Trigonometric ratios :- In right angled triangle ABC right angled at point B.
Q
Trigonometric ratios of some specific angles :-
Angle of Elevation
Q
Angle of depression
Problems:
Type-I
Example 1.)
formed?
If the observer is at point B and the object is at point C. What is the angle
(Elevation /depression)
Ans.) As the observer is at B and the object is at C. So the observer has to look downwards.
So the angle formed is the angle of depression.
Example 2.)
In the given fig: find BC
Q
Ans.) In the given fig : angle is 30°.
= Sin 30°
therefore: BC = therefore BC = 3cm
EXERCISE
Q.1) In the following figure identify the angle formed: (elevation/ depression)
( Ans Elevation)
Q.2) In the adjoining figure; angle of elevation is 60° and the distance AB = 10 m. Find the
height of tower.
( Ans 30 m)
Q.3) The kite is flying at a height of 40 m from the level ground, attached to a string inclined
at 60° to the horizontal. Find the length of the string.
(Ans 80 m)
Q
Q.4) The angle of elevation of the top of tower from a point on the ground and at a distance
of 30m from its foot is 30°. Find the height of tower.
(Ans 20 √3m )
Q.5) From the point on the a ground, which is 15m away from the foot of a tower, the angle
of elevation is found to be 60°. Find the height of the tower. (Ans 15√3 m)
Type-II
Example 1:- A tower is 100√ m high. Find the angle of elevation of its top from a point
100m away from the foot.
B
Sol. Let θ be the angle of elevation of the top of the tower from point C.
↑
In ∆CAB, tan θ =
Tan θ =
√
100√3 m
= √3 = tan 60
θ = 60
↓
ᶿ
So angle of elevation is 60
C
←
100 m
→
Example 2 :- A circus artist is climbing from the ground along a rope stretched from the top of a
vertical pole and tied at the ground. The height of the pole is 12m and the angle made by the
Q
A
rope with the ground level is 30 . Calculate the distance
to the top of the pole.
covered by the artist in climbing
A
E
Sol. Distance covered by the artist = length of the rope AC
P
= 30
In right-angled triangle ABC, we have
AB = 12 m,
O
R
= 30
↑
1
2
m
30
Sin 30 =
C
=
AC = 24 m.
So, the distance covered by the circus artist is 24 m.
Exercise:
Q.1 : The height of a tower is 10m.What is the length of its shadow when sun’s
altitude is 45 ?
(Ans 10 m)
Q.2: The ratio of the height of a tower and the length of its shadow is √3 : 1. What is the
angle of elevation of the sun?
(Ans 600 )
Q.3: From a point 20m away from the foot of a vertical tower, the angle of elevation of the
top of tower is 60 . What is the height of the tower?
(Ans 20√3 m )
Q.4: If the angle of elevation of the top of a tower from two points at a distance 4m and
9m from the base of the tower and in the same straight line with it are complementary.
Find the height of the tower.
(Ans 6 m )
Q.5: A tree is broken by the wind. The top struck the ground at an angle 30 and at a
distance 30m from the foot. Find the whole height of the tree. (Ans 30 √3 m )
Q
B
↓
Type-III
Example 1)
From a point P on the level ground the angle of elevation of top of tower is 30°.
If the tower is 100m high; how far is P from the foot of the tower?
Ans.)
Let P is the point on the ground. Let AB is the height of the tower which is 100m. Let the angle
of elevation is 30°.
In right angled triangle ABC:AB/PB = tan 30°.
( Perpendicular/Base = tan ϴ)
100/x = 1/√3
therefore x = 100√3 m
Example 2)
From a top of 7m high building, the angle of elevation of cable tower is 30°. And
the angle of depression of its foot is 45°. Determine the height of the tower.
Ans.)
Let AB is the building and CD is the tower. Let BC is the level ground as shown in the fig.
In right triangle ABC , AB/BC = tan 45°.
Q
Therefore 7/BC = 1
Therefore BC = 7m.
Hence AE = 7m.
In triangle AED , DE/AE
Therefore
=
tan 30°.
DE/7 = 1/ √3
Hence height of the tower = DE + EC = (7/√3 + 7)m
Exercise
Q.1
From the point P on the ground, the angle of elevation of the top of 10m tall building
and the helicopter hovering over the building are 30° & 60° respectively. Find the height of the
helicopter above the ground. How the helicopters can be helpful at the time of natural
climates?
(Ans 30 m)
Q.2
A window in the building is at a height of 10m from the ground. The angle of depression
of a point P on the ground from window is 30°. The angle of elevation of the top of the building
from point P is 60°. Find the height of the building.
(Ans 30 m )
Q.3 From a point on a bridge across a river the angle of depression of the banks on opposite
sides of the river are 30° and 45°. If the bridge is at a height of 3m from the bank, find the width
of the river.
(Ans 3 (1 + √3 ) m )
Q.4
Two Pillars of equal height stands on either sides of a roadway which is 80m wide. At a
point on the road between pillars the elevations of the pillar are 60° & 30°. Find the heights of
the pillars and the position of the point.
(Ans 20 √3 m , 20 m
Q.5 From a point on the ground, the angles of elevation of the bottom and the top of a tower
fixed at the top of a 20m high building are 45° and 60° respectively. Find the height of the
tower.
(Ans 20 (√3– 1 ) m )
Q
Chapter 10
CIRCLE
Gist Of the Chapter
# Secant of The Circle:- It is a line which intersect the circle at two different points.
# Tangent :- The line which intersect the circle at only one point. And this point is called the
POINT OF CONTACT.
# Infinite many tangents can be drawn to a circle.
# Only two parallel tangents can be drawn to a circle.
# At most only two tangents can be drawn from an external point to a circle.
# The length of the tangents drawn from an external point to a circle are equal.
# Radius and the tangent are perpendicular to each other at the point of contact.
# Pythagoras Theorem
(Hypotenuse)2 = (Base)2 + (Perpendicular)2
Q
Solved Examples
Type-1
1) A point R is 5cm apart from the centre of circle. The length of the tangent from R to
the circle is 4cm. Find the radius of the circle.
Sol.
O
4cm
R
Q
5cm
In Right angled triangle QOR
(RQ)2 = (OR)2 + (OQ)2
( By PGT)
52 = 42+ (OQ)2
25 = 16 + (OQ)2
25 – 16 =(OQ)2
9 = (OQ)2
√9 = OQ
OQ = 3cm
2) A tangent XY at a point X of a circle of radius 5cm meets a line through the center O at
a point Y and OY = 13cm. Find the length of XY.
SOL
Y
5cm
X
13cm
O
Q
In Right angled triangle XYO
(XO)2 = (OY)2 + (XY)2
( By PGT)
132 = 52 + (XY)2
169 = 25 +(XY)2
169 – 25 =(XY)2
144 =(XY)2
XY = √144
= 12cm
Unsolved Exercise for practice
1) PT is a tangent to the circle with center O . Find the length of the tangent PT if a point
P is 5cm from the center and the radius of the circle is 4cm.
( Ans – 3cm)
2) From a point Q , the length of the tangent to the circle is 24 cm and the distance of the
point Q from the center is 25cm. Find the radius of the circle. ( Ans – 7cm)
3) AB is a tangent to the circle with center O of length 8cm .If the radius OA of the circle
is 6cm find OB.
(Ans- 10cm)
Type- 2
1. A Circle touches all the four sides of a ABCD with sides AB = 6cm, BC= 7cm and CD
=4cm .Find AD.
SOL : AS AD + BC = AB +CD
AD + 7 = 6 +4
AD = 10 – 7
AD = 3cm.
A
B
D
C
Q
2. A quadrilateral PQRS is drawn to circumscribe a circle with sides PQ=5cm, PS=7cm,
and QR=6cm.Find RS.
P
Q
S
R
Sol : AS PQ + RS = PS +QR
5 + RS = 7 + 6
RS = 13-5
RS = 8cm.
Unsolved Exercise for practice
1
A quadrilateral ABCD is drawn to circumscribe a circle (see Fig. 10.12). Prove that AB +
CD = AD + BC
2
Prove that the parallelogram circumscribing a circle is a rhombus.
3
A quadrilateral HOTS is drawn to circumscribe a circle with sides HO=7cm, OT=5cm,
and TS=6cm.Find HS.
Type 3
1) In Fig. if TP and TQ are the two tangents to a circle with centre O so that
110°, then find ∟ PTQ .
P
T
O
Q
∟POQ =
Sol:- In Quardrilateral POQT
∟P = ∟Q = 900 ( Radius and Tangent are perpendicular)
∟P + ∟Q + ∟O + ∟T = 3600
900 + 900 + 1100 + ∟T = 3600
2900 + ∟T = 3600
∟T = 3600 – 2900
∟T = 700
2
In Fig. if AB and BC are the two tangents to a circle with centre O so that
60°, then find ∟ AOC.
A.
B
O
C
Sol:- In Quardrilateral AOCB
∟A = ∟C = 900 ( Radius and Tangent are perpendicular)
∟A + ∟C + ∟B + ∟O = 3600
900 + 900 + 600 + ∟O = 3600
240 0 + ∟O = 3600
∟O = 3600 – 2400
∟O = 1200
UNSOLVED EXERCISE FOR PRACTICE
Q
∟ABC =
3. If PQ and PR are tangents to the circle with centre O .If ∟QPR = 30 0,Find ∟PRQ and
∟QOR.
4. Show that the tangents at the ends of any chord makes equal angles with the chord..
5. If two concentric circles, prove that a chord of large circle, which is tangent to the smaller
circle is bisected at point of contact.
Q
CONSTRUCTIONS
SECTION A (KEY CONTENTS)
(1)
Division of a line segment in the given ratio (Internally).
(2)
Tangent to a circle from a point outside it.
(3)
Construction of a triangle similar to a given triangle.
Section B
Requirement for Construction
(1) Use of instruments to perform construction.
(2) Idea for tangent.
(3) How to construct the triangle.
(4) Sum of the angles of the triangles.
(5) Adjacent sides and adjacent angles.
(6) Angles which can be made with the help of Ruler and Compass.
(7) Criteria for construction of a triangle.
(8) What do you understand by similar triangles .
TYPE --1
Example1.
Draw a line segment AB= 7cm. and divide it internally in the ratio 3:2
Solution:- Steps of construction:1. Draw AB =7cm.
2. Draw any ray AX making an acute angle
3. Mark 3+2 = 5 points A1, A2, A3, A4, A5 on AX such that AA1=A1A2=A2A3= A3A4 =A4 A5.
4. Join A5 to B.
5. Through A3 draw a line parallel to A5B intersecting AB at C
6. Now C divides AB in the ratio 3:2.
EXCERCISE
1.
2.
3.
4.
Divide a line segment of 7cm internally in the ratio 2:3.
Draw a line segment AB=8cm and divide it in the ratio 4:3.
Draw a line segment AB and locate a point C on AB such that AC=2/7AB.
Draw a line segment MN=9.5cm and locate a point P on MN such that MP=3/8 X MN.
Q
TYPE 2
Example 1- Draw a triangle ABC with BC=6cm, ˂B=60 and ˂C=45.construct another triangle
whose side are 2/3 of the corresponding sides of given triangle.
Sol. Steps of construction1.
2.
3.
Draw a triangle ABC with BC=6cm , ˂B=60, ˂C=45
Draw a ray BX making an acute angle CBX .
On BX mark three points B1, B2 ,B3
Such that BB1=B1B2=B2B3.
Join B 3C
Through B 2 draw B 2C’ parallel to B 3C.
Through C’ draw C’A’ parallel to CA.
Therefore A’BC’ is the required triangle.
4.
5.
6.
Example 2.Draw triangle ABC with BC=6cm, AB=4.5cm and ˂B=30.construct a triangle whose
sides are 3/2 the corresponding sides of given triangle.
Steps of construction1.
2.
3.
4.
Draw a triangle ABC with BC=6 cm , ˂B=30, AB=4.5cm.
2.Draw a ray BX making an acute angle CBX .
On BX mark three points B1, B2,B 3 Such that BB1=B1B2=B2B3.
Join B2C.
5. Through B 3 draw B 3C’ parallel to B 2C.
6. Through C’ draw C’A’ parallel to CA.
Therefore A’BC’ is the required triangle
EXCERCISE
1. Draw a triangle ABC in which AB=5cm, AC=6cm and BC=7cm. Then, draw another
triangle whose sides are 3/5 times the corresponding sides of triangle ABC.
2. Draw a right triangle in which the sides (other than hypotenuse) are of lengths 8cm and
6cm. Then construct another triangle whose sides are 3/5 times the corresponding sides
of the given triangle.
3. Construct a triangle PQR in which QR=6cm, angle Q=60 and angle R=45. Construct
another triangle similar to triangle PQR such that its sides are 5/6 of the corresponding
sides of triangle PQR.
4. Draw a right triangle ABC in which angle B=90, AB=5cm, BC=4cm then construct another
triangle whose sides are 5/3 times the corresponding sides of triangle ABC.
Q
TYPE -3
Example 1. Draw a circle of radius 4cm. and construct a pair of tangents to the circle which are
inclined to each other at 300
Solution Steps of construction
1. Draw a circle with centre O and radius 4 cm.
2. Draw a radius OA draw an angle AOB = 1500.
3. At A and B draw perpendiculars intersecting at P.
4. Now ˂ APB = 300.
Example 2. Draw circle of radius 3cm. From a point P outside the circle at a distance of 5cm
from the Centre of the circle, draw two tangents to the circle measure the length of these
tangents
Solution:- 1. Draw a circle of radius of 3cm with center O.
2. Take point P outside the circle such that OP equal to 5cm.
3. Draw a perpendicular bisector of OP at M.
4. With M as a center and radius MP equal to MO, draw a circle intersecting the first
circle at Q and Q’.
5. Join PQ and PQ’,
6. Measure PQ and PQ’. We have PQ= PQ’ =4cm.
EXERCISE
1. Draw a circle of radius 5cm from a point 8cm away from its centre. Construct the pair of
tangents to the circle and measure their lengths.
2. Draw a pair of tangent to a circle of radius 5cm which are inclined to each other at an
angle of 60 degree.
3. Draw a circle of radius of 6cm. From a point 10cm away from its centre, construct the
pair of tangents to the circle and measure their lengths.
4. Construct a tangent to a circle of radius 4cm from a point on concentric circle of radius
6cm and measure its length. Also verify the measurement by actual calculation.
5. Draw a circle of radius of 3cm. Take two points P & Q on one of its extended diameter
each at a distance of 7cm from its center. Draw tangents to the circle from these two
points P & Q.
Q
Chapter 12
AREAS RELATED TO CIRCLES
Main contents:
Perimeter and area of a circle.
Sector of a circle and its area.
Segment of a circle and its area.
Areas of combinations of plane figures
IMPORTANT FORMULAE
Area of circle = πr2
Circumference of circle = 2 πr
Area of semi – circle = 1/2 πr2
Area of quadrant = 1/4 πr2
Area of a Sector of angle
Length of an arc of angle
LEVEL-1
Question 1:
The radii of two circles are 19 cm and 9 cm respectively. Find the radius of the circle which
has circumference equal to the sum of the circumferences of the two circles.
Circumference of 3rd circle = Circumference of 1st circle + Circumference of 2nd circle
2πR = 2πr1 + 2πr2
2πR= 38π + 18π = 56π
R
Question 2:
The radii of two circles are 8 cm and 6 cm respectively. Find the radius of the circle having
area equal to the sum of the areas of the two circles.
Area of 3rd circle = Area of 1st circle + Area of 2nd circle
Q
Question 3:
Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60°.
Let OACB be a sector of the circle making 60° angle at centre O of the circle.
Area of sector of angle
Area of sector
θ=
OACB =
Question 4:
Find the area of a quadrant of a circle whose circumference is 22 cm.
Let the radius of the circle be r.
Circumference = 22 cm
2πr = 22
Quadrant of circle will subtend 90° angle at the centre of the circle.
Q
Area of such quadrant of the circle
PRACTICE QUESTIONS
Q1: The circumference of the circle exceeds the diameter by 16.8 cm. Find the radius of the
circle.
Q2: Find the area of quadrant of circle whose circumference is 22cm.
Q3. Find the area of a circle whose circumference is 44cm.
LEVEL 2
Question 1:
The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in
5 minutes.
We know that in 1 hour (i.e., 60 minutes), the minute hand rotates 360°.
In 5 minutes, minute hand will
rotate =
Therefore, the area swept by the minute hand in 5 minutes will be the area of a sector of 30° in
a circle of 14 cm radius.
Q
Area of sector of angle θ =
Area of sector of 30°
Question 2:
A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the
corresponding:
(i) Minor segment
(ii) Major sector
Area of major sector OADB =
Area of minor sector
OACB =
Q
Area of ΔOAB =
= 50 cm2
Area of minor segment ACB = Area of minor sector OACB −Area of ΔOAB
= 78.5 − 50 = 28.5 cm2
PRACTICE QUESTIONS
Q1: Find the area of the sector of a circle with radius 4cm and angle of 300, also find the area of
the corresponding major sector. Use [ π=3.14]
Q2: The perimeter of a certain sector of a circle of radius 5.6 m is 27.2m.Find the area of the
sector.
Q3: In a circle of radius 21 cm, an arc subtends angle of 600 at the center. Find
(i) The length of arc
(ii) Area of sector formed by the arc
(iii) Area of the segment formed by the corresponding chord.
Q 4: A chord of radius 12cm subtends an angle of 1200 at the center. Find the area of the
corresponding segment of the circle. Use (π=3.14 & √3=1.73)
LEVEL-III
Question 1:
A horse is tied to a peg at one corner of a square shaped grass field of side 15 m by means of
a 5 m long rope (see the given figure). Find
(i) The area of that part of the field in which the horse can graze.
(ii) The increase in the grazing area of the rope were 10 m long instead of 5 m.
Q
Area that can be grazed by horse = Area of sector OACB
Area that can be grazed by the horse when length of rope is 10 m long
Increase in grazing area = (78.5 − 19.625) m2
= 58.875 m2
Question 2:
An umbrella has 8 ribs which are equally spaced (see figure). Assuming umbrella to be a flat
circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.
There are 8 ribs in an umbrella. The area between two consecutive ribs is subtending
at the centre of the assumed flat circle.
Area between two consecutive ribs of circle =
Q
PRACTICE QUESTIONS
Q1: A car has wipers which do not overlap. Each wiper has a blade of length 25cm sweeping
through angle of 450. Find total area cleaned at each sweep of the blade.
Q2. A cow is tied with a rope of length 14m at the corner of a rectangular field of dimensions
20m x 16m. Find the area of the field in which the cow cannot graze.
Q3. A sheet of paper is in the form of a rectangle ABCD in which AB = 40 and AD = 28 cm. A
semi-circular portion with BC as diameter is cut off .Find the area of the remaining paper.
LEVEL-IV
Question 1:
Find the area of the shaded region in the given figure, if PQ = 24 cm, PR = 7 cm and O is the
centre of the circle.
It can be observed that RQ is the diameter of the circle. Therefore, RPQ will be 90º.
By applying Pythagoras theorem in ΔPQR,
RP2 + PQ2 = RQ2
(7)2 + (24)2 = RQ2
Radius of circle,
Q
Since RQ is the diameter of the circle, it divides the circle in two equal parts.
Question2:
Find the area of the shaded region in the given figure, if radii of the two concentric circles
with centre O are 7 cm and 14 cm respectively and AOC = 40°.
Radius of inner circle = 7 cm
Radius of outer circle = 14 cm
Area of shaded region = Area of sector OAFC − Area of sector OBED
Q
Question3:
From each corner of a square of side 4 cm a quadrant of a circle of radius 1 cm is cut and also
a circle of diameter 2 cm is cut as shown in the given figure. Find the area of the remaining
portion of the square.
Each quadrant is a sector of 90° in a circle of 1 cm radius.
Area of each quadrant
Area of square = (Side)2 = (4)2 = 16 cm2
Area of circle = πr2 = π (1)2
Area of the shaded region = Area of square − Area of circle − 4 × Area of quadrant
Question 4:
In the given figure, OACB is a quadrant of circle with centre O and radius 3.5 cm. If OD = 2 cm,
find the area of the
(i) Quadrant OACB
(ii) Shaded region
Q
i)
Since OACB is a quadrant, it will subtend 90° angle at O.
Area of quadrant OACB
(ii)
Area of ΔOBD
Area of the shaded region = Area of quadrant OACB − Area of ΔOBD
Q
PRACTICE QUESTIONS
Question.1
Find the perimeter of the shaded region where ADC, AEB and BFC are semi-circles on diameters
AC, AB and BC, respectively.
Question 2:
In the given figure, AB and CD are two diameters of a circle (with center O) perpendicular to
each other and OD is the diameter of the smaller circle. If OA = 7 cm, find the area of the
shaded region.
Question 3:
On a square handkerchief, nine circular designs each of radius 7 cm are made (see the given
figure). Find the area of the remaining portion of the handkerchief.
Q
Question 4:
In the given figure, a square OABC is inscribed in a quadrant OPBQ. If OA = 20 cm, find the area
of the shaded region.
Question5 :
AB and CD are respectively arcs of two concentric circles of radii 21 cm and 7 cm and center O
(see the given figure). If AOB = 30°, find the area of the shaded region
Question6:
In the given figure, ABC is a quadrant of a circle of radius 14 cm and a semicircle is drawn with
BC as diameter. Find the area of the shaded region.
Q
Q7.
The area of an equilateral triangle is 49√3. Taking each angular point as center a circle is
described with radius equal to half the length of the side of triangle as shown in figure.
Find the area of triangle not included in circle.
Q
Chapter 13
Surface area and volume
Section-A
Contents
1. Introduction
2. Surface Area of the combination of solids (any two of the solids)- cubes, cuboids,
spheres, hemispheres and right circular cylinders/cones.
3. Volumes of a combination of solids such as cubes, cuboids, spheres, hemispheres, right
circular cylinder/cones.
4. Conversion of solids from one shape to another.
5. Formulae related to this chapter
6. Solved Examples and Exercises based on them.
7. Frustum of a cone . Surface area and volume of frustum of a cone.
INTRODUCTION
Dear students let us recall the work done earlier in the previous classesGEOMETRICAL SHAPES
Plane
figures
Solid
Figures
Triangles
Cuboids/cubes
Quadrilaterals
Cylinder
Polygons
Cone
Circles
Sphere
Frustum of a cone
Q
Plane figures
1. Plane figures are two dimensional
having only length and breadth
2. Plane figures have only areas.




Solid Figures
1. Solids have definite shape and size and
it occupies a space.
2. A solid has three dimension- length,
breadth and height.
3. Solids have volumes and surface areas.
Surface areas of a combination of basic solids = sum of the curved surface areas of the
constituent basic solids.
Volume of the solid before conversion = Volume of the solid after conversion.
The volume of the solid formed by joining two basic solids will actually be the sum of the
volumes of the constituents.
It may be noted that in calculating the surface area, we have not added the surface
areas of the two constituents because some part of the surface areas disappeared in the
process of joining them.
SECTION – B
Formulae related to this chapter:Surface areas and Volume of Solids
1. Cuboid
Q
Let a, b, c be the length, breadth, and height of a cuboid respectively, then
Lateral Surface Area= 2c (a+b) square unit
Total Surface area = 2(ab+bc+ca) square unit
Volume = abc
cubic unit
2. Cube
Let a be the edge of the cube, then
Lateral surface area=
square unit
Total Surface area=
square unit
Volume of cube=
cubic unit
3. Solid Cylinder:Let r be the radius and h be the height of a solid cylinder, then
Curved (lateral) Surface Area= 2 rh square unit
Total surface area= 2 r(r+h)
square unit
Volume =
h cubic unit
4. Hollow cylinder
Let R and r be the external and internal radii, h be the height of a hollow cylinder,
then
Thickness of cylinder= R-r
External curved Surface area=2 Rh
square unit
Internal curved surface area = 2 rh
square unit
Total Surface area= 2 Rh+2 rh+2
-2
=2
Rh+rh+
Area of the base= (
)
square unit
-
Volume of the material= (
)
square unit
-
)h
cubic unit
5. Cone
let r, h, l be the radius, height and the slant height respectively of a
circular cone, then
Q
Slant height, l= √
Curved surface area= rl
square unit
Total Surface Area= r(l+r)
Volume=
h
square unit
cubic unit
6. Solid sphere
Let r be the radius of the solid sphere, then
Surface area=4
Volume=
square unit
cubic unit
7. Spherical Shell
Let R and r be the radii of the outer and inner spheres, then
Thickness of the shell= R2-R1
Volume of material =
(
) cubic unit
8. Solid Hemi- Sphere
Let r be the radius of a hemi sphere, then
Curved Surface Area= 2
square unit
Total Surface Area= 3
Volume=
square unit
cubic unit
9. Hemispherical Shell
Let R and r be the radii of the outer and inner hemispherical, then
Thickness of the shell= R-r
External Curved Surface Area=2
square unit
Internal Curved Surface Area= 2
square unit
Volume of the material=
(
)
cubic unit
Q
10 .Frustum of a cone:
Let r1 and r2 are the radii of the frustum and h&l be its height and slant height respectively
Curved Surface Area of frustum of a cone = π (r1+ r2) ℓ square unit
Where ℓ=√
T.S.A of frustum of a cone = π ℓ (r1+ r2)+ π r12+π r22 square unit
Volume of frustum of a cone = πh(r12+r22+r1r2)
cubic unit
SECTION –C
Problems on finding Surface Area and Volume of a Combination of two solids- Cubes, Cuboid,
Sphere, Right circular Cylinders and cones.
Type 1
Very short and short answers
EXAMPLE 1. If two cubes each of volume 64
surface area of the resulting cuboid .
are joined face to face, then find the
SOLUTION- Volume of a cube= 64
Volume of a cube =
Edge of the cube= 64=
=4 cm
Hence the edge of cube is 4 cm.
Now, Two cubes of same volume are joined face to face i.e. another cube has the edge of
length 4 cm. On joining them we get a cuboid whose dimensions are
Length,
l=8cm
Breadth, b=4 cm
Height,
h= 4 cm
Therefore, SURFACE AREA of the resulting cuboid= 2( l b + b h+ hl) square unit
=2 {(8×4)+(4×4)+(4×8)}
Q
=2 (32+16+32)
=2 (80)
=160
ANS
EXAMPLE 2. The diameter of a right circular cylinder is 10 cm and height is 4 cm find it’s total
Surface Area .
Solution- Diameter of the base of the right circular cylinder= 10 cm
Therefore , radius of the circular base =
=
= 5 cm
Height of the cylinder= 4cm
Total Surface Area of the right circular cylinder
= Curved Surface Area of cylinder + area of two plane circular ends
Total Surface Area
= 2 rh+
+
= 2 rh+2
=2 r(h+r)
=2 5(4+5)
=10 (9)
90
EXERCISE 1. Find the Surface Area of a cuboid whose dimensions are 8 m× 6m ×5 m .
ANS: 236 sq. m
EXERCISE 2. Edge of the cube is 5 cm.Find it’s volume in
).
ANS :125 cubic cm.
EXERCISE 3. The perimeter of one face of a cube is 20 cm, then find it’s surface area.
ANS : 150 sq.cm.
Q
EXERCISE 4. The curved surface area of a right circular cylinder is 1760
21 cm, then find it’s height .
Ans:
and it’s radius is
cm
EXERCISE 5. If a rectangular piece of paper of dimension is 60 cm × 88 cm is rolled to form a
hollow circular cylinder of height 60 cm, then find the radius of the cylinder .
Ans: 14 cm
Example 1:-Find the volume of a right circular cylinder of radius 7 cm and height 20 cm .
Solution:- radius of the circular base of the cylinder,
r = 7cm
Height of the cylinder,
h= 20 cm
Volume of the cylinder,
V= ( Area of the base× height)
,
V=
V=
h
× 7×7×20
V= 22× 140 cm3
3080
ans
Example2:Findt the volume of a right circular cone of radius 3 cm and height 14 cm .
Solution :Radius of the base of the one= 3 cm
Height ,h= 14 cm
Volume ,V=
V= ×
h
×3×3×14
V =22 × 6
Q
=132
EXERCISE1:- A cone of height 8 cm has base diameter 12 cm, then find it’s curved surface area .
ANS: 188.4 cm2
EXERCISE 2- If the base area of a cone is 51
height.
and its volume is 85
then find its vertical
ANS: 5 cm
EXERCISE3 . If the height and radius of cone are doubled then how many times will be the
volume of the cone .
ANS:8 times
EXERCISE4 .Find the volume of a cone of height 8 cm and diameter of the base 6 cm
ANS: 24π cm
EXERCISE 5. If two cones have their heights in the ratio 1:3 and radii in the ratio 3:1 then find
the ratio of their volumes.
ANs: 3:1
--------------------------------------------------------------------------------------------------EXAMPLE1 . Find the volume of a sphere of diameter d .
Solution- Volume of sphere=
Diameter = d
Radius= d/2
Volume of the sphere =
=
cubic unit
EXAMPLE 2.Find the Volume of a hemisphere of radius 15 cm .
Solution- r= 15 cm
Q
Volume of hemisphere = ½ (Volume of a Sphere)
× ×
=2250
EXERCISE 1: The ratio of the surface area of the two spheres is 9:4.Find the ratio of their
volumes
ANS:- 27:8
EXERCISE2. If the Surface Area of a sphere is 324
ANS:
, then find it’s volume is
972π
EXERCISE 3. If a solid of one shape is converted to another then the volume of new solid
a)
b)
c)
d)
Remains same
Increases
Decreases
Can’t say
Sol- a) is the correct option
EXERCISE4. The radius of a sphere is r cm. It is divided into two equal parts. The whole
Surface area of each part will be:a)
b)
c)
d)
Sol- d) is the correct option
EXERCISE5: :-An ice-cream cone consisting of a cone surmounted by a hemisphere. The radius
of the hemisphere is 3.5 cm and height of the ice-cream cone is 12.5 cm. Calculate the volume
of the ice-cream in the cone.
Q
Ans: 205.33cm3
EXAMPLE 1. If the volume and Surface Area of a solid sphere are numerically equal, then find
it’s radius .
Solution:
volume of the solid sphere= surface area of the sphere
=
r =1
r= 3 unit
EXAMPLE 2. If a solid right circular cone of height 24 cm and base radius 6 cm is melted and
recast in the shape of a sphere, then find the radius of the sphere .
Solution –
Volume of cone= Volume of Sphere
=
6×6×24 =
or
=
=
r = 6 cm
ANS
EXERCISE 1. Find the length of the diagonal of a cube that can be inscribed in a sphere of
radius 7.5 cm .
Ans: 15cm
EXERCISE 2. If a cone, a hemisphere and a cylinder have equal bases and have same height,
then find the ratio of their volumes .
ANS:1:2:3
EXERCISE 3. The surface areas of two spheres are in the ratio of 1:4. Then find the ratio of
their volumes is
Q
ANS: 1:8
EXERCISE4- The ratio of the radii of two spheres is 4:5, find the ratio of their total surface
area .
ANS: 16:25
EXERCISE 5. Find the total Surface area of a solid hemisphere of radius r .
ANS:
EXAMPLE1-If a solid sphere of radius 6 cm is melted and drawn into a wire of radius 0.2 cm,
then find the length of the wire.
Solution – radius ,r = 6 cm
Volume of the sphere ,
V=
Volume of sphere V1= Volume of wire V2
Volume of wire
R= 0.2 cm
L=?
V2=
l
= (0.2)(0.2)l
(216)=0.4l
7200=l
Length of wire= 7200cm =72 m
Q
EXAMPLE 2– Find the number of balls of radius 1 cm that can be made from sphere of radius
10 cm .
Solution- R of the sphere =10 cm
V of the sphere=
r of 1 ball=1cm
V of 1 ball =
=
Let n be no. of balls can be made from the sphere
n volume of 1 ball = Volume of sphere
n
=
Or,
n= 1000
EXERCISE1- If three solid spheres of radii 6 cm, 8cm and 10 cm are melted to form a sphere,
then the radius of the sphere so formed is
a)
b)
c)
d)
12 cm
14 cm
16 cm
24 cm
Option (a) is correct
EXERCISE2- The no. of balls of radius 2 cm that can be made from a cube of side 44 cm Is____
A)
B)
C)
D)
2500
2525
2541
2580
Option (c) is correct
Q
EXERCISE3- A cylindrical rod of iron, whose height is 4 times its radius, is melted and cast into
spherical balls of the same radius. The no. of balls cast is ____
a)
b)
c)
d)
1
2
3
4
Option (c) is correct
EXERCISE4- How many lead balls of radius 2 cm can be made from a ball of radius 4 cm
a)
b)
c)
d)
1
2
4
8
Option (d) is correct
EXERCISE 5 – A hollow cylinder of height 3 cm is melted and cast into a solid cylinder of height
9 cm. If the internal and external radii of the hollow cylinder are 1 cm and 2 cm respectively,
then radius of solid cylinder
a) 1 cm
b) 2 cm
c) 3 cm
d) 4 cm
Option (a) is correct
TYPE -2
EXAMPLE1 – The figure given below, shows a cuboidal block of wood through which a circular
cylindrical hole of the biggest size is drilled. Find the volume
of the wood left in the block ?(dimensions Cuboid:- l=70 cm,
b= 30 cm, h=30 cm )
Sol- Volume of cuboidal block of wood= l b h
Q
V1=30 30
V1=63000
Hole :- Cylindrical hole is to be drilled. Hole is of the biggest size.
V of the cylindrical hole=
R of the hole (circular end)=
V2 =
30= 15 cm
15
=49500
Volume of wood left in the block= (V1-V2)
63000-49500 =
13500
EXAMPLE2- The figure given below shows a solid trophy made of shining glass. If one cub.cm
of glass costs Rs. 0.75, find the cost of the glass for making the trophy?
(Dimensions of cylinderH= 28 cm,
Edge of the cube=28 cm)
Solution :- Volume of the solid glass trophy
=Vol. of cube + Vol. of Cylinder
= (28 28
+
)
Cost of glass for making the trophy
=
= Rs. 29400
ANS
Q
EXERCISE1- From a cube of edge 14 cm, a cone of maximum size is carved out. Find the
volume of the cone and the remaining material?
Ans:2025 cm3
EXERCISE2- A 20m deep well with diameter 7m is dug and the earth from digging evenly
spread out to form a platform 24 m by 14 m. Find the height of the platform.
Ans: 2.5 m
EXERCISE3-A cone of maximum volume is carved out of a block of wood of size 20 cm
find the volume of the cone carved out correct to one decimal place.
Ans: 523.6 cm3
EXAMPLE1 – Find what length of canvas 2m in width is required to make a conical tent of 20
m in diameter and 42 m in slant height. Also find the cost of the canvas at the rate of Rs. 60
per meter.
Sol- Slant height, l=42 m
Diameter of the circular base= 20 m
Radius= 10 m
Curved surface area of the Conical Tent=
= 1320
As the width of the canvas = 2m
Length of the canvas required =
= 610 m
Area of the canvas = lb= 610 2
1220
Cost of the canvas = 610 60=
Rs. 36,600
Q
EXAMPLE2 – A vessel is in the form of a hollow hemisphere mounted by hollow cylinder. The
diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm, find the inner
surface area of the vessel?
Sol- Diameter of the hollow hemisphere =14 cm
Radius = 7 cm
Total height of the vessel = 13 cm
Height of the hollow cylinder = 13-6 = 6 cm
Inner Surface area of the vessel = Surface area of the hollow
hemisphere+ inner Surface area of the hollow cylinder
=
+
(7)(6)
=98
572
EXERCISE1- A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of the
same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.
Ans:: 214.5 cm3
EXERCISE2- A solid metal cylinder of radius 14 cm and height 21 cm is melted and recast into
spheres of radius 3.5 cm. Calculate the no. of sphere that can be made?
Ans: 72
Exercise 3. The diameter of a metallic sphere is 6 cm. The sphere is melted and drawn into a
wire of uniform cross- section .If the length of the wire is 36 m, find its radius.
Ans: 1 mm.
EXAMPLE1- Marbles of diameter 3 cm are dropped into a cylindrical beaker containing some
water and are fully submerged. The diameter of the beaker is 12 cm. Find how many marbles
have been dropped in it if the water rises by 10 cm?
Q
Sol- Diameter of the marble = 3 cm
Radius= cm
Volume of 1 marble =
=
Diameter of the cylindrical beaker= 12 cm
Radius of the cylindrical beaker = 6cm
Height of the water raised= 10 cm
Volume of water raised =
=
Let n be the no. of marbles dropped into the beaker.
Volume of n marbles=
( ) ]=
n[
Or, n =
n= 80
=
[
( ) ]
No. of marbles = 80
TYPE 3 (Frustum of a cone)
EXAMPLE 1: If the radii of the circular bases of the frustum of a cone are 6cm , 8cm and its
slant height is 5cm, then the curved surface area of the frustum is ------(a) 1100 cm2
sol: r1 =6cm
r2 =8 cm
l= 5cm
(b)440 cm2
(c) 220 cm2
(d) 110 cm2
8 cm
5 cm
Q
curved surface area =πl(r1 +r2)
6 cm
cm2
=
=220 cm2
Example 2:A drinking glass is in the shape of frustum of a cone of height 14 cm the diameters
of its end are 4 cm and 2 cm. find the capacity of glass
Solution : radius of upper end = 2 cm
Radius of lower end = 1 cm
Height of glass
= 14 cm
Therefore,
Volume of drinking glass = 1/3π (R2 + r2 + Rr) x h
= 1/3 x 22/7 (4 +1 + 2) x 14
= 102.67 cm3
Exercise 1: : If the radii of the circular ends of a bucket 12 cm high are 15 cm and 10 cm,
then its curved surface area is
(a)
180 π cm2
(b) 240π cm2
(c) 300π cm2
(d) 325π cm2
Option (d) is correct.
Exercise 2: : If the radii of the circular ends of a bucket 45 cm high are 28 cm and 7 cm, then
the capacity of the bucket is
(a)48150 cm3
(b)48510 cm3
(c)97020 cm3
(d)24255 cm3
Option (b) is correct.
Exercise3: A cylindrical bucket 32 cm high and with radius of base 18 cm is filled with sand
.This bucket is emptied on the ground and a conical heap of sand is formed .If the height of
conical heap is 24 cm ,find the radius and slant height of the heap.
ANS: radius =36cm
slant height = 12√
Q