sheniblog-EM-SSLC Worksheets - Questions and Answers - John P A

2020-21 Academic year Worksheets

Mathematics X
Geometry and Algebra

Concepts

⋆ The inclination of a line with x axis is very important for higher level learning coordinate geometry.
There is a slope to a line. It is the measure of inclination. But mere inclination cannot determine a
perticular line. There are more more one line with same slope.

⋆ Let A(x1, y1), B(x2, y2)be two lines. Slope of this line is .y2 −y1

x2 −x1

⋆ The tan measure of the angle made by a line with the positive direction of x axis is also the slope

of this line

1) Consider the pairs in the order (1, 3), (2, 5), (3, 7), (4, 9) · · · . These are the coordinates of points

a) Write the sequence of x coordiantea and y coordinates seperately.
b) Take the term of the first sequence as x and write the algebraic form of the second sequence .
c) Take two points and find the slope of the line joining these points
d) Prove that these are the points on a line.

begintcolorbox[colback=blue!5!white,colframe=blue!75!black,title=Answers]

a) Sequence of x coordinates is 1, 2, 3, 4 · · ·
Sequence of y coordinates is 3, 5, 7, 9 · · ·

b) The sequence of y coordinates is an arithmetic sequence .xn = dn + (f − d) → y = 2x + 1

c) Slope of the line joining (1, 3), (2, 5) is = y2 −y1 = 5−3 = 2 =2
x2 −x1 2−1 1

d) Let (a, b), (c, d)be two points . Slope of the line joining these points is y2 −y1 = d−b =
x−2−x−1 c−a

(2c+1)−(2a+1) = 2(c−a) = 2
c−a c−a

e) Consider the points on the line A(a, b), B(c, d), C(e, f ). Slope of AB is = d−b = 2, Slope of BC
c−a
f −d
is = e−c = 2. These are on a line.

2) Consider the points A(2, −3), B(−5, 1), C(7, −1), D(0, 3)

a) Find the slope of the line AB
b) Find the slope of the line CD
c) Are these points the vertices of a parallelogram ?

Answers

a) Slope of AB = y− 2−y1 = 1−− 3 = 4 = − 4
x2 −x1 −5−2 −7 7

b) Slope of CD = 3−− 1 = − 4
0−7 7

c) Slope of AB and slope of CD are equal .
3−− 3
Slope of AD = 0−2 = −3

slope of BC= −1−1 = −2 = − 1
7−− −5 12 6

Side AD is not parallel to BC.Opposite sides are not parallel. ABCD is not a

parallelogram.

1

visit www.shenischool.in or whatsapp 7012498606

3) A(1, −2), B(x, 4) are the points on a line of slope x.

a) Find x
b) Write the coordinates of another point on this line
c) Find the point at which the line cut x axis
d) Find the point at which the line cut y axis

Answers

a) Slope : y2 −y1 = 3, 4−− 2 = 3, 6 = 3, 3x − 3 = 6, 3x = 9, x = 3 , B(3, 4)
x2 −x1 x−1 x−1

b) Since 3is the slope another point is C(3 + 1, 4 + 3) → C(4, 7)

c) ycoordinate of the point on x axis is 0 The point is P (x, 0) Consider A(1, −2) ,and (x, 0)

−2−0 = 3, x = 5 , P ( 5 , 0)
1−x 3 3

d) x coordinate of the point on y axis is 0. Point is Q(0, y). y−− 2 = 3, y = −5 Q(0, −5)
0−1

4) A(−4, 2), B(2, 6), C(8, 5), D(9, −7)are the vertices of a quadrilateral

a) Find the coordinates of the mid point of the sides.
b) Prove that the quadrilateral formed by the mid points is a parallelogram
c) Find the coordinaters of the point where the diagonals intersect.

Answers

a) Mid point of AB is P ( −4+2 , 2+6 )→ P (−1, 4)
2 2
2+8 ,65++222+5−2−)7 7)→→) →QR((S51,(271252,1,−)−152))
Mid point of BC is Q( 2 ,
Mid point of CD is
Mid point of AD is SR((−8+422+9 9,

b) Find slope of P Q and RS. These are found to be equal. Both are 1 . PQ is parallel to
4

RS.

Slopes of P S and QRare equal . Slopes are equal. It is a parallelogram

c) Try yourself

5) A(−4, 3), B(7, 3), C(5, 1), D(−2, 1) are the vertices of a quadrilateral.

a) Find the slope of the sides ABand CD
b) Prove that ABCD is an isosceles trapezium
c) Find the area of the trapezium
d) Calculate the perimetre of the trapezium

2

visit www.shenischool.in or whatsapp 7012498606

Answers

a) y coordinates of A, Bare equal .AB is parallel to x axis. Slope of AB is 0
y coordinates of C, Dare equal . CD is parallel to x axis. slope 0

b) Since A√B is parallel to CD , ABCD is√a parallelogra√m. √
AD = √(−2 −− 4))2 + (1 − 3)2√= 22 + 22 = 8 =√2 2 √
BC = (5 − 7))2 + (1 − 3)2 = (−2)2 + (−2)2 = 8 = 2 2
AD = BC, AB is parallel to CD. It is isosceles trapezium

c) AB =| 7 −− 4 |= 11, CD =| 5 −− 2 |= 7, Distance between the parallel sides

=| 3 − 1 |= 2

Area 1 ×2× (11 + 7) = 18
2
√√ √
d) Perimetre = 11 + 7 + 2 2 + 2 2 = 18 + 4 2

3

visit www.shenischool.in or whatsapp 7012498606

2020-21 Academic year Worksheets

Mathematics X
Geometry and Algebra

Concepts

a) The relation between the coordinates of points in a line which is common in all points is the
characteristic property of that line.The relation between x and y coordinates is named as the
equation of the line.

b) y coordinates of all points in x axis is 0. The equation y = 0 is considered as the equation of x
axis.

c) The y coordinates of all points on a line parallel to x axis are equal. For example in the line parallel
to x axis and passing through (0, 5) , y coordinates of all points are 5. Therefore y = 5 is the
equation of this line.

d) x coordinates of all points on y axis is 0. Therefore x = 0 is considered as the equation of y axis.
e) x coordinates of all points on a line parallel to y axis are equal. For example , in the line parallel to

y axis and passing through (5, 0), the x coordinates of all points are 5 . x = 5 is the equation of
this line.
f) Sometimes it is difficult to find the linear relation between the coordinates for getting the sequation.
We can use the measure slope for writing the equation successfully.
g) There are three types of lines. The line parallel to x axis, line parallel to y axis and inclined lines .
The inclination may be either to the left or right . The line inclined to the right has positive slope,
the line inclined to the left have negative slope.

1) A line is drawn parallel to x axis and passing through (0, 4).Another line is drawn parallel to y axis and
passing through (4, 0).
a) Write the equation of these lines.
b) What are the coordinates of points intersecting these lines?
c) There is a line passing through the intersecting point and the origin. Write the equation of this line

Answers

a) The equation of line passing through (0, 4) and parallel to x axis is y = 4
The equation of the line passing through (4, 0) and parallel to y axis is x = 4

b) The intersecting point of the lines x = 4 and y = 4 is (4, 4)
c) All points on the line passing through (4, 4) and (0, 0) have x coordinates and y coordinates

equal.The equation of the line is x = y

2) (1, 5), (2, 7), (3, 9), (4, 11) are the points on a line.
a) Write the property common in all points
b) Find the slope of this line
c) Take (x, y) as a point on this line and write the equation of the line using slope and another given
point
1

visit www.shenischool.in or whatsapp 7012498606

Answers

a) y coordinates of all points is 3 more than two times its x coordinates.

b) Slope = y2 −y1 = 7−5 = 2 = 2
x2 −x1 2−1 1

c) Let (x, y) be a point on the line. Take (x, y) and another point (1, 5) . Slope = y−5 = 2,
x−1
y − 5 = 2(x − 1), y − 5 = 2x − 2

The equation of the line is 2x − y − 2 + 5 = 0, 2x − y + 3 = 0

3) The sum of the x coordinates and y coordinates of points on a line is 0. Write the equation of this line

a) Write the coordinates of some points on this line . Write the equation of this line.
b) Find the point where this line cut x axis .
c) Find the coordiantes of the point where this line cut y axis
d) Calculate the slope of the line.

Answers

a) (1, −1), (−3, 3), (2, −2)
Equation of the line is x + y = 0

b) When the line cut x axis , y = 0. x + 0 = 0, x = 0 . The point is (0, 0)

c) When the line cut y axis , x = 0. 0 + y = 0, y = 0 . The point is (0, 0)

d) Consider two points (1, −1), (2, −2). Slope = y2 −y1 = −2−− 1 = −1 = −1
x2 −x1 2−1 1

4) Two measures are necessary for finding the equation of a line.Knowing the slope of a line and and a point
through which the line passes we can write the equation of the line easily.
The slope of a line is 3. The line passes through (1, 2).

a) If (x, y) is a point on this line write the statement connecting the slope and the given point (1, 2)
b) Write the equation of this line.
c) At what point the line cut x axis ?
d) At what point the line cut y axis ?

Answers

a) Slope = y2 −y1 , y−2 =3
x2 −x1 x−1

b) y − 2 = 3(x − 1), y − 2 = 3x − 3, 3x − y − 3 + 2 = 0, 3x − y − 1 = 0

c) When the line cut x axis , y = 0

3x − 0 − 1 = 0, 3x = 1, x = 1 . The point is ( 1 , 0)
3 3

d) When the line cut y axis , x = 0
3 × 0 − y − 1 = 0, y = −1. The point is (0, −1)

5) Consider the lines x = 3, x = 7, y = 4, y = 8

a) Draw a rough diagram showing the lines
b) Write the name of the geometric figure enclosing these lines .
c) Write the coordinates of the vertices.
d) Calculate area and perimetre

2

visit www.shenischool.in or whatsapp 7012498606

Answers
a) See the diagram

b) AB =| 7 − 3 |= 4, BC =| 8 − 4 |= 4 ABCD is a square.
c) A(3, 8), B(7, 8), C(7, 4), D(3, 4)
d) Area = 42 = 16, Perimetre = 16

3

visit www.shenischool.in or whatsapp 7012498606

2020-21 Academic year Worksheets

Mathematics X
Geometry and Algebra

Concepts

a) The relation between the coordinates of points in a line which is common in all points is the
characteristic property of that line.The relation between x and y coordinates is named as the
equation of the line.

b) y coordinates of all points in x axis is 0. The equation y = 0 is considered as the equation of x
axis.

c) The y coordinates of all points on a line parallel to x axis are equal. For example in the line parallel
to x axis and passing through (0, 5) , y coordinates of all points are 5. Therefore y = 5 is the
equation of this line.

d) x coordinates of all points on y axis is 0. Therefore x = 0 is considered as the equation of y axis.

e) x coordinates of all points on a line parallel to y axis are equal. For example , in the line parallel to
y axis and passing through (5, 0), the x coordinates of all points are 5 . x = 5 is the equation of
this line.

f) Sometimes it is difficult to find the linear relation between the coordinates for getting the sequation.
We can use the measure slope for writing the equation successfully.

g) There are three types of lines. The line parallel to x axis, line parallel to y axis and inclined lines .
The inclination may be either to the left or right . The line inclined to the right has positive slope,
the line inclined to the left have negative slope.

1) A line cut x axis at (12, 0) , cut y axis at (0, 5).

a) What is the slope of this line ?
b) Write the equation of this line.
c) Calculate the area of the triangle formed by the line and coordinate axes
d) What is the length of the line in between coordinate axes ?
e) What is the perpendicular distance from origin to the line?

Answers

a) Slope = y2 −y1 = 5−0 = − 5
x2 −x1 0−12 12

b) Let P (x, y) be a point on this line. Consider the points (0, 5), (x, y).
y−5
x−0 = − 5
12
12(y − 5) = −5x, 12y − 60 = −5x, 5x + 12y = 60

c) This is a right triangle with perpendicular sides 12 and 5. Area of the triangle is 1 × 12 × 5 =
2

30

√
d) Length of the line in between the axes is = 122 + 52 = 13

e) Let h be the perpendicular distance from origin to the line. It is the altitude of the triangle .

Equating the area ,

1 × 13 × h = 30, h = 60 1
2 13

visit www.shenischool.in or whatsapp 7012498606

2) Consider the equation 2x + 3y = 6

a) What is the point at which the line cut x axis ?
b) What is the point at which the line cut y axis ?
c) what is the slope of this line?
d) How does the slope is related to the coefficients of x and y in the equation?

Answers

a) When the line cut x axis y = 0.
2x + 3 × 0 = 6, x = 3. The point is (3, 0)

b) When the line cut y axis x = 0.
2 × 0 + 3y = 6, y = 2
The point is (0, 2)

c) Slope of the line is y2 −y1 = 2−0 = −2
x2 −x1 0−3 3

d) If ax + by = c is the equation of the line then slope is − a
b

3) Consider the equation 3x + 2y = 6

a) What is the slope of this line?
b) Write the equation of another line having the same slope . Suggest a method to write the equation of

lines parallel to the given line.
c) Write the equation of the line parallel to the line 3x + 2y = 6 and passing through (1, 1)

Answers

a) Slope = −3
2

b) The slope of a line is determined by the coefficient of x and coefficient of y in the equation
ax + by = c.That is the slope is independent of the constant number c. By giving different
values for c we can write equation of lines parallel to eachother.
3x + 2y = 1, 3x + 2y = 2, 3x + 2y = −1 etc are the lines parallel to the given line.Note
that the multiples of a and b also possible instead of a and b keeping slope contant. Example
6x + 4y = 5

c) Slope is = −3 .
2
y−1 −3
Let P (x, y) be a point on this line. x−1 = 2 .

(y − 1) × 2 = −3 × (x − 1), 3x + 2y = 5

4) Consider the equation x + y + 1 = 0

a) Find the point at which the line cut x axis.
b) Find the point at which the line cut y axis .
c) Find the slope of the line
d) Calculate the area of the triangle enclosed by the line with the coordinate axes .
e) What is the length of median of the triangle from the origin.

2

visit www.shenischool.in or whatsapp 7012498606

Answers

a) When the line cut x axis , y = 0.x + y + 1 = 0 → x + 0 + 1 = 0, x = −1 . The point is
(−1, 0). This is on x axis.

b) When the line cut y axis , x = 0.x + y + 1 = 0 → 0 + y + 1 = 0, y = −1 . The point is
(0, −1). This is on y axis.

c) Slope of this line is 0−− 1 = −1
−1−0

d) The triangle is a right triangle with perpendicular sides 1. Area 1 ×1×1= 1
2 2

e) Mid point of the hypotenuse is (− 1 , − 1 ). Length of median is √1
2 2 2

5) A line makes 45◦ with the positive direction of x axis .

a) What is the slope of this line?
b) The line passes through (3, 5). Write the equation of this line
c) At what point the line cut x axis ?
d) At what point the line cut y axis .

Answers

a) Slope= tan 45◦ = 1

b) Let (x, y) be a point on the line. y−5 = 1 , x − 3 = y − 5, x − y = −2
x−3

c) When the line cut x axis , y = 0, x = −2. The point is (−2, 0)

d) When the line cut y axis , x = 0, y = 2. The point is (0, 2)

3

visit www.shenischool.in or whatsapp 7012498606

2020-21 Academic year Worksheets

Mathematics X
Geometry and Algebra

Concepts

a) The relation between the coordinates of points in a line which is common in all points is the
characteristic property of that line.The relation between x and y coordinates is named as the
equation of the line.

b) y coordinates of all points in x axis is 0. The equation y = 0 is considered as the equation of x
axis.

c) The y coordinates of all points on a line parallel to x axis are equal. For example in the line parallel
to x axis and passing through (0, 5) , y coordinates of all points are 5. Therefore y = 5 is the
equation of this line.

d) x coordinates of all points on y axis is 0. Therefore x = 0 is considered as the equation of y axis.

e) x coordinates of all points on a line parallel to y axis are equal. For example , in the line parallel to
y axis and passing through (5, 0), the x coordinates of all points are 5 . x = 5 is the equation of
this line.

f) Sometimes it is difficult to find the linear relation between the coordinates for getting the sequation.
We can use the measure slope for writing the equation successfully.

g) There are three types of lines. The line parallel to x axis, line parallel to y axis and inclined lines .
The inclination may be either to the left or right . The line inclined to the right has positive slope,
the line inclined to the left have negative slope.

1) Consider the numbers (1, 2), (5, 7)

a) What is the slope of the line passing through these points ?
b) Write the equation of this line
c) At what point the line cut x axis ?
d) At what point the line cut y axis ?

Answers

a) Slope = y2 −y1 = 7−2 = 5
x2 −x1 5−1 4

b) Let(x, y) be a point on the line. Take (x, y) and (1, 2) , y−2 = 5 ,
x−1 4
(y − 2) × 4 = (x − 1) × 5, 5x − 4y + 3 = 0

c) When the line cut x axis , y = 0. 5x − 4 × 0 + 3 = 0, 5x = −3, x = −3 . The point is
5
−3
( 5 , 0)

d) When the line cut y axis , x = 0. 5 × 0 − 4y + 3 = 0, −4y = −3, y = −3 = 3 . The
−4 4

point is (0, 3 )
4

2) A line cut the coordinate axes at (4, 0) and (0, 4)
a) What is the slope of this line ?
b) What is the angle made by this line with the1 positive direction of x axis ?
c) Write the equation of this line.
d) What is the perpendicular distance from origin to the line ?

visit www.shenischool.in or whatsapp 7012498606

Answers

a) Slope = y2 −y1 = 4−0 = −1
x2 −x1 0−4

b) The line encloses an isosceles right triangle with coordinate axes. Acute angle of this triangle
is 45◦. So the line makes 180 − 45 = 135◦ with the positive direction of x axis .

c) Let (x, y) be a point on this line. y−0 = −1, y = −1(x − 4), x + y = −4
x−4

d) Let h be the perpendicular dist√ance from origin to the hypotenuse of the triangle formed by
line and axes. H12 y×po4te√nu2s×e is 4 2
h, h = √
1 × 4 × 4 = 1√6 = 22
2 42

3) A line passes through (4, 0) and (0, 5)

a) What is the slope of this line?

b) Show that the equation of this line is x + y = 1.
4 5

c) What is the equation of the line passing through (a, 0) and (0, b)?

Answers

a) Slope= y2 −y1 = 5−0 = − 5
x2 −x1 0−4 4

b) Let (x, y) be a point on this line. y−0 = − 5
x−4 4
4y = −5(x − 4), 4y = −5x + 20, 5x + 4y = 20

Dividing both sides by 20 we get x + y = 1
4 5

c) The equation will be x + y =1
a b

4) Consider the points (6, 0) and (0, 6)

a) What are the coordinates of the mid point of the line joining these points?
b) The line passing through the mid point makes 60◦ angle with the positive direction of x axis.What is

the slope of this line?
c) Write the equation of this line.
d) At what point this line cut y axis ?

Answers

a) A(6, 0) on x axis and B(0, 6) on y axis. The mid point of AB is (3, 3) √
3
b) The slope of the line making 60◦ with the positive direction of x axis is tan 60 =

y−3 √√ √
c) Le√t (x√, y) be the po√int on the line . x−3 = 3, y − 3 = 3(x − 3), −3 = 3x −

3 3, 3x − y = 3 3 − 3 √√

d) When the line cut y axis x = 0. y = 3 − 3 3. The point is (0, 3 − 3 3)

5) A line passes through (a, 0) and (0, b).

a) If the mid point of the line in between the coordinate axes is (x1, y1) then show that the equation of
y
this line is x + y1 = 2
x1

b) If this line passes through (6, 0) and (0, 6)then write the equation of the line .

2

visit www.shenischool.in or whatsapp 7012498606

Answers

a) x1 = a , y1 = b . Mid point is ( a , b ), a = 2x1, b = 2y1
2 2 2 2
−b
Since (a, 0) and (0, b) are the given points , slope= a . Let (x, y) be the point on the line,

y−0 = − b
x−a a
ay = −b(x − a) , bx + ay = ab.

Dividing both sbiedceosmbeysa2bxx,w1 e+ge2tyy1xa = y = 1
The equation b +
x y
=1, x1 y1 = 2

b) Equation is x + y = 2 → x+y = 6
3 3

3

visit www.shenischool.in or whatsapp 7012498606

2020-21 Academic year Worksheets

Mathematics X
Geometry and Algebra

Concepts

a) The relation between the coordinates of points in a line which is common in all points is the
characteristic property of that line.The relation between x and y coordinates is named as the
equation of the line.

b) y coordinates of all points in x axis is 0. The equation y = 0 is considered as the equation of x
axis.

c) The y coordinates of all points on a line parallel to x axis are equal. For example in the line parallel
to x axis and passing through (0, 5) , y coordinates of all points are 5. Therefore y = 5 is the
equation of this line.

d) x coordinates of all points on y axis is 0. Therefore x = 0 is considered as the equation of y axis.

e) x coordinates of all points on a line parallel to y axis are equal. For example , in the line parallel to
y axis and passing through (5, 0), the x coordinates of all points are 5 . x = 5 is the equation of
this line.

f) Sometimes it is difficult to find the linear relation between the coordinates for getting the sequation.
We can use the measure slope for writing the equation successfully.

g) There are three types of lines. The line parallel to x axis, line parallel to y axis and inclined lines .
The inclination may be either to the left or right . The line inclined to the right has positive slope,
the line inclined to the left have negative slope.

1) There are some cows and hens in a farm. Total number of heads is 15 and total number of legs is 50

a) If the number of cows is x and number of hens is y then form equations
b) These are the equation of lines. Find the intersecting point of these lines .
c) Calculate the number of cows and hens

Answers (1)
(2)
a) x + y = 15, 4x + 2y = 50
b)

4x + 2y = 50
x + y = 15
1 × 2 → 2x + 2y = 30 → eqn3

2x + 2y = 30 (3)

1 − 3 gives

2x = 20, x = 10, y = 5

The intersecting point is (10, 5)

c) Number of cows is 10 , number of hens is 5.
1

visit www.shenischool.in or whatsapp 7012498606

2) Following are the equation of lines
x − y = −1, x − y = 1, x + y = 7, x + y = 3

a) Make the pairs of parallel lines
b) Find the intersecting points of non parallel sides
c) Prove that the quadrilateral enclosed by the lines is a square .
d) Calculate the area of the square.

Answers

a) (x − y = −1, x − y = 1) and (x + y = 7, x + y = 3) are the pairs of parallel lines.

b) x − y = −1, x + y = 7 → 2x = 6, x = 3, y = 4 . The intersecting point is (3, 4)
x − y = −1, x + y = 3 → 2x = 2, x = 1, y = 2. The intersecting point is (1, 2)
x − y = 1, x + y = 7 → 2x = 8, x = 4, y = 3. The point is (4, 3).
x − y = 1, x + y = 3 → 2x = 4, x = 2, y = 1. The point is (2, 1)

c) Conside√r the quadrilaterals A(1, 2)√, B(2, 1), C(4, 3),√D(3, 4).
AB = √(2 − 1)2 + (1 − 2)2 = √12 + (−1)2 = √2
CD = √(4 − 3)2 + √(3 − 4)2 = 12 + (−1)2 = 2 , AB = CD
BC = 2 2, AD√= 2 2. opp√osite sides are equal.
Diagonal AC = 10, BD = 10. Diagonals are equal . It is a rectangle .
√

d) Area = AB × BC = 2 × 22 = 8sq.unit

3) Three lines x = 3, y = 4, 4x + 3y = 36encloses a polygon.

a) Suggest a suitable name to this polygon.
b) Find the vertices of this polygon.
c) Calculate the area
d) What is the radius of the circle passing through the vertices of the polygon.
e) What are the coordinates of the circumcentre.

Answers

a) x = 3 is the equation of the line parallel to yaxis passing through (3, 0). y = 4 is the
equation of the line passing through (0, 4) and parallel to x axis . The line 4x + 3y = 36
intersect these two lines . The lines encloses a right angled triangle.

b) Vertices are A(3, 4) right anged at A.
Let B be the intersecting point of x = 3 and 4x + 3y = 36. This implies 4 × 3 + 3y =
36, 3y = 24, y = 8 , B(3, 8).
Let C be the intersecting point of y = 4 and 4x + 3y = 36. This implies 4 × x + 3 × 4 =
36, 4x = 24, x = 6 , C(6, 4).

c) AB =| 8 − 4 |= 4, AC =| 6 − 3 |= 3

Area = 1 × 4× 3 = 6sq.cm
2
√
d) Length of hyptenuse is 32 + 42 = 5. Radius of circumcircle is 2.5

e) Centre of the circumcircle is at the mid point of the hypotenuse. It is ( 3+6 , 8+4 ) = ( 9 , 6)
2 2 2

4) Consider the points A(3, 4), B(−1, 2)

a) The points equdistant from A and B are on a line. Form the equation of that line.
b) Write the equation of line passing through A(3, 4) and B(−1, 2)
c) Find the mid point of AB.

2

visit www.shenischool.in or whatsapp 7012498606

Answers

a) L√et (x, y) be the point equid√iatant from A and B.
(x − 3)2 + (y − 4)2 = (x + 1)2 + (y − 2)2

Squaring on both sides
(x − 3)2 + (y − 4)2 = (x + 1)2 + (y − 2)2 . On expanding and simplifying we get

2x + y = 5.This is a straight line.

b) Let (x, y) be a point on the line. y−4 = 2−4 , y−4 = 1 , x − 2y = −5. This is the line
x−3 −1−3 x−3 2
passing through A(3, 4), B(−1, 2).

c) Here we find the mid point by solving the equation of lines x − 2y = −5 and 2x + y = 5.
Solving these equations we get x = 1, y = 3. Mid point is (2, 3)
(Note Mid point can be determined by the method discussed earlier)

5) Consider the lines x + y = 4, x + y = −2, x − y = 2, x − y = −2

a) Write the pairs of parallel lines.
b) Find the intersecting points of the lines (draw brough figure)
c) Check whether this is a square or not.

Answers

a) Parallel pairs are x + y = 4, x + y = −2 and x − y = 2, x − y = −2
b) Point of intersection of x + y = 4 and x − y = 2 . 2x = 6, x = 3, y = 1. Point is (3, 1)

x + y = 4 and x − y = −2 . 2x = 2, x = 1, y = 3. Point is (1, 3)
x + y = −2 and x − y = 2 . 2x = 0, x = 0, y = −2. Point is (0, −2)
x + y = −2 and x − y = −2 . 2x = −4, x = −2, y = 0. Point is (−2, 0)

c) Draw rough diagram.Do yourself .You can refer the diagram drawn in Geogebra

3

visit www.shenischool.in or whatsapp 7012498606

2020-21 Academic year Worksheets

Mathematics X
Geometry and algebra

Concepts

⋆ The equation of the circle with centre at the origin and radius r is x2 + y2 = r2
⋆ If the centre is (a, b)and radius r then equation of the circle is (x − a)2 + (y − b)2 = r2

1) There is a circle with centre at the origin and radius 4

a) What are the points where the circle cut the axes?
b) If P (x, y) is a poi√nt on t√he circle then write the equation of the circle
c) Chec√k whe√ther (2 2, 2 2) a point on this cicle.
d) If (2 2, 2 2) is a point on the circle what are the other three points on the circle make the vertices

of a square

Answers

a) The points where the circle cut axes are A(4, 0), B(0, 4), C(−4, 0), D(0, −4)
√
b) Distance from O(0, 0) to P (x, y) is (x − 0)2 + (y − 0)2 = 4

x2 + y2 = 42, x2 + y2 = 16

√√ √√
c) When x = 2 2 and y = 2 2, x2 + y2 = 16 become√s (2 √2)2 + (2 2)2 = 8 + 8 = 16.
The equation satisfies for these coordinate values. (2 2, 2 2) is a point on this circle
√√ √√ √ √√ √
d) Points are (2 2, 2 2),(−2 2, 2 2),(−2 2, −2 2),(2 2, −2 2)

2) Following are the equations of circle
x2 + y2 = 1, x2 + y2 = 2, x2 + y2 = 3, x2 + y2 = 9, x2 + y2 = 17

a) Find the centre and radius of each circle
b) Find the coordinates of the point where the circle cut the axes
c) Calculate the area and perimetre of the square joining these points

Answers

a) x2 + y2 = 1 → x2 + y2 = 1√√232C22eCCneetrnnettrr(ee0((,000,,)00,))ra,, drraaiuddsiiuurss =1 √
x2 + y2 = 2 → x2 + y2 = r= √2
x2 + y2 = 3 → x2 + y2 = r=
3

x2 + y2 = 917→→xx2 2++yy2 2==3√2 C1e7n2trCee(n0tr,e0()0,,r0a)di,ursardiu=s 3 = √
x2 + y2 = r 17

b) Circle x2 + y2 = 1 cut the axes at (√1, 0), (0, 1)√, (−1, 0)√, (0, −1) √
Circle x2 + y2 = 2 cut the axes at (√2, 0), (0, √2), (−√2, 0), (0, −√2)
Circle x2 + y2 = 3 cut the axes at ( 3, 0), (0, 3), (− 3, 0), (0, − 3)

Circle x2 + y2 =√9 cut the ax√es at (3, 0√), (0, 3), (−3,√0), (0, −3)
cut the axes at ( 17, 0), (0, 17), (− 17, 0), (0, − 17)
√ √
c) Side of the square is 2. Area = 2 sq1 uare unit.perimetre 42 unit

Do other cases yourself

visit www.shenischool.in or whatsapp 7012498606

√
3) There is a semicircle with diametre on x axis and centre ar the origin.One end of the diametre is (2 5, 0)

a) What are the points where the semicircle cut the axes
b) What is the side of the square ABCD
c) What are the coordinates of the vertices of the square?
Answers

√√ √
a) The points are (2 5, 0), (−2 5, 0), (0, 2 5)

√
b) Joi√n OC. If OB = x, BC = 2x, OC = 2 5

(2 5)2 = x2 + (2x)2, 5x2 = 20, x2 = 4, x = 2. Side of the square is 4.
c) A(−2, 0), B(2, 0), C(2, 4), D(−2, 4)
4) The circle x2 + y2 = 3touches the sides of a square.

a) What are the points where the circle cut the axes?
b) What are the vertices of the square ?
c) Calculate the area and perimetre

2

visit www.shenischool.in or whatsapp 7012498606

Answers

a) x2 + y2 = 3 √→ x2 + y2 =√√32 √ √
Points are A( 3, 0), B(0, 3), C(− 3, 0), D(0, − 3)
√√ √√ √√ √√
b) P ( 3, 3), Q(− 3, 3), R(− 3, − 3), S( 3, − 3)
√√ √ √
c) P Q =| 3 −− √3 |= 2 √3. Area = (2 3)2 = 12
Perimetre = 4 × 2 3 = 8 3

5) AP is the tangent from an exterior point A to the circle x2 + y2 = 36.∠OAP = 30◦

a) What is the radius of the circle ?
b) What are the points the circle cut the axes ?
c) Find the coordinates of A
d) What is the length of tangent?
e) Find the coordinates of P

3

visit www.shenischool.in or whatsapp 7012498606

Answers

a) x2 + y2 = 36 → x2 + y2 = 62
Radius 6

b) Points are (6, 0), (0, 6), (−6, 0), (0, −6)

c) △OP A is a 30◦ − 60◦ − 90◦ triangle. Since OP = 6, OA = 12.

A(12, 0)

√
d) P A = 6 3
√
e) △OM√P is a 30◦ − 60◦ − 90◦ .Since OP = 6 , OM = 3, P M = 3 3.

P (3, 3 3)

4

visit www.shenischool.in or whatsapp 7012498606

2020-21 Academic year Worksheets

Mathematics X
Geometry and Algebra

Concepts

⋆ The equation of the circle with centre at the origin and radius r is x2 + y2 = r2
⋆ If the centre is (a, b)and radius r then equation of the circle is (x − a)2 + (y − b)2 = r2

1) Centre of a circle is (3, 2) and radius 1.
a) Write the equation of the circle.
b) What are the coordinates of the end points of the diametre parallel to x axis ?
c) What are the coordinates of the ends of the diametre parallel to y axis.
d) Find the area and perimetre of the square formed by joining the ends of these two perpendicular
diametres .
Answers
a) (x − 3)2 + (y − 2)2 = 12
x2 − 6x + 9 + y2 − 4y + 4 = 1, x2 + y2 − 6x − 4y + 12 = 0
b) Look at the rough diagram

AB is parallel to x axis and mid point is (3, 2).
A(2, 2), B(4, 2)

c) CD is parallel to y axis and mid point is (3, 2).

C(3, 3), B(3, 1)

d) Since diagonal is 2 Side of the square is √2 = √ Area = √22 = 2, perimetre
√2 2.

=4 2

1

visit www.shenischool.in or whatsapp 7012498606

2) Consider the circles (x − 3)2 + (y − 4)2 = 1, (x + 3)2 + (y − 4)2 = 1, (x + 3)2 + (y + 4)2 = 1,
(x − 3)2 + (y + 4)2 = 1
a) With the help of coordiante axes draw a rough diagram
b) Write the equation of line joining the centres
c) Suggest a suitable name to the geometric figure formed by enclosed by these lines .
d) Calculate the area of the quadrilateral so fomed .
Answers
a) Centres are A(3, 4), B(−3, 4), C(−3. − 4), D(3, −4)

b) The equation of the line passing through A and B is y = 4
The equation of the line passing through B and C is x = −3
The equation of the line passing through C and D is y = −4
(The equation of the line passing through A and D is x = 3
You can write the equation of lines passing through A an C , B and D . Try yourself)

c) Rectangle.
d) AB =| 3 −− 3 |= 6, AD =| 4 −− 4 |= 8. Area = 6 × 8 = 48, Perimetre = 28.
3) The end points of the diametre of a circle are (−3, 7) and (7, 13)
a) Find the coordinates of its centre.
b) What is the diametre of the circle?
c) Write the equation of the circle.

2

visit www.shenischool.in or whatsapp 7012498606

Answers

a) Centre ( −3+7 , 7+13 )
2 2

Centre (2, 10)

√ √√
b) D√iametre of the circle = (7 −− 3)2 + (13 − 7)2 = 102 + 62 = 136 =

2 34

√
c) Radius of the circle is 34.
Equation of the circle is (x − 2)2 + (y − 10)2 = √342

x2 − 4x + 4 + y2 − 20y + 100 − 34 = 0

x2 + y2 − 4x − 20y + 70 = 0

4) The end points of the diametre of a circle are (−1, 2), (3, 2)

a) Write the coordinates of the centre
b) Write the equation of the circle
c) What is the point where the circle touches x axis?
d) What are the points where the circle cut y axis?

Answers

a) C(1, 2)

b) Radius of the circle is 2
Equation is (x − 1)2 + (y − 2)2 = 22,
x2 − 2x + 1 + y2 − 4y + 4 = 4, x2 + y2 − 2x − 4y + 1 = 0

c) At x axis , y = 0. x2+02−2x−4×0+1 = 0, x2−2x+1 = 0, (x−1)2 = 0, x = 1
Note this second degree equation has only one solution. The line x axis is a tangent
to the circle. It touches only one point (1, 0)

d) At y axis , x = 0

Equation o√f the circle bec√omes y2−4y+1 = 0. Solving this second degree equation,
y = 2 + 3, y = √2 − 3
√
Points are (0, 2 + 3) and (0, 2 − 3)

5) The end points of the diametre of a circle are (4, 2), (2, 4)

a) Find the centre and radius of the circle .
b) Write the equation of the circle.
c) If the tangent through (4, 2)to the circle make 45◦ with the positive direction of x axis. Write

the equation of tangent .
d) At what point the tangent intersect x axis ?

3

visit www.shenischool.in or whatsapp 7012498606

Answers

a) Centre √(3, 3), √
Radius (4 − 3)2 + (2 − 3)2 = 2

b) (x − 3)2 + (y − 3)2 = √22, x2 − 6x + 9 + y2 − 6y + 9 = 2
x2 + y2 − 6x − 6y + 16 = 0

c) Slope of tangent is tan 45◦ = 1
y−2
Equation of tangent is x−4 = 1, y − 2 = x − 4, x − y = 2

d) At x axis y = 0, x = 2 . The point is (2, 0)

1

1Prepared by John P.A, mob- 9847307721, email- [email protected], [email protected]

4

visit www.shenischool.in or whatsapp 7012498606

2020-21 Academic year Worksheets

Mathematics X
Geometry and Algebra

Concepts

⋆ The equation of the circle with centre at the origin and radius r is x2 + y2 = r2
⋆ If the centre is (a, b)and radius r then equation of the circle is (x − a)2 + (y − b)2 = r2

1) Consider the equation of the circle x2 + y2 − 6x − 10y − 15 = 0
a) write this equation in the form (x − a)2 + (y − b)2 = r2
b) Write the centre and radius of this circle.
c) Write coordinates of four points in this circle.
Answers
a) x2 + y2 − 6x − 10y − 15 = 0 → x2 + y2 − 6x − 10y = 15
x2 − 6x + 32 + y2 − 10y + 52 = 15 + 32 + 52
(x − 3)2 + (y − 5)2 = 49
(x − 3)2 + (y − 5)2 = 72.
b) Centre(3, 5), Radius = 7
c) Consider the diametre parallel to x axis.The end points of the diametre are on the
circle.
These are A(3 − 7, 5), B(3 + 7, 5) → A(−4, 5), B(10, 5)
Similarly we can write the end points of the vertical diametre
C(3, 5 + 7), D(3, 5 − 7) → C(3, 12), D(3, −2)

2) Consider the equation of the circle x2 + y2 − 6x − 10y + 9 = 0
a) Write this equation in the form (x − a)2 + (y − b)2 = r2
b) Write the centre and radius of this circle.
c) Write the coordinates of points at which the circle touches the x axis
d) Write the coordinates of the end points of the diametres parallel to x axis and parallel to y
axis

1

visit www.shenischool.in or whatsapp 7012498606

Answers
a) x2 + y2 − 6x − 10y + 9 = 0 → x2 + y2 − 6x − 10y = −9
x2 − 6x + 32 + y2 − 10y + 52 = −9 + 32 + 52
(x − 3)2 + (y − 5)2 = 52
b) Centre (3, 5), radius 5
c) At the point on x axis , y = 0
(x − 3)2 = 52 − 52, x − 3 = 0, x = 3. The point at which the circle touches x
axis is (3, 0)
d) Let A and B be the ends of the diametre parallel to x axis . A(3 − 5, 5) →
A(−2, 5), B(3 + 5, 5) → A(8, 5)
Let C and D be the ends of the diametre parallel to y axis . C(3, 5+5) → A(3, 10),
D(3, 0) → D(3, 0)

3) Consider the quation x2 + y2 + 6x + 8y = 0
a) Write the equation in the form (x − a)2 + (y − b)2 = r2
b) Write the centre and radius of the circle.
c) What are the points where the circle intersect the coordinate axes ?
d) What is the distance between centre and origin of coordinates ?

Answers
a) x2 + 6x + 32 + y2 + 8y + 42 = 0 + 32 + 42
(x + 3)2 + (y + 4)2 = 52, (x −− 3)2 + (y −− 4)2 = 52
b) Centre (−3, −4) , Radius 5
c) When the circle cut x axis y = 0
(x + 3)2 + (0 + 4)2 = 25, (x + 3)2 = 9, x + 3 = 3 or −3
If x + 3 = 3, x = 0, If x + 3 = −3, x = −6
Points on x axis are (0, 0) and (−6, 0)
When the circle cut y axis x = 0
(0 + 3)2 + (y + 4)2 = 25, (y + 4)2 = 16, y + 4 = 4 or −4
If y + 4 = 4, y = 0, If (y + 4 = −4, y = −8
Points on y axis are (0, 0) and (0, −8)

√
d) The distance from origin to the center is 32 + 42 = 5

4) The centre of a circle is (3, 0) and radius 6
a) Write the equation of the circle .
b) What are the points the circle intersect the axes ?
c) Is (0, 5) a point of this circle. How can we realize it .

2

visit www.shenischool.in or whatsapp 7012498606

Answers

a) (x − 3)2 + (y − 0)2 = 62, x2 − 6x + 9 + y2 = 36, x2 + y2 − 6x − 27 = 0

b) When the circle cut x axis, y will be 0.(x − 3)2 = 36, x − 3 = 6, x = 9. Points

are (9, 0), (−3, 0)

When the circle cut y axis. x = 0. √ Points are √ √
(0 − 3)2 + y2 = 36, y2 = 27, y = ±3 3. (0, 3 3), (0, −3 3)

c) x2 + y2 − 6x − 27 = 0 becomes 02 + 52 − 6 × 0 − 27= 25 − 27 ̸= 0. This is
not a point on the circle.

√
5) The vertices of a triangle are (4, 2), (8, 2), (6, 2 3 + 2).

a) What kind of triangle is this ?
b) Find the centre of its circumcircle.
c) What is the radius of its circumcircle?

Answers

√
a) A(4, 2), B(8, 2), C(6, 2 3 + 2)
√√
AB is parallel to x axis. AB =| 8 − 4 |= 4, AC = (6 − 4)2 + (2 3 + 2 − 2)2 =
√
22 + 12 = 4, BC = 4. This is an equilateral triangle.

b) Since it is an equilateral triangle centroid itself is the circumcentre.It is the point which divides

the median in the ratio 2 : 1 √

Mid point AB is (6, 2).Centroid is 6, 6+2 3 .This is the centroid of the circle.
3

c) Radius of the circle is 2 √ = √4
3 ×2 3 3

Note : If a is the side of an equilateral triangle , its circumradius will be √a . Here r = √4

33

1

1Prepared by John P.A, mob- 9847307721, email- [email protected], [email protected]

3

visit www.shenischool.in or whatsapp 7012498606

10

Polynomials

ബഹുപദങ്ങൾ

visit www.shenischool.in or whatsapp 7012498606

2020-21 Academic year Worksheets

Mathematics X
Polynomials

Concepts

⋆ If the polynomial p(x) is the product of the polynomials q(x) and
r(x)then we say that the polynomilals q(x) and r(x) are the factors
of the polynomial p(x)

⋆ If the first degree polynomial (x − a) is a factor of the polynomial
p(x) then p(a) = 0: that is a is a solution of the equation p(x) = 0

⋆ If the polynomial p(x) can be split into first degree factors as p(x) =
(x − a1)(x − a2)(x − a3) · · · (x − an) then a1, a2, a3 · · · anare the
solutions of the equation p(x) = 0.

⋆ In the second degree polynomial p(x) if p(a) = 0 then x − a will be
the factor of p(x).
If p(−a) = 0 then x + a will be a factor.

1) The sides of a rectangle are (x − 3) and (x + 1)
a) Find the area a(x)
b) If x = 4then what is its area ?
c) If x = 0 is it possible to get a rectangle ? Why?
d) What is the condition for x to get a rectangle?
Answers
a) a(x) = (x − 3)(x + 1) = x(x + 1) − 3(x + 1) = x2 − 2x − 3
b) If x = 4 then a(4) = 42 − 2 × 4 − 3 = 16 − 8 − 3 = 5
c) If x = 0 side becomes a negative number. Square cannot be
drawn
d) x > 3 .

1

visit www.shenischool.in or whatsapp 7012498606

2) Sides of a rectangular box are x − 1, x + 2 and x + 3.

a) Calculate the volume v(x).
b) If x = 2 then what is its volume?
c) Is it possible to make the box for x = 1
d) What is the condition for x to make the recctangular box?

Answers
a) v(x) = (x − 1)(x + 2)(x + 3)
v(x) = (x2 + x − 2)(x + 3) = x3 + 4x2 + x − 6
b) v(2) = 23 + 4 × 22 + 2 − 6 = 8 + 16 + 2 − 6 = 20

c) If x = 1then side becomes 0.Box cannot be made .

d) x > 1

3) The difference between breadth ,length and length of diagonal of a rectangle
differ by 1

a) If breadth is x what is its length and diagonal of the rectangle?
b) Make an equation connecting breadth ,length and diagonal in the form

p(x) = 0

c) Find the solution of the equation p(x) = 0
d) Find the length , breadth and diagonal.

Answers
a) If breadth is x then length is x + 1, diagonal x + 2

b) (x + 2)2 = x2 + (x + 1)2, x2 + 4x + 4 = x2 + x2 + 2x + 1,

x2 − 2x − 3 = 0

√
−b± b2−4ac
c) x =
2a √
x = −(−2)± (−2)2−4×1×−3
2

x = 3, −1

d) If x = 3then bradth = 3, length = 4,diagonal= 5

4) Consider the polynomial p(x) = x2 − 7x + 12
a) Write p(x) = (x − a)(x − b).

2

visit www.shenischool.in or whatsapp 7012498606

b) Write p(x)as the product of two first degree polynomials.
c) Find the solution of the equation p(x) = 0
Answers

a) x2 − 7x + 12 = (x − a)(x − b) = x2 − (a + b)x + ab
a + b = 7, ab = 12

(a − b)2 = (a + b)2 − 4ab
(a − b)2 = (7)2 − 4 × 12 → a − b = ±1
If a−b = 1 then , a−b = 1, a+b = 7 → 2a = 8, a = 4, b = 3
(What change occur in the answer corresponding to a − b =
−1.Try yourself )
b) p(x) = (x − 4)(x − 3)
c) p(x) = 0 → (x − 4)(x − 3) = 0
x = 3, 4

5) Consider the polynomial p(x) = x3 − 4x2 + 2x + k
a) If x is a factor then find x.
b) x − 1 is a first degree factor of p(x)then what is k?
c) Use k for becoming x − 1 a factor and write the polynomial
d) Is (x + 1 a factor of this polynomial .

Answers
a) k = 0
b) If x − 1 = 0 then p(1) = 0
13 − 4 × 12 + 2 × 1 + k = 0, k = 1
c) p(x) = x3 − 4x2 + 2x + 1
d) p(−1) = (−1)3 − 4(−1)2 + 2(−1) + 1 = −1 − 4 − 2 + 1 ≠ 0
x + 1 is not a factor

1

1Prepared by John P.A, mob- 9847307721, email- [email protected], [email protected]

3

visit www.shenischool.in or whatsapp 7012498606

2020-21 Academic year Worksheets

Mathematics X
ബ പദ ൾ

Concepts

⋆ If the polynomial p(x) is the product of the polynomials q(x) and r(x)then we say that the
polynomilals q(x) and r(x) are the factors of the polynomial p(x)

⋆ If the first degree polynomial (x − a) is a factor of the polynomial p(x) then p(a) = 0: that is a is
a solution of the equation p(x) = 0

⋆ If the polynomial p(x) can be split into first degree factors as p(x) = (x − a1)(x − a2)(x −
a3) · · · (x − an) then a1, a2, a3 · · · anare the solutions of the equation p(x) = 0.

⋆ In the second degree polynomial p(x) if p(a) = 0 then x − a will be the factor of p(x).
If p(−a) = 0 then x + a will be a factor.

1) Consider the polynomial p(x) = x2 − 8x + 12
a) If p(x) = (x − a)(x − b)then what is a + b and ab
b) Find a, b and write p(x) as the product of two first degree factors
c) Find the solution of the equation p(x) = 0

Answers
a) x2 − 8x + 12 = (x − a)(x − b) = x2 − (a + b)x + ab,a + b = 8, ab = 12
b) (a − b)2 = (a + b)2 − 4ab
(a − b)2 = 82 − 4 × 12 = 16, a − b = 4.
a + b = 8, a − b = 4 → 2a = 12, a = 6, b = 2
p(x) = (x − 6)(x − 2)
c) p(x) = 0 → (x − 6)(x − 2) = 0, x = 6, x = 2

2) If p(x) = x3 − 4x2 + 6x − kthen
a) Find k such that x − 1 a factor of p(x)
b) Write the polynomial . Is (x + 1) a factor of p(x)
c) What is the speciality of the coefficients of p(x) having x − 1 a factor
d) Write three polynomials having x − 1 a factor

Answers

a) Since (x − 1)a factor p(1) = 0.
13 − 4 × 12 + 6 × 1 − k = 0, 1 − 4 + 6 − k = 0, k = 3

b) p(x) = x3 − 4x2 + 6x − 3
p(−1) = (−1)3 − 4 × (−1)2 + 6 × (−1) − 3 = −1 − 4 − 6 − 3 = −14 ̸= 0
p(−1) ̸= 0. Therefore (x + 1)not a factor.

c) Sum of the coefficients will be zero (x − 1)
d) It can be ant polynomial with sum of th1e coefficients is zero .

x3 − x2 + x − 1,2x3 − 4x2 + 5x − 3, x3 − 4x2 + 2x + 1

visit www.shenischool.in or whatsapp 7012498606

3) Consider the polynomials p(x) = x3 + 1 , q(x) = x3 + x2 + x + 1
a) Find p(−1) and q(−1)
b) What is the factor common to both the polynomials
c) Find r(x) = p(x) + q(x)
d) what is the first degree factor of r(x)
Answers
a) p(−1) = (−1)3 + 1 = −1 + 1 = 0
q(−1) = (−1)3 + (−1)2 + (−1) + 1 = −1 + 1 − 1 + 1 = 0
b) p(−1) = 0, q(−1) = 0 implies (x − 1) is a factor of both .(x − 1)is the common factor
c) r(x) = (x3 + 1) + (x3 + x2 + x + 1) = 2x3 + x2 + x + 2
d) r(−1) = 2(−1)3 + (−1)2 + (−1) + 2 = −2 + 1 − 1 + 2 = 0
x + 1 is the factor of r(x)

4) x2 − 1 is the factor of p(x) = a3 + bx2 + cx + d
a) Find p(1), p(−1)
b) Show that a = −c, b = −d
c) Write a polynomial having x2 − 1 a factor
Answers
a) x2 − 1 = (x − 1)(x + 1)
(x − 1), (x + 1) are the factors of p(x)
p(−1) = 0, p(1) = 0
b) p(1) = 0 → a + b + c + d = 0
p(−1) = 0 → a − b + c − d = 0, a + c = b + d
a + b + c + d = 0 → 2(a + c) = 0, a + c = 0, a = −c, b = −d
c) The condition a = −c, b = −d should be satisfied. Example 3x3 − 4x2 − 3x + 4

5) If p(x) = x3 − 8 then
a) Check whether x − 2 a factor of p(x)
b) Write a first degree factor of x3 − 27
c) What is the second degree factor of x3 − 27

2

visit www.shenischool.in or whatsapp 7012498606

Answers
a) p(2) = 23 − 8 = 8 − 8 = 0
x − 2 is a factor of p(x)
b) q(x) = x3 − 27 implies q(3) = 33 − 27 = 27 − 27 = 0
x − 3 is a factor of x3 − 27
c) x3 − 27 = x3 − 33 = (x − 3)(ax2 + bx + c)
.ax2 + bx + c is the second degree factor .
x3 − 27 = (x − 3)(ax2 + bx + c
x(ax2 + bx + c) − 3(ax2 + bx + c) = ax3 + bx2 + cx − 3ax2 − 3bx − 3c =
ax3 + (b − 3a)x2 + (c − 3b)x − 3c
Equating the coefficients a = 1, (b − 3a) = 0, (c − 3b = 0), −27 = −3c, c = 9
c − 3b = 0 → 9 − 3b = 0, b = 3,
x3 + 3x + 9is the second degree factor

1

1Prepared by John P.A, mob- 9847307721, email- [email protected], [email protected]

3

visit www.shenischool.in or whatsapp 7012498606

2020-21 Academic year Worksheets

Mathematics X
Polynomials

Concepts

⋆ If the polynomial p(x) is the product of the polynomials q(x) and r(x)then we say that the
polynomilals q(x) and r(x) are the factors of the polynomial p(x)

⋆ If the first degree polynomial (x − a) is a factor of the polynomial p(x) then p(a) = 0: that is a is
a solution of the equation p(x) = 0

⋆ If the polynomial p(x) can be split into first degree factors as p(x) = (x − a1)(x − a2)(x −
a3) · · · (x − an) then a1, a2, a3 · · · anare the solutions of the equation p(x) = 0.

⋆ In the second degree polynomial p(x) if p(a) = 0 then x − a will be the factor of p(x).
If p(−a) = 0 then x + a will be a factor.

1) Consider the polynomial p(x) = 3x2 + 4x + 1

a) Write p(x)as the product of two first degree factors
b) Find the solution of the equation p(x) = 0

Answers

a) p(x) = 3x2+4x+1 = k(x−a)(x−b) = k(x2−(a+b)x+ab) = kx2−k(a+b)x+kab

k = 3, a +b = − 4 , ab = 1
3 3
(a − b)2 = (a + b)2 − 4ab → ( −4 )2 − 1 4
3 4 3 = 9

a − b = 2
a − b =
3 −4 −−31−,3b1
2 3 → = −1
3 , a + b = a = )(x −− 1)

p(x) = k(x − a)(x − b) → 3(x = 3( 3x+1 )(x + 1) = (x + 1)(3x + 1)
3

b) x + 1 = 0 → x = −1, 3x + 1 = 0 → x = −1
3

2) Consider the equation p(x) = x3 + 4x2 + x − 7

a) Check whetherx − 1 a factor of this polynomial or not
b) If not what should be subtracted from p(x) to get another polynomial q(x) in which x − 1 is a factor
c) Write q(x) as the product of three first degree factors
d) Write the solution of the equation q(x) = 0.

1

visit www.shenischool.in or whatsapp 7012498606

Answers

a) p(1) = 13 + 4 × 12 + 1 − 7 = 6 − 7 = −1 ≠ 0
x − 1 is not a factor

b) Since p(1) = −1 ,on subtracting −1 from p(x) we get (x − 1) a factor .
q(x) = x3 + 4x2 + x − 6

c) x3 + 4x2 + x − 6 = (x − 1)(ax2 + bx + c) .
Equating the constant terms both sides −6 = −c, c = 6
Equating coefficients of x on both sides 1 = c − b → 1 = 6 − b, b = 5
equating coefficients of x2on both sides −a + b = 4, −a + 5 = 4, a = 1
ax2 + bx + c = x2 + 5x + 6 = (x + 2)(x + 3)
q(x) = (x + 1)(x + 2)(x + 3)

d) q(x) : (x + 1)(x + 2)(x + 3) = 0, x = −1, −2, −3 are the solutions.

3) Consider the second degree polynomial x2 − 20x + 91.This is the area of a rectangle with the sides the
first degee polynomials.

a) What are the sides of this rectangle.
b) What is the condition for getting a rectangle .
c) What is the polynomial representing the perimetre of the rectangle.

Answers

√
p(x) : x2 − 20x + 91 = 0, x = −b± b2 −4ac
a) 2a

x = 13, 7

p(x) = (x − 7)(x − 13)

sides are x − 7, x − 13

b) Side should be greater than 0.
x − 13 > 0 → x > 13

c) Perimetre = 2(x − 13 + x − 7) = 2(2x − 20) = 4x − 40

4) When a first degree polynomial is added to x3 + 2x2 we get a polynomial p(x) such that x2 − 1 is a factor.

a) Write two first degee factors of p(x)
b) What is the first degree factor to be added?
c) What are the solutions of the equation p(x) = 0

Answers

a) x2 − 1 = (x − 1)(x + 1) implies x − 1, x + 1 are the factors of p(x)

b) Term to be added is ax + b.
p(x) = x3 + 2x2 + ax + b
Since (x − 1) is a factor sum of coefficients is 0.
1 + 2 + a + b = 0, a + b = −3
Since x + 1 is a factor p(−1) = 0
−1 + 2 − a + b = 0, a − b = 1
a + b = −3, a − b = 1 → 2a = −2, a = −1, b = −2
േ പദം ax + b = −x − 2
p(x) = x3 + 2x2 − x − 2

c) One factor is x2 − 1 . This is a second degree factor.Other factor is px + q.
x3 + 2x2 − x − 2 = (x2 − 1)(px + q)
we get q = 2, p = 1by equating the coefficients .Third factor is x + 2

2

visit www.shenischool.in or whatsapp 7012498606

5) Consider the equation p(x) = x2 + 4x + k
a) If k = 0 write the first degree factors of p(x)
b) If k = 4 what are the facors of this polynomial ?
c) What is the condition for k to get two first degree factors of p(x)?
Answers
a) If k = 0 , p(x) = x2 + 4x. x is a factor. x + 4 is also a factor.
b) If x = 4 then x2 + 4x + k = x2 + 4x + 4 = (x + 2)(x + 2).Both factors are x + 2.
c) p(x) = x2 + 4x + k .
(x − a), (x − b) are the factors x2 + 4x + k = (x − a)(x − b) = x2 − (a + b)x + ab
a + b = −4, ab = k . (a − b)2 = 42 − 4k. If k is greater than 4 , (a − b)2become a
negative number. It is meaningless. kshould be less than or equal to 4.

1

1Prepared by John P.A, mob- 9847307721, email- [email protected], [email protected]

3

visit www.shenischool.in or whatsapp 7012498606

2020-21 Academic year Worksheets

Mathematics X
Polynomials

Concepts

⋆ If the polynomial p(x) is the product of the polynomials q(x) and r(x)then we say that the
polynomilals q(x) and r(x) are the factors of the polynomial p(x)

⋆ If the first degree polynomial (x − a) is a factor of the polynomial p(x) then p(a) = 0: that is a is
a solution of the equation p(x) = 0

⋆ If the polynomial p(x) can be split into first degree factors as p(x) = (x − a1)(x − a2)(x −
a3) · · · (x − an) then a1, a2, a3 · · · anare the solutions of the equation p(x) = 0.

⋆ In the second degree polynomial p(x) if p(a) = 0 then x − a will be the factor of p(x).
If p(−a) = 0 then x + a will be a factor.

1) Write a polynomial p(x) such that p(1) = p(−2) = p(0) = 0

a) What are the first degree factors of p(x)
b) What should be added to p(x) to get a polynomial in which x + 1 is a factor?

Answers

a) Since p(1) = 0 , x − 1 is a factor. Since p(−2) = 0, x + 2 is a factor .Since p(0) = 0, x
is a factor .
p(x) = x(x − 1)(x + 2), p(x) = x3 + x2 − 2x First degree factors are x − 1, x + 2, x

b) Since x + 1 is a factor p(−1) = 0 .
p(−1) = (−1)3 + (−1)2 − 2 × (−1) = −1 + 1 + 2 = 2 ̸= 0. −2should be subtracted .

2) Consider the polynomial p(x) = 4x2 − 16x + 15

a) Write p(x) in the form (x − a)(x − b) and find a + b, ab
b) Find a − b
c) Write p(x)as the product of two first degree factors
d) Find the solution of the equation p(x) = 0

Answers

a) 4x2 − 16x + 15 = k(x − a)(x − b) = k(x2 − (a + b)x + ab) = kx2 − k(a + b)x + kab

Equating the coefficients of x2 on both sides k = 4

16 = k(a + b) → a+ b = 16 = 4
4
→ 15
kab = 15 ab = 4

b) (a − b)2 = (a + b)2 − 4ab = 42 − 4 × 15 = 1, a − b = 1
4

c) a + b = 4, a − b = 1 → 2a = 5, a = 522x,2−b 3=) 3

p(x) = 4(x − 5 )(x − 3 ) = 4( 2x−5 )( 2 (2x − 5)(2x − 3)
2 2 2
=

d) p(x) = 0 → 2x − 3 = 0, x = 3 .2x − 5 = 0 → x = 5
2 2

1

visit www.shenischool.in or whatsapp 7012498606

3) Consider the polynomial p(x) = xn + 1
a) What are the numbers suitable for n getting x + 1 a factor of p(x)?
b) Can x − 1 a factor of p(x) for any n ?
c) Can x2 − 1 a factor of p(x)?

Answers
a) n should be odd .If n is odd p(−1) = (−1)n + 1 = −1 + 1 = 0.Therefore (x + 1) is a
factor.
b) Whether n is odd or even P (1) ≠ 0.x − 1 is not a factor
c) x2 − 1 = (x + 1)(x − 1).
x + 1, x − 1 are not factors .x2 − 1 is not a factor

4) Consider the polynomial p(x) = x2 + 6x + k
a) If k = 0then what are the first degree factors of p(x)?
b) What is the value of k to get two equal first degree factors ?
c) What are the values of k not for occuring a first degree factor to this polynomial?
d) If k = 8 what are the first degree factors of p(x)?

Answers

a) If k = 0 , p(x) = x2 + 6x → x(x + 6). First degree factors are x, x + 6
b) x2 + 6x + k = x2 + 2 × 3 × x + 32 → (x + 3)2.For this k = 9 . x + 3, x + 3 are equal

the factors
c) x2 + 6x + k = (x − a)(x − b) → a + b = −6, ab = k

(a − b)2 = (a + b)2 − 4ab → (a − b)2 = (−6)2 − 4 × k
(a − b)2 = 36 − 4k.
If k > 9 then (a − b)2become a negative number,it is not possible. If k > 9 no first degee
factor exist.
d) If k = 8then p(x) = x2+6x+8 = x2+4x+2x+8 = x(x+4)+2(x+4) = (x+4)(x+2)
First degree factors are (x + 4), (x + 2)

5) Consider the polynomial p(x) = ax2 − 2bx + c
a) If x − 1 is a factor of p(x) prove that a, b, c are in an arithmetic sequence.
b) Write twp polynomials in the form ax2 − 2bx + c such that a, b, c are in an arithmetic sequence .
c) If x2 − 1 is a factor of p(x) then what is a + b ?

Answers

a) If x − 1 is a factor then p(1) = 0.
a × 12 − 2b × 1 + c = 0, a − 2b + c = 0
a + c = 2b, a + c = b + b → b − a = c − b → a, b, c are in an arithmetic sequence .

b) If a = 4, b = 3, c = 2then 4x2 − 6x + 2.
c) x2 − 1 = (x − 1)(x + 1), x − 1, x + 1 are the factos .p(1) = 0 → a − 2b + c = 0

p(−1) = 0 → a + 2b + c = 0
adding these equations 2a + 2c = 0, a + c = 0

1

1Prepared by John P.A, mob- 9847307721, email- [email protected], [email protected]

2

visit www.shenischool.in or whatsapp 7012498606

2020-21 Academic year Worksheets

Mathematics X
Polynomials

Concepts

⋆ If the polynomial p(x) is the product of the polynomials q(x) and r(x)then we say that the
polynomilals q(x) and r(x) are the factors of the polynomial p(x)

⋆ If the first degree polynomial (x − a) is a factor of the polynomial p(x) then p(a) = 0: that is a is
a solution of the equation p(x) = 0

⋆ If the polynomial p(x) can be split into first degree factors as p(x) = (x − a1)(x − a2)(x −
a3) · · · (x − an) then a1, a2, a3 · · · anare the solutions of the equation p(x) = 0.

⋆ In the second degree polynomial p(x) if p(a) = 0 then x − a will be the factor of p(x).
If p(−a) = 0 then x + a will be a factor.

1) Consider the polynomial p(x) = x3 + 4x2 + x − 6
a) Find p(1).Is x − 1 a factor of this polynomial ?
b) What is the quotient when p(x) is divided by x − 1?
c) Write the quotient as the product of two first degree factors
d) Find the solution of the equation p(x) = 0
or

Write x3+4x2+x−6 as the product of three first degree factors .Also solve the equation x3+4x2+x−6 =
0.

Answers
a) p(1) = 13 + 4 × 12 + 1 − 6 = 1 + 4 + 1 − 6 = 0
Since p(1) = 0 we can write (x − 1) is a factor
b) Let ax2 + bx + c is the quotient.x3 + 4x2 + x − 6 = (x − 1)(ax2 + bx + c)
x3 + 4x2 + x − 6 = x(ax2 + bx + c) − (ax2 + bx + c) = ax3 + (b − a)x2 + (c − b)x − c
Equating the coefficients a = 1, b − a = 4 → b = 4 + a = 4 + 1 = 5, c − b = 1 → c =
1+b=1+5=6
Quotient is x2 + 5x + 6
c) x2 + 5x + 6 = x2 + 2x + 3x + 6 = x(x + 2) + 3(x + 2) = (x + 2)(x + 3)
d) p(x) = (x + 1)(x + 2)(x + 3) , p(x) = 0 → (x + 1) = 0or(x + 2) = 0or (x + 3) = 0
x = −1, −2. − 3

2) Consider the polynomial p(x) = ax4 + bx3 + cx2 + dx + e
a) Write a fourth degree polynomial having x − 1 a factor
b) If x + 1 is a factor what is the relation between the coefficients of p(x)
c) Using these conditions write a fourth degree polynomial having x2 − 1 as a factor.

1

visit www.shenischool.in or whatsapp 7012498606

Answers

a) If x − 1 is a factor then p(1) = 0.That is sum of the coefficients is 0.
p(x) = x4 + 2x3 + 3x2 + 4x − 10

b) Since x + 1 is a factor p(−1) = 0.
a − b + c − d + e = 0 That is a + c + e = b + d
Choose five numbers as coefficients . a = 1, b = 2, c = 3, d = 4, e = 2
q(x) = x4 + 2x3 + 3x2 + 4x + 2

c) Combining the conditions a + b + c + d + e = 0 and a + c + e = b + d, we can write
2(b + d) = 0, b + d = 0, b = −d
Also a + c + e = 0
Example r(x) = 4x4 + 5x3 + 3x2 − 5x − 7.

3) In the second degree polynomial p(x) p( 1 ) = 0, p( 1 ) = 0, constant term 4.
2 3

a) If p(x) = ax2 + bx + 4 then find a, b

b) Write the polynomial .What should be added to this polynomial to get another polynomial in which x−1
a factor.

Answers

a) p( 1 ) = 0, a( 1 )2 + b( 1 ) + 4 = 0
2 2 2
a b
4 + 2 = −4, a + 2b = −16

a( 1 )2 + b( 1 ) + 4 = 0, a + b = −4 a + 3b = −36
3 3 9 3
Solving the equations a + 2b = −16, a + 3b = −36 we get a = 24, b = −20

b) p(x) = 24x2 − 20x + 4
p(1) = 8, −8 should be added

4) Consider the polynomial x2 + kx + 6
a) If x − 1 is a factor what is k?
b) Write other first degree factor.
c) Find the solution of x2 − 7x + 6 = 0

Answers

a) If x − 1 is a factor sum of the coefficients will be 0 . k = −7
b) x2 − 7x + 6 = x2 − x − 6x + 6 = x(x − 1) − 6(x − 1) = (x − 1)(x − 6). Other factor

is x − 6
c) x = 1, x = 6

5) Consider the polynomial p(x) = (x + 1)(x + 2)(x + 3) + k
a) If p(−1) = 10then what is k ?
b) Write the polynomial using the number k
c) Change the constant term only to make the x − 1 a factor of the polynomial

Answers

a) p(−1) = 10 → k = 10
b) p(x) = (x + 1)(x + 2)(x + 3) + 10 = x3 + 6x2 + 11x + 6
c) 6 should be changed to −18

2

visit www.shenischool.in or whatsapp 7012498606

11

Statistics

സ്ഥിതിവിവരക്കണക്ക്

visit www.shenischool.in or whatsapp 7012498606

2020-21 Academic year Worksheets

Mathematics X
statistics

Concepts
a) Statistics is the study of numerical data collected from a group , its analysis and interpretation .
b) Mean is an average calculated by dividing the sum of the numerical data by the number of

observations.It touches each and every data collected from the group.
c) The middle observation of the arrangement in the increasing or decreasing order is called

median.Comparing to mean it touches the observations comes in the middle.
d) We use the concept of arithmetic sequence for calculating median .
1) The table shows the consumption of elecricity in houses in a region.

a) If the houses are arranged according to the increasing order of consumption which house comes in
the middle .In which class it belongs to?

b) If the consumptions in the median class are in an arithmetic sequence what is the consumption of 17
th house?

c) What are the consumption amount of the houses comes in the middle?
d) Calculate the median consumption of electricity?

1

visit www.shenischool.in or whatsapp 7012498606

Answers
a) Look at the table given below

Total number of houses is 40. Therfore 20 th and 21 houses comes in the middle.These
houses belongs to the class 110 − 120

b) There are 10 houses in the class and the width consumption is 10 unit. If 10 unit is divided

equally among 10 houses each one’s share is 1.

Cosnsumption of 17 th house is 110 + 1 = 110 + 0.5 = 110.5
2

c) 20th and 21st consumption units comes in the middle.These are 4 th and 5 th terms of the
arithmetic sequence sequence with first term 110.5 and common difference 1.
x4 = 110.5 + 3 × 1 = 113.5, x5 = 114.5

d) Median = 113.5+114.5 = 114
2

2) The table gives the distribution of marks obtained by the students in a class in an examination

a) Prepare a table for calculating the median.
b) In which calss the middle marks occur
c) What will be the mark of 13th student according to the assumption of calculating median .
d) What are the marks comes in the middle?
e) Calculate median .

2

visit www.shenischool.in or whatsapp 7012498606

Answers
a) Look at the table given below

b) Since n = 36,even , 18 th and 19 th marks comes in the middle. These marks are in the
class 20 − 30

c) 10 marks are divided equally among 10 students. Each one’s ahare is 1.

Mark of 13 th student is 20 + 1 = 20.5
2

d) 18 th mark is the 8 th term of the arithmetic sequence with first term 20.5and common
difference 1.
x6 = f + 5d = 20.5 + 5 × 1 = 25.5, x7 = 26.5

d) Median = 25.5+26.5 = 26
2

3) (SSLC 2019 Model )
The table shows the marks obtained by the students in a class.

a) If the scores are arranged in the ascending order what are the positions of scores comes in the
middle.

b) What is the mark of 15 th student?

c) Calculate the median.

3

visit www.shenischool.in or whatsapp 7012498606

Answers
a) Look at the table.

Sine n = 41, odd 41+1 = 21 st score comes in the middle.
2

b) Middle score belongs to the class 20 − 30.There are 10 marks in this class and 10 students

in it. Dividing marks equally among 10 children each one’s share is 1.

15th score is 20.5

c) Marks in the median class are in an arithmetic sequence. f = 20.5, d = 1
Seventh term of this sequence is f + 6d = 20.5 + 6 × 1 = 26.5
Median = 26.5

4) (SSLC 2020 )
marks obtained by students in a class are given below

a) If the scores are arranged in the increasing order what is the position of the mark comes in the middle.
b) What is the mark obtained by 12 th student?
c) Calculate median.

4

visit www.shenischool.in or whatsapp 7012498606

Answers
a) Table

Since n = 41, odd, 21 mark comes in the middle.It is in the class 20 − 30.
b) 12 th mark is 20 + 0.5 = 20.5
c) In the arithmetic sequence with f = 20.5, d = 1 twentyfirst mark will be the median. It is

10th term of the sequence . x10 = f + 9d = 20.5 + 9 × 1 = 29.5
Median is 29.5

1

1Prepared by John P.A, mob- 9847307721, email- [email protected], [email protected]

5

visit www.shenischool.in or whatsapp 7012498606