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

a) What is the area of the outer square?
b) What is the area of inner square?
c) What is the area of the shaded triangle ?
d) If a fine dot is placed into the figure then what is the probability of falling the dot in the shaded triangle?

Answers

a) One side of outer square = 28 = 7cm
4
Area= 72 = 49sq.cm

b) One side of inner square 20 = 5cm
4
Area 52 = 25sq.cm

c) Sum of the area of the triangles in between the squares is

49 − 25 = 24sq.cm

Area of shaded part= 24 = 6sq.cm
4

d) Probability= 6
49

3) The mid points of the two sides and one vertex of a square are joined in such a way as to get a triangle
which is coloured in the picture.

a) If the side of the square is a, what is are of unshaded triangles ?

b) What is the area of the shaded triangle?

c) If a fine dot is placed into the figure then what is the probability of falling the dot in the coloured
traingle?

Answers

a) Total area of unshaded triangles = ( 1 ×a× a ) × 2 + 1 × a × a
2 2 2 2 2
a2 a2 5a2
= 2 + 8 = 8

b) Area of shaded part= a2 − 5a2 = 3a2
8 8

c) Probability= 3a2 ÷ a2 = 3
8 8

4) O is the center of the circle of diametre AB.
There is another circle with diametre OB.If r is the radius of the small circle

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a) What is the radius of the big circle ?
b) Find the area of big circle and small circle.
c) If a fine dot is placed into the figure what is the probability of falling the dot in the shaded part.

Answers

a) 2r

b) Area of small semicircle πr2
2
π(2r)2
Area of big semicircle 2 = 2πr2

c) Area of coloured part 2πr2 − πr2 = 3 πr2
2 2
3
Probability= 4

5) Square ABCD is drawn in triangle P QR.Also QD = DC = CR .If the side of the square is a then

a) What is the altitude of the triangle P QR to the side P Q
b) What is the area of the triangle P QR?
c) If a fine dot is placed into the figure then what is the probability of falling the dot in the shaded square?

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Answers

a) Triangle P M Q, Triangle ADQ are similar triangles .Height P M
a
h = a+ 2
a
3aa
2
h =

b) Area of triangle P QR is = 1 × QR × PM = 1 × 3a × 3a = 9a2
2 2 2 4

c) Probability= 4
9

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2020-21 Academic year Works

Mathematics X
mathematics of Chance

32

Concepts

a) There is a concept known as the Fundamental principle of counting.If a process can be completed
in m different ways, another concept can be completed in n ways then both processes can be
completed in m × n ways.

b) Let us see an example. The journey from Kochi to Mumbai can be completed in 4 ways. Journey
from Mumbai to Delhi can be completed in 3 ways.The journey fom Kochi to Delhi can be completed
in the following ways .
(Rail,Rail),(Rail ,Road), (Rail , Air)
(Road,Rail),(Road, Road), (Road, Air)
(Air,Rail),(Air ,Road), (Air , Air)
(Sea ,Rail),(Sea ,Road), (Sea, Air)
Total number of ways is 4 × 3 = 12.

Worksheet 32

1) A boc contains three paper slips carrying numbers 2, 3, 4 . Another box contains paper slips carrying

fractions 1 , 1 , 1 .One is taken from each box at random
2 3 4

a) How many pairs are possible?

b) What is the probability of getting the product of numbers in each pair a natural number?

c) Wha is the probability of not getting the numbers in the pair whose product is not a natural number?

Answers

a) Number of pairs= 3 × 3 = 9

(2, 1 ), (2, 1 ), (2 1 )
2 3 4
1 1 1
(3, 12 ), (3, 31 ), (3 14 )
(4, 2 ), (4, 3 ), (4 4 )

b) Pairs with the product a natural number are (2, 1 ), (3, 1 ), (4, 1 ), (4, 1 )
2 3 4 2

There are four such pairs.

Probability of getting the product a natural number is = 4
9

c) Probability of not getting the product a natural number is 1 − 4 = 5
9 9

2) Manju has three ornaments :Green , Red and Blue ear rings and chains.She ware it in different ways.

a) How many ways she can ware the ornaments?
b) What is the probability of waring ornaments of same colour?
c) What is the probability of wearing the ornaments of different colours?

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Answers

a) Number of pairs 3 × 3 = 9
(Green, Green),(Green ,Red),(Green,Blus)
(Blue, Green),(Blue ,Red),(Blue ,Blue)
(Red, Green),(Red,Red),(Red,Blue)

b) (Green,Green),(Red,Red),(Blue,Blue)

Probability= 3 = 1
9 3

c) Probability of wearing different colours is 1 − 1 = 2
3 3

3) A box contains 4 black balls and 3 white balls. Another box contains 5 black balls and 3 white balls. One
from each box is taken at random.

a) How many pair of balls are possible ?
b) What is the probability of getting both balls black?
c) What is the probability of getting both balls white?
d) What is the probability of getting balls of different colours?

Answers

a) Total number of possible selections = (3 + 4) × (5 + 3) = 7 × 8 = 56

b) Probability of getting both black 4×5 = 20
56 56

c) Probability of getting both white 3×3 = 9
56 56

d) Probability of getting balls of different colours (4×3)+(3×5) = 27
56 56

4) A box contains four paper slips carrying numbers 1, 2, 3, 4.Another box contains paper slips carrying numbers
1, 2, 3. One from each box is taken at random and entered as pairs.

a) How many pairs are possible ?
b) What is the probability of getting a pair with the product of the digits odd?
c) What is the probability of getting a pair with the product of the digits even?

Answers

a) Number of pairs4 × 3 = 12
(1, 1), (1, 2), (1, 3)
(2, 1), (2, 2), (2, 3)
(3, 1), (3, 2), (3, 3)
(4, 1), (4, 2), (4, 3)

b) Pairs of getting product odd are (1, 1)(1, 3), (3, 1)(3, 3)

Probability 4 = 1
12 3

c) Probability of getting product even = 1 − 1 = 2
3 3

5) There are 30boys and 20girls in 10A. There are 15boys and 25girls in 10B.
One student is selected from each class at random.

a) How many ways the selections can be made ?
b) What is the probability of getting both boys?
c) What is the probability of getting both girls?

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Answers

a) Total number of pairs (20 + 30) × (15 + 25) = 50 × 40 = 2000

b) Probability of selecting both boys= 30×15 = 450 = 9
2000 2000 40

c) Probability of getting both girls= 20×25 = 500 = 1
2000 2000 4

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04

Second Degree
Equations

രണ്ടാംകൃ‍തി
സമവാക്യങ്ങൾ

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2020-21 Academic year Worksheets

Mathematics X
Second Degree Equations

33
Concepts

Let us learn second degree equations.This is a tool for solving problems in various mathematical
situations.Theory of second degree equation is not the scope of this unit. Here we discuss the
methods to solve equations and its applications.

Worksheet 33

1) Form the equations in the following cases.

a) The sum of a number and its square is 12

b) When a number is subtracted from its square results 20

c) The sum of the square of a number and two times that number is 63

d) Product of two consecutive odd numbers is 63.

e) The sum of a number and its reciprocal is 10 .
3

Answers

a) If the number is x then x2 + x = 12

b) If the numebr is x then x2 − x = 20

c) If the numebr is x then x2 + 2x = 63

d) Numbers are x, x + 2then x(x + 2) = 63, x2 + 2x = 63

e) If the number is xthen x + 1 = 10
x 3
x2 +1
x = 10
3
3(x2 + 1) = 10x,

3x2 − 10x + 3 = 0

2) The square of a number is 16.

a) What are the numbers ?
b) Take the number as x and form an equation
c) Can the square of a real number −16? Explain.

Answers

a) Numbers are 4, −4
b) If the number is x then x2 = 16

c) No real number exist with its square a negative number. The square of −4 and the square
of +4 is 16.
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3) The sum of a number and its square is 30.
a) If the number is x, form an equation.
b) What is the positive numberxന്?
c) Can more than one number satisfying this condition?

Answers
a) If the number is x then x2 + x = 30
b) x = 5
c) 52 + 5 = 30, (−6)2 + (−6) = 30. Numbers are 5, −6

4) x is an odd number greater than 1.
a) What are the odd numbers nearer to x
b) If the product of those numbers is 45, form an equation.
c) Find the numbers.

Answers
a) Odd number is x. The numbers nearer to it are x − 2, x + 2
b) (x − 2)(x + 2) = 45
x2 − 4 = 45, x2 = 49
c) x = 7
Numbers are 5, 9

5) If the sides of a square are reduced by 1 , the area becomes 100.
a) If the side of the first square before reducing is x, form an equation.
b) Find the side of the square.
c) What will be the perimetre of the new square.

Answers
a) If one side is xthen (x − 1)2 = 100
√
b) x − 1 = 100 = 10,x = 11
c) Perimetre decreases by 4

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2020-21 Academic year Works

Mathematics X
Second Degree Equations .

34
Concepts

Let us learn second degree equations.This is a tool for solving problems in various mathematical
situations.Theory of second degree equation is not the scope of this unit. Here we discuss the
methods to solve equations and its applications.

Worksheet 34
1) The chords AB and CD meet at a point P inside the circle.

CD = 21cm, P C = 5cm.

a) What is P D?
b) If P A = x + 1 and P B = x − 1then form an equation
c) Find the lenght P A and P B.
Answers

a) P D = 21 − 5 = 16cm
b) P A × P B = P C × P D

(x + 1)(x − 1) = 5 × 16 = 80
x2 − 12 = 80, x2 − 1 = 80
c) x2 − 1 = 80 → x2 = 81, x = 9
d) P A = 9 + 1 = 10cm,P B = 9 − 1 = 8cm

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2) The product of two consecutive even numbers is 360
a) If the odd number in between these numbers is x then write the numbers .
b) Form an equation using the given condition.
c) Find the numbers.
Answers
a) Numbers are x − 1, x + 1
b) (x − 1)(x + 1) = 360,x2 − 1 = 360
√
c) x2 = 361, x = 361 = 19. Numbers 19 − 1 = 18, 19 + 1 = 20

3) Consider the arithmetic sequence 5, 9, 13, 17, 21 · · · .
a) Write the algebraic form of this sequence.
b) What is the position of the term in the sequence whose square is 625?
c) Is 36 a term of this sequence . How can you realize it ?
d) What is the position of 49in this sequence ?
Answers
a) xn = dn + (f − d) = 4n + (5 − 4) = 4n + 1
√
b) (4n + 1)2 = 625,4n + 1 = 625 = 25, 4n = 24, n = 6
c) All terms are odd numbers . The even number 36cannot be a term of this sequence
d) 4n + 1 = 49, 4n = 48, n = 12.
12 th term is 49

4) Three boxes in which dates of a calandar are given.

a) If B = x find A, C
b) If A × C = 120 form an equation.
c) Find B
d) Find the days A and C

Answers
a) A = x − 1, C = x + 1
b) (x − 1)(x + 1) = 120,x2 − 1 = 120, x2 = 121
√
c) x = 121 = 11,B = 11
d) A = 10, C = 12

5) Sum of the areas of two rectangles is 130. Side of one square is 2 more than the side of the other square .
a) If the side of the small square is x then what is the side of the big square ?
b) Form a second degree equation using the condition.

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Answers
a) Side of the big square is x + 2
b) x2 + (x + 2)2 = 130
x2 + x2 + 4x + 4 = 130, 2x2 + 4x + 4 − 130 = 0, 2x2 + 4x − 126 = 0
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2020-21 Academic year Works

Mathematics X
Second Degree Eqautions

35
Concepts

Let us learn second degree equations.This is a tool for solving problems in various mathematical
situations.Theory of second degree equation is not the scope of this unit. Here we discuss the
methods to solve equations and its applications.

Worksheet 35

1) Consider two adjacent even numbers

a) If one of them is x then what is the other?
b) If the product is 120 then write a second degree equation.
c) Convert this equation as a completed square by suitable changes
d) Find the numbers .

Answers

a) x + 2

b) x(x + 2) = 120
x2 + 2x = 120

c) Add 1on both sides x2 + 2x + 1 = 120 + 1
(x + 1)2 = 121
√

d) x + 1 = 121 = 11, 11, x + 1 = 11, x = 10
Even numbers are 10, 12

2) Length of a rectangle is 8 more then its breadth.

a) If the breadth is x then what is its length?
b) If the area is 240 sq.cm form a second degree equation.
c) Calculate the lenght and breadth

Answers

a) Length= x + 8

b) x(x + 8) = 240, x2 + 8x = 240

c) Add ( 8 )2 on both sides. It is 16
2
x2 + 8x + 16 = 240 + 16√
(x + 4)2 = 256, x + 4 = 256 = 16, x = 16 − 4 = 12

Breadth 12 cm,Length 12 + 8 = 20cm

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3) In the figure AB is the diametre of the semicircle. AB is perpendicular to P C.Also,AP = BP + 5,
P C = 6.

a) Write the relation between the lenghts P A, P B and P C

b) If P B = x then write an equation connecting the lenghts P A, P B and P C

c) What is the length of P B?

d) What is the radius of this circle. ിെ ആരെമ ?

Answers

a) P A × P B = P C2

b) (x + 5) × x = 62, x2 + 5x = 36

x2 + 5x + ( 5 )2 = 36 + ( 5 )2
2 2
5 )2 25
(x + 25 )2 = 36 + 4
(x + 2 =√1649

(x + 5 ) = 169 = 13
2 4 2

x = 13 − 5 = 4
2 2

c) P B = 4
AP = 4 + 5 = 9, AB = 9 + 4 = 13
Radius = 12cm

4) Consider the sequence of even numbers 2, 4, 6, 8 · · · .

a) What is its algebraic form?
b) How many terms from the beginning in the order makes the sum 210?

Answers

a) xn = 2n

b) n(n + 1) = 210, n2 + n = 210

n2 + n + 1 = 210 + 1
4 4
1 )2 841
(n + 2 √= 4

n + 1 = 841 = 29
2 4 2

n = 29 − 1 = 14
2 2

The sum of the first 14 even numbers is 210

5) The smallest side of a right angled triangle is 4 less than its hypotenuse.Third side is 2 more than the
smallest side.

a) If the smallest side is x what are the other two sides.
b) Write an equation connecting the length of the sides .
c) What is the length of the smallest side?

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d) Find the length of other sides of the triangle.
Answers

a) If the smallest side is x then hypotenuse is = x + 4, third side is x + 2
b) (x + 4)2 = (x + 2)2 + x2, x2 + 8x + 16 = x2 + 4x + 4 + x2

x2 − 4x − 12 = 0
c) x2 − 4x = 12, x2 − 4x + 4 = 12 + 4

(x − 2)2 = 16, x − 2 = 4, x = 6
Smallest side is 6
d) Sides are 6, 8, 10

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2020-21 Academic year Works

Mathematics X
Second Degree Equations

36
Concepts

Let us learn second degree equations.This is a tool for solving problems in various mathematical
situations.Theory of second degree equation is not the scope of this unit. Here we discuss the
methods to solve equations and its applications.

Worksheet 36
1) In traingle ABC, AB = AC

AD is the perpendicular from A to BD. This perpendicular distance from A to BC is 2 cm more than
BC. Area of the triangle is 60 sq.cm

a) If BC = x then what is the langth AD?
b) Form an equation connecting the lengths BC, AD and area of the rectangle
c) Find the length of BC.
d) What is the lenghth of AD?
e) Calculate the perimetre of the triangle ABC

Answers

a) AD = x + 2

b) 1 × x × (x + 2) = 60
2
x(x + 2) = 120, x2 + 2x = 120

√
c) x2 + 2x + 1 = 121, (x + 1)2 = 121, (x + 1) = 121 = 11, x = 11 − 1 = 10cm

d) AD = 10 + 2 = 12

e) AB2 = BD2 + AD2 √

AB2 = 52 + 122 = 169, AB = 169 = 13cm

Perimetre = 13 + 13 + 10 = 36cm

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2) Length of a rectangle is 4 more than its breadth .Area of the rectangle is 357 sq.cm
a) If the breadth is x then what is its length?
b) Write an equation connecting length , breadth and area
c) Find the lenghth and breadth of the rectangle .
Answers
a) Length x + 4
b) x(x + 4) = 357, x2 + 4x = 357
√
c) x2 + 4x + 4 = 357 + 4 = 361,(x + 2)2 = 361,x + 2 = 361 = 19, x = 19 − 2 = 17
d) Breadth 17cm ,length 17 + 4 = 21cm

3) Given picture is the dates marked in a calandar .
A, B, C, D denotes the dates.

a) If A = x write B, C, D?
b) If A × C = 84form a second degree equation.
c) Find the number corresponding to A.
d) Write the numbers in the boxes A, B, C, D
Answers

a) B = x + 1, D = x + 7, C = x + 8
b) x(x + 8) = 84, x2 + 8x = 84
c) x2 + 8x + 42 = 84 + 42 √

(x + 4)2 = 100, x + 4 = 100, x = 10 − 4 = 6
A=6
d) A = 6, B = 7, C = 14.D = 13

4) Sum of the areas of two squares is 468sq.cm.The difference between the perimetres is 24cm.
a) If the small side is x then what is the length of the big side ?
b) What is the perimetre of the gig square?
c) Write the length of the sides the squares in x
d) Form a second degree equation and find the length of the small square.
e) Find the length of the big square.

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Answers

a) 4x

b) 4x + 24

c) Side of the small square is x, Side of the big square is 4x+24 =x+6
4

d) x2 + (x + 6)2 = 468, x2 + x2 + 12x + 36 = 468,2x2 + 12x = 432
x2 + 6x = 216, x2 + 6x + 9 = 225, (x + 3)2 = 225

(x + 3) = 15, x = 12

Side of the small square is 12cm

e) Side of the big square is 12 + 6 = 18cm

5) Hypotenuse of a right angled triangle is 1 less than twice its small side.Third side is 1 more than its small
side

a) If the small side is x what is the length of other two sides .
b) Form an equation connecting the length of the sides .
c) Calculate the length of the sides of the triangle.

Answers

a) Hypotenuse = 2x − 1, Third side = x + 1
b) (2x − 1)2 = x2 + (x + 1)2, 4x2 − 4x + 1 = x2 + x2 + 2x + 1

2x2 − 6x = 0

c) x = 3.Sides are :
Hypotenuse2x − 1 = 6 − 1 = 5cm.
Other two sides are 3cm ,4cm.

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2020-21 Academic year Works

Mathematics X
Second Degree Equations.

37
Concepts

We are discussing the process of solving a second degree equation using the colpleting the square
method.As the generalization of this method we can establish a formula to solve the second degree
equation. It is not necessary to use the formula for solving the second degree equation.
In the equation ax2 + bx + c = 0

√
−b ± b2 − 4ac
x=

2a

Worksheet 37

1) The sum of the squares of two consecutive natural numbers is313.

a) If one number is x then what is the other?
b) Form an equation using this condition.
c) Find the numbers .

Answers

a) x + 1

b) x2 + (x + 1)2 = 313

x2 + x2 + 2x + 1 = 313

2x2 + 2x − 312 = 0

x2 + x − 156 = 0

√
−b± b2−4ac
c) x= √2a

x = −1± 12−4×1×−156
√ 2×1
−1± 1+624
x =
x = −1±252 x = 12, −13, Numbers are 12, 13
2

2) The sum of two numbers is 15. The sum of its reciprocals is 3 .
10

a) If one number is x then what is the other?

b) Form an equation.

c) Find the numbers.

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Answers

a) 15 − x

b) 1 + 1 = 3
x 15−x 10
15−x+x
x(15−x) = 3
10

150 = 45x − 3x2

x2 − 15x + 50 = 0

√
−b± b2−4ac
c) x=
2a √
x = −(−15)± (−15)2 −4×1×50
2×1
x = 10, 5

Numbers are 10, 5.

3) The sum of a number and its reciprocal is 2 1 .
30

a) If one number is x what is the other.

b) Find the numbers.

Answers

a) x + 1 = 2 1
x 30
x2 +1
x = 61
30
30x2 + 30 = 61x,

30x2 − 61x + 30 = 0

√
−b± b2−4ac
b) x=
2a √
x = −(−61)± (−61)2 −4×30×30
2×30
5 6
x = 6 , 5

4) Two chords AB and CD intersect at P inside the circle.If AB = 13cm
and P C = 12 cm ,P D = 3 cm

a) Write the relation between P A, P B, P C and P D
b) If P A = x form an equation
c) Find the lengths of P A and P B

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Answers

a) P A × P B = P C × P D

b) x(13 − x) = 12 × 3

13x − x2 = 36

−x+13x − 36 = 0, x2 − 13x + 36 = 0

√
−b± b2−4ac
c) x=
2a √
x = −(−13)± (−13)2 −4×1×36
2×1
x = 9, 4

If x = 9 then 13 − 9 = 4. P A = 9, P B = 4

If x = 4then P A = 4, P B = 9

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5) The sum of the first n terms of the arithmetic sequence 7, 9, 11, 13 · · · is 40.

a) Form a second degree equation using this condition.
b) How many terms make the sum 40 ?
c) Find n in another method .

Answers

a) Algebraic form of the sequence is 2n + 5

The sum of the first n terms is 40 .

(7 + 2n + 5) × n = 40
2
n2 + 6n = 40, n2 + 6n − 40 = 0,

√
−b± b2−4ac
b) n= √2a

n = −6± 62−4×1×−40
√ 2×1
−6± 196
x = 2

n=4

Sum of the first 4terms is 40

c) The sum of the first 7 odd numbers is = 72 = 49.40 is obtained on subtracting the sum of
first three odds from 49. The number of remaining numbers is 4.

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2020-21 Academic year Works

Mathematics X
ര ാം തിസമവാക ൾ

38
Concepts

We are discussing the process of solving a second degree equation using the colpleting the square
method.As the generalization of this method we can establish a formula to solve the second degree
equation. It is not necessary to use the formula for solving the second degree equation.
In the equation ax2 + bx + c = 0

√
−b ± b2 − 4ac
x=

2a

Worksheet 38

1) The product of the digits of a two digit number is 18.When 63 is subtracted from the number we get the
two digit number with digits in the reversed order.

a) If the digit in the tens place is x when what will be the digit in the one’s place ?
b) Write the number using the place value of the digits.
c) Form a second degree equation using the given condition.
d) Find the number.

Answers

a) ഒ െട ാനെ അ ം 18
x

b) സംഖ = 10x + 18
x

c) 10x + 18 − 63 = 10 × 18 + x
10x + 1x8 − =
x 180 − x x
x
63

9x2 − 162 = 63x

9x2 − 63x − 162 = 0

ഇ വശ ം 9െകാ ് ഹരി ാൽ

x2 − 7x − 18 = 0.

√
−b± b2−4ac
d) x=
2a √
x = −(−7)± (−7)2 −4×1×−18
2×1
√
7± 121
x = 2

x = 9, −2

അ ം ഒ ന നസംഖ ആ ി . x = 9

e) പ ിെ ാനെ അ ം9, ഒ െട ാനെ അ ം 18 = 2
9
സംഖ = 10 × 9 + 2 = 92

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2) The difference between the squares of two numbers is 45.The square of the small number is 4 times the
large number.

a) Write an equation by taking x as the large number.
b) Calculate the numbers.

Answers

a) വലിയ സംഖ x ആയാൽ െചറിയ സംഖ െട വർ ം = 4x
x2 − 4x = 45, x2 − 4x − 45 = 0

√
−b± b2−4ac
b) x=
2a √
x = −(−4)± (−4)2 −4×1×−45
2×1
√
4± 196
x = 2

x = 9, −5

x = 9ആയാൽ െചറിയ സംഖ െട വർ ം= 4x = 36, െചറിയ സംഖ 6, −6.

സംഖ കൾ9, 6 അെ ിൽ9, −6

ഒ സംഖ െട വർ ം ന നസംഖ ആകി . അതിനാൽ x = −5 ആ ി .

സംഖ കൾ9, 6അെ ിൽ9, −6എ ിവയാണ് .

3) A rod of 16cm length is cut into two pieces. Two times the square of the length of the larger piece is equal
to 164 more than the square of the smaller piece.

a) If the length of the larger piece is x then what is the length of the smaller piece.
b) Form an equation using the given conditions.
c) Find the length of the pieces .

Answers

a) െചറിയ ഭാഗ ിെ നീളം16 − x

b) 2x2 = (16 − x)2 + 164

x2 + 32x − 420 = 0

√
−b± b2−4ac
c) x= 2√a

x = −32± 322−4×1×−420
√ 2×1
−32± 1024+1680
x = √2

x = −32± 2704 , x = 10
2

c) ര ഭാഗ ൾ10, 16 − 10, 10െസ ീമീ ർ ,6 െസ ീമീ ർ

4) The sum of the squares of two positive numbers is 208.18 times the small number is equal to the square
of the large number.

a) Form an equation by taking x as the small number.
b) Find the numbers.

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Answers

a) െചറിയ സംഖ = x,വലിയ സംഖ െട വർ ം18x

x2 + 18x = 208

x2 + 18x − 208 = 0

√
−18± 182−4×1×−208
b) x = √ 2×1

x = −18± 324+832
√2×1
−18± 1156
x =
x = −18±234 ,x = 8
2
വലിയ സംഖ െട വർ ം= 18 × 8 = 144. വലിയസംഖ = 12, സംഖ കൾ8, 12

5) A two digit number is 4 times the sum of the digits.Also the number is 3 times the product of the digits.

a) Form an equation by taking x, y as the digits.
b) Make a second degree equation using the given condition.
c) Find the numbers.

Answers

a) പ ിെ ാനെ അ ംx, ഒ െട ാനെ അ ംyആയാൽ
സംഖ 10x + y

10x + y = 4(x + y) (1)

10x + y = 3xy (2)
ാനം 0ആകി . പ ിെ
b) 10x + y = 4x + 4y, 6x = 3y, y = 2x
10x + y = 3xy → 10x + 2x = 3x × 2x
12x = 6x2

c) x = 0, x = 2. ര സംഖ ആയതിനാൽ പ ിെ
ാനം= 2, ഒ െട ാനം 2x = 4

സംഖ = 24

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2020-21 Academic year Works

Mathematics X
Second Degree Equations.

39
Concepts

We are discussing the process of solving a second degree equation using the completing the
square method.As the generalization of this method we can establish a formula to solve the second
degree equation. It is not necessary to use the formula for solving the second degree equation.
In the equation ax2 + bx + c = 0

√
−b ± b2 − 4ac
x=

2a

Worksheet 39

1) The difference between the length of the perpendicular sides of a right triangle is 10, area of the triangle is
600 square cm. ആണ്.

a) One of the perpendicular sides is x then what is the length of the other?
b) Form an equation using the given condition.
c) What is the length of the perpendicular sides?
d) Calculate the perimetre of the triangle.

Answers

a) x + 10

b) 1 × x × (x + 10) = 600
2
x2 + 10x = 1200

x2 + 10x − 1200 = 0

√
−b± b2−4ac
c) x= 2√a

x = −10± 102−4×1×−1200
√ 2×1
−10± 4900
x = 2×1

x = 30, −40. Length of the side cannot be a negative quantity. x = 30cm.

Parpendicular sides are 30cm, 40cm.

√√
d) Hypotenuse = 302 + 402 = 2500 = 50cm Perimetre= 30 + 40 + 50 = 120cm.

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2) Perimetre of a rectangle is 82cm, area 400 sq.cm

a) What is the total length of the adjacent sides?
b) If one side is x then what is the length of the other side?
c) Form a second degree equation using the given condition.
d) Calculate the length of the sides.

Answers

a) 2(length + breadth) = 82, l + b = 41

b) If one side is x , the other side will be 41 − x

c) x(41 − x) = 400,−x2 + 41x = 400,x2 − 41x + 400 = 0

√
−b± b2−4ac
d) x=
2a √
x = −(−41)± (−41)2 −4×1×400
2×1
√
41± 1681−1600
x = √ 2×1

x = 41± 81
2×1
41±9
2 = 25, 16

If one side is 16 cm and other side will be 41 − 16 = 25cm

3) In triangle ABC, AB = AC = 13cm, area of the triangle is 60sq.cm .The perpendicular distance from A
to BC is AD.

a) If BD = xthen what is AD?
b) Form a second degree equation using BC, AD, and area .
c) What is the length of BC?
d) What is the perimetre of the triangle?

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Answers
√

a) AD = 132 − x2

b) 1 × BC ×√AD = 60
2 2x × 132 − x2
1
2 √× = 60

x 132 − x2 = 60

Squaring on both sides , x2(169 − x2) = 3600

If x2 = y, y(169 − y) = 3600, y2 − 169y + 3600 = 0.

c) Solving y = 144, 25.
If x2 = 144 ,x = 12, −12.
If x2 = 25, x = 5, −5

If x = 12, BC = 24cm .

If x = 5, BC = 10cm

d) Perimetre of the triangle 13 + 13 + 10 = 36cm or 13 + 13 + 24 = 50cm

4) The perimetre of a right triangle is 60 cm, hypotenuse is 25cm

a) What is the total length of the perpendicular sides ?
b) If the length of one perpendicular side is x then what will be the length of the other ?
c) Form an equation using the length of the sides
d) Calculate the area of the triangle.

Answers

a) Length of the perpendicular sides is 60 − 25 = 35cm

b) If one perpendicular side is x then the other perpendicular side is 35 − x

c) 252 = x2 + (35 − x)2
x2 − 35x + 300 = 0

d) Solving , x = 20, 15.

If x = 20 other perpendicular side is 35 − 20 = 15. Area = 1 × 20 × 15 = 150sq.cm
2

5) The difference between the length of the sides of two squares is 4cm. The sum of the areas is 400sq.cm

a) If the side of the small square is x then what is the side of the other square?
b) Form an equation using the given condition.
c) Calculate the side of the squares.

Answers

a) x + 4
b) x2 + (x + 4)2 = 400

x2 + x2 + 8x + 16 = 400
2x2 + 8x − 384 = 0
x2 + 4x − 192 = 0

c) Solving x = 12, −16.
Length of the side of one square is 12, length of the side of other square is 12 + 4 = 16cm

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2020-21 Academic year Works

Mathematics X
Second Degree Equations

40
Concepts

We are discussing the process of solving a second degree equation using the completing the
square method.As the generalization of this method we can establish a formula to solve the second
degree equation. It is not necessary to use the formula for solving the second degree equation.
In the equation ax2 + bx + c = 0

√
−b ± b2 − 4ac
x=

2a

Worksheet 40

1) The age of a man after 15 years would be the square of his age before 15 years.

a) If the present age is x , form a second degree equation
b) Find the present age.
c) Without using algebra find the present age .

Answers

a) (x − 15)2 = (x + 15)

x2 − 30x + 225 = x + 15

x2 − 31x + 210 = 0

√
−b± b2−4ac
b) x=
2a √
x = −(−31)± (−31)2 −4×1×210
2×1
√
x = 31± 961−840

31±11 = 2
2
21, 10

c) x values are 21, 10.The value of x = 10 cannot be admissible.Present age is 21years.

d) While considering the age before and after 15 years there will be a gap of 30years.The
perfect square just above 30 is 36.
15 years below 36 is 21, 15years below 21 is 6. The square of 6 is 36. The present age
is 21

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2) Manju’s present age is the square of Laya’s age.After 5years Manju’s age become 3 times Laya’s age.

a) If Laya’s present age is x form a second degree equation.
b) Find the present age of both.
c) How many years later the age of Manju become two times Laya’s age?

Answers

a) Laya’s present age is = x. manju’s present age is = x2.

Laya’s age after five years is = x + 5,Manju’s age after five years is = x2 + 5.

x2 + 5 = 3(x + 5), x2 + 5 = 3x + 15, x2 − 3x − 10 = 0

√
−b± b2−4ac
b) x=
2a √
x = −(−3)± (−3)2 −4×1×−10
2×1
3±7
x = 2 = 5, −2

Laya’s present age is 5 years .Manju’s present age is 25 years

c) Suppose that after n years Manju’s age become two times Laya’s age .
(5 + n) × 2 = 25 + n, n = 15
After 15 years Manju’s age become two times Laya’s age .

3) One year ago, Ajayan’s age was 8 times his son’s age.Present age of Ajayan is the square of his son’s
present age.

a) If son’s age before 1 year is x what was Ajayans age one year ago.
b) Form a second degree equation using the given condition.
c) Calculate their present age.

Answers

a) Son’s age before one year is x, Ajay’s age is 8x.

b) Present age of son = x + 1, Ajay’s present age = 8x + 1
(x + 1)2 = 8x + 1, x2 + 2x + 1 − 8x − 1 = 0, x2 − 6x = 0,

c) x(x − 6) = 0, x = 6, 0
We can take x value 6. Son’s present age x + 1 = 7, Ajay’s present age is 8 × 6 + 1 = 49

4) The sum of the ages of a father and son is 45. 5 years ago the product of their ages was 124.

a) If father’s present age is x what is son’s present age?
b) Form a second degree equation using the given condition.
c) Find the their present age.

Answers

a) Son’s present age = 45 − x

b) Fathers age before fice years = x − 5, Sons’s age before five years = 40 − x

(x − 5)(40 − x) = 124,

x2 − 45x + 324 = 0

√
−b± b2−4ac
c) x=
2a √
x = −(−45)± (−45)2 −4×1×324
2×1
√
45± 729
x = 2

x = 36, 9

Fathers age = 36, Son’s age = 9

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5) Nasrin’s age is two times Riswan’s age. Four years hence the product of their ages become 160.

a) If Riswan’s present age is x what is Nasrin’s present age?
b) Form a second degree equation using the given condition.
c) Calculate their present age.

Answers

a) Riswan’s age = x,Nasrin’s age = 2x

b) Nasrin’s age after four years = x + 4, Riswan’s age after four years = 2x + 4

(x + 4)(2x + 4) = 160

2x2 + 12x + 16 = 160

2x2 + 12x − 144 = 0

x2 + 6x − 72 = 0

√
−b± b2−4ac
c) x= √2a

x = −6± 62 −4×6×−72
2×1

x = 6, −12.Riswan’s age = 6, Nasrin’s age = 12.

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2020-21 Academic year Works

Mathematics X
Second Degree Equations

41
Concepts

We are discussing the process of solving a second degree equation using the completing the
square method.As the generalization of this method we can establish a formula to solve the second
degree equation. It is not necessary to use the formula for solving the second degree equation.
In the equation ax2 + bx + c = 0

√
−b ± b2 − 4ac
x=

2a

Worksheet 41

1) The speed of a boat in still water is 8 km in an hour.The boat travels 15 kilometre in upstream and 22
kilometre in downstream in5 hours.

a) If the speed of the stream is x what will be the speed attained by the boat in the downstream.
b) If the speed of the stream is x what will be the resulting speed inn the upstream?
c) Form an equation using the given condition.
d) Calculate the speed of the stream.

Answers

a) 8 + x

b) 8 − x

c) Speed of boat in the downstream journey= 8 + x. Time taken for the downward journey is

22

8+x

Speed of boat in the upstream journey= 8 − x. Time taken for the downstream journey 15
8−x
15 22
8−x + 8+x = 5

15(8+x)+22(8−x) = 5
82 −x2
5x2 − 7x − 24 = 0.

√
−b± b2−4ac
d) x=
2a √
x = −(−7)± (−7)2 −4×5×−24
√ 2×5

x= 7± 529

7+23 10
10
=3

speed of stream 3 km/h

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2) A train travels with uniform speed in 300 km.If the speed of the train is increased by 5 km per hour, the
journey would have taken 2 hours less.

a) If the usual speed is x what will be the time taken for the journey.
b) If the speed is increased by 5 km per hour what will be the time taken for the journey?
c) Form an equation using the given condition.
d) Calculate the speed of the train .

Answers

a) 300
x

b) 300
x+5

c) 300 − 300 = 2
x x+5

1 − 1 = 2
x x+5 300

1 − 1 = 1
x x+5 150

x2 + 5x − 750 = 0

x = −5± √ 52 −4×1×−750
2a
d)

x = 25, −30.Usual speed of train 25km/h.

3) There are 64 small squares in a chess board.The area of one small square is 6.25 sq.cm.There is a boarder
of width 2cm around the chess board squares.

a) If the length of the board is x what will be the total area of small squares?
b) Form a second degree equation using the given condition.
c) Calculate the length of the chess board.

Answers

a) On subtracting 2 from both sides area of the square formed by 64 small squares is (x − 4)2

b) (x − 4)2 = 6.25 × 64

x2 − 8x + 16 = 400

x2 − 8x − 384 = 0

√
−b± b2−4ac
c) x=
2a √
x = −(−8)± (−8)2 −4×1×−384
2×1
x = 24, −16.

Side of the chess board is 24cm

4) In a group of children each child gives a gift to every other child.If the total number .of gifts is 132, then

a) If the number of children is n then how many gifts each child give other children.
b) Form an equation using the given condition.
c) calculate the number of children in the group.

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Answers

a) n − 1 gifts should be given by a child

b) Total number of gifts is n(n − 1) = 132

n2 − n − 132 = 0

√
−b± b2−4ac
c) x=
2a √
x = −(−1)± (−1)2 −4×1×−132
2×1
√
1± 529

2

x = 12, −11. Number of children = 12

5) Teacher asked the children to draw a rectangle of area 5 sq.cm and perimetre 8 cm. Manju , a good student
made a comment that it is possible to draw such a square after some algebraic calculations.

a) If one side of the reactangle is x then what will be the other.
b) Form a second degree equatio.
c) Prove that it is not possible to construct such a rectangle.

Answers

a) 2(l + b) = 8, l + b = 4
ഒ വശം xആയാൽ മേ വശം4 − x

b) x(4 − x) = 5,−x2 + 4x − 5 = 0, x2 − 4x + 5 = 0

√
−b± b2 −4ac
c) x = 2a
√
b2 − 4ac = −4. b2 − 4acis not a real number. Side is not real.Rectangle cannot be

constructed.

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2020-21 Academic year Worksheets

Mathematics X
Second Degree Equations

41
Concepts

We are discussing the process of solving a second degree equation using the completing the
square method.As the generalization of this method we can establish a formula to solve the second
degree equation. It is not necessary to use the formula for solving the second degree equation.
In the equation ax2 + bx + c = 0

√
−b ± b2 − 4ac
x=

2a

Worksheet 41

1) The participants of a meeting gave hanshakes to eachother. It is found that there are 190handshakes in
total.

a) If there are n participants , what is the number of handshakes given by a participant to others?
b) Form a second degree equation using the number of participants and the number of handshakes.
c) Calculate the number of participants of the meeting.

Answers

a) n − 1

b) n(n−1) = 190
2
n2 − n − 380 = 0

√
−b± b2 −4ac
c) n=
2a√
x = −(−1)± (−1)2 −4×1×−380
2×1
√
1± 1521
x = 2×1

n = 20, −14, Number of participants is 20

2) If the price of a book is reduced by 5 rupees,a person can buy 5 more books for 300 rupees.

a) If the original price of the book is x, how many books can be purchased for 300 rupees?
b If the price is decreased by 5 how many books can be purchased in 300 rupees
c) Form a second degree equation using the given condition.
d) Calculate the original price of the book

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Answers

a) 300
x

b) If the price is x − 5, the number of books = 300
x−5

c) 300 − 300 = 5
x−5 x
300x−300x+1500
x(x−5) = 5

x2 − 5x − 300 = 0

√
−b± b2−4ac
d) x=
2a √
x = −(−5)± (5)2 −4×1×−300
2×1
x = 20, −15. The original price of the book is 20rupees

3) The perimetre of a rectangle is 82cm, area 400 sq.cm

a) If the length of one side is x then what is the length of other side ?
b) Form a second degree equation
c) Find the sides of the rectangle.

Answers

a) 2(l + b) = 82,l + 2 = 41. If one side is x the other side will be 41 − x

b) x(41 − x) = 400, −x2 + 41x − 400 = 0, x2 − 41x + 400 = 0

√
−b± b2−4ac
c) x=
2a √
x = −(−41)± (−41)2 −4×1×400
2×1
x = 25, 16

Sides are 25cm and 16cm

4) The hypotenuse of a right triangle is 25cm,the difference between other two sides is 5cm

a) If one of the perpendicular sides is x what is the length of other perpendicular side?
b) Form a second degree equation .
c) Calculate the length of its sides.
d) Calculate the area of the triangle.

Answers

a) x + 5

b) x2 + (x + 5)2 = 252

x2 + x2 + 10x + 25 = 625

2x2 + 10x − 600 = 0

x2 + 5x − 300 = 0

√
−b± b2−4ac
c) x= √2a

x = −5± 52 −4×1×−300
2×1
x = 15, −20. perpendicular sides are 15cm and 20cm

d) Area = 1 × 15 × 20 = 150 sq.cm
2

5) The denominator of a fraction is 1 more than two times its numerator.The sum of the fraction and its reciprocal

is 2 16 .
21

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a) If the numerator is x what is its denominator.
b) Write the fraction in x
c) Form a second degree equation using the given condition.
d) Find the fraction.

Answers

a) 2x + 1

b) Fraction is x .
2x+1

c) x + 2x+1 = 2 16
2x+1 x 21

x + 2x+1 = 58
2x+1 x 21

11x2 − 26x − 21 = 0

d) Solving , x = 3.

Fraction is 3 = 3
2×3+1 7

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2020-21 Academic year Worksheets

Mathematics X
Second Degree Equations

42
Concepts

We are discussing the process of solving a second degree equation using the completing the
square method.As the generalization of this method we can establish a formula to solve the second
degree equation. It is not necessary to use the formula for solving the second degree equation.
In the equation ax2 + bx + c = 0

√
−b ± b2 − 4ac
x=

2a

Worksheet 42

1) First term of an arithmetic sequence is 7 and common difference 3.

a) What is its algebraic form?
b) Find the sum of the first n terms
c) How many terms of this sequence beginning from the first term makes the sum 710?

Answers

a) xn = dn + (f − d) = 3n + (7 − 3) = 3n + 4

b) Sn = (x1 + xn) × n
2
× n
Sn = (7 + 3n + 4) 2

(11 + 3n) × n = 11n + 3n2
2 2 2

c) 3n2 + 11n = 710
2 2
3n2 +11n
2 = 720. 3n2 + 11n − 1420 = 0

√
−b± b2 −4ac
x = 2√a

x = −11± 112−4×3×−1420
√ 2×3
−11+ 6 17161 , 120 −142
x= n = 6 , 6

n = 20.Twenty terms make the sum 720

2) On joining two vertices of a polygon we get either a side or a diagonal.Consider a polygon of n sides.

a) How many diagonals can be drawn from a vertex?
b) How many diagonals are there in a polygon of n sides ?
c) Find the number of sides of a polygon having 35 diagonals.
d) Name the polygon having number of sides and diagonals equal.

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Answers

a) n − 3
(Joining a vertex to other vertices on either side of it forms a side of the polygon. That is ,
n − 3 diagonals can be drawn from a vertex)

b) n(n−3)
2

(Diagonal from the vertex A to B and the diagonal from B to A are treated as a single diagonal.

So we should take half of n(n − 3)

c) n(n−3) = 35
2
n(n − 3)√= 70, n2 − 3n = 70, n2 − 3n − 70 = 0.
−b± b2 −4ac
x =
2a √
x = −(−3)± (−3)2 −4×1×−3
2×−3
√
3± 289
2 , n = 10, −7. The number of sides is 10

d) Pentagon.

3) The points A1, A2, A3 · · · An are marked in a circle. On joining two points we get a chord .

a) How many chords can be drawn from a given point to other points?
b) What is the total number of chords?
c) How many points should be marked on the circle to get 120 chords.

Answers

a) n − 1

b) n(n−1)
2

c) n(n−1) = 120
2
n(n − 1)√= 240, n2 − n − 240 = 0,
−b± b2 −4ac
n =
2a √
n = −(−1)± (−1)2 −4×1×−240
2×1
√
1± 961
n = 2 , n = 16, −15. The number of points marked on the circle is 16

4) Consider the sequence of numbers which gives the remainder 3 on dividing by 4.

a) Write the algebraic form of this sequence ?
b) What is the sum of first n terms of this sequence ?
c) How many terms from the beginning make the sum 820?
d) Can the sum of any 25 terms of this sequence 2020?

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Answers

a) The sequence is 3, 7, 11 · · ·
xn = dn + (f − d) = 4n − 1

b) Sn = (x1 + xn) × n = (3 + 4n − 1) × n
2 2
n + 2n2

c) 2n2 + n =√ 820, 2n2 + n − 820 = 0
−b± b2−4ac
n = √2a

n = −1± 12−4×2×−820
√ 2×2
−1± 6561
n =
n = −1±841 ,n = 20.
4

The sum of 20 terms is 820

d) All terms are odd numbers. The sum of 25 odd numbers cannot be an even number.

5) The sum of a number and its positive square root is 6 .
25

a) If x is the number , write an equation using the given conditions.

b) Write the equation in the form ax2 + bx + c = 0

c) Find the number.

Answers

a) √ 6
x+ x= 25

b) √ 6265−2−5xx
√x =
25
x=

Squaring on both sides and solving we get in the form ax2 + bx + c = 0. It is 625x2 −

925x + 36 = 0

c) Solving we get x = 1
25

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2020-21 Academic year Worksheets

Mathematics X
ര ാം തിസമവാക ൾ

43
Concepts

We are discussing the process of solving a second degree equation using the completing the
square method.As the generalization of this method we can establish a formula to solve the second
degree equation. It is not necessary to use the formula for solving the second degree equation.
In the equation ax2 + bx + c = 0

√
−b ± b2 − 4ac
x=

2a

Worksheet 43

1) An aeroplane takes 1 hour less for a journey of 1200km. If the speed is increased by 100 km per hour,
from its usual speed.

a) If the usual speed is x then what is the time taken for the journey?
b) If the speed is increased by 100 km/h what will be the time taken for the journey?
c) Form a second degree equation using the given condition.
d) Calcualte the usual speed.

Answers

a) 1200
x

b) 1200
x+100

c) 1200 − 1200 = 1
x x+100

1 − 1 = 1
x x+100 1200
100+x−x
x(x+100) = 1
1200

100 = 1
x2 +100x 1200

120000 = x2 + 100x

x2 + 100x − 120000 = 0

√
−b± b2−4ac
d) x= 2a√

x = −100± 1002−4×1×100
√ 2×1
−100± 490000 −100±700
x = 2 , 2 ,x = 300 speed300 km/h

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2) The length of a rectangular hall is 5m more than its width and the area of the hall is 84 sq.

a) If the length is x , what will be its width?
b) Form a second degree equation using the given condition.
c) Find the length and breadth of the hall.

Answers

a) If x is the length , width will be= x − 5

b) x(x − 5)√= 84,x2 − 5x − 84 = 0
−b± b2 −4ac
x =
2a √
x = −(−5)± (−5)2 −4×1×−84
2×1
x = 12, −7

c) Length12m , width = 12 − 5 = 7cm

3) A two digit number is 4 times the sum of the digits and twice the product of the digits.

a) If the digit in the one’s place is y and the digit in ten’s place is x, write two equations using the given
conditions.

b) Form a second degree equation.
c) Find the digits and write the number.

Answers

a) 10x + y = 4(x + y), 10x + y = 2xy
10x + y = 4x + 4y, 6x − 3y = 0, 2x = y

b) 10x + y = 2xy, 10x + 2x = 2 × x × 2x
12x = 4x2, 12 = 4x, x = 3

c) Digits are x = 3, y = 6
Number = 3 × 10 + 6 = 36

4) The area of a rectangular plot is 528 sq.m.Length of the plot is 1 more than twice its breadth.

a) If the breadth is x what will be its length?
b) Form a second degree equation with the given condition.
c) Find the length and beradth of the plot.

Answers

a) breadth x, length = 2x + 1
b) x(2x + 1) = 528, 2x2 + x − 528 = 0,
c) x = 16

Langth = 2 × 16 + 1 = 33m , breadth = 16m

5) In copying a second degree equation to solve it,the term without x was written as 24 instead of −24.The
answers found were 4 and 6.

a) If the equation wrongly written is ax2 + bx + 24 = 0, write two equations using the wrong answers
given

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b) Find a and b by solving the equations.
c) Write the correct equation and find its solution.

Answers

a) Since x = 4 , a × 42 + b × 4 + 24 = 0,4a + b = −6
Since x, a × 62 + b × 6 + 24 = 0,6a + b = −4

b) Solving the equations 4a + b = −6, 6a + b = −4, a = 1, b = −10
c) Correct equation is x2 − 10x − 24 = 0

x = 12, −2



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05

Trigonometry

ത്രികോണമിതി

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2020-21 Academic year Works

Mathematics X
Trigonometry

44

Concepts

a) There are some special right triangles. The diagonal of a square makes two right triangles of angles

45◦, 45◦, 90◦

b) If the si√de opposite to 45◦ is 1 then the side opposite to 90◦ will be √ The sides are in the ratio
2.

1:1: 2

c) The altitude of an equilateral triangle makes two right triangles.The angles of these triangles are

30◦, 60◦, 90◦. √

If the side opposite to 30◦ is 1, the side opposite to 90◦ will be 2, side opposite tos 60◦ will be 3

Worksheet 44

1) Consider a square of perimetre 40cm
a) What is the length of its side?
b) What is the length of its diagonal
c) What is the area of the square drawn on its diagonal?

a) Length of one side = 40 = 10cm
4

b) Two sides and the diagonal form a 45◦, 45◦, 90◦ right triangle .

The side opposite to 45◦ is 10cm√.
∴ the side opposite to 90◦ is 10 2cm

√
c) Area = (10 2)2 = 100 × 2 = 200 sq.cm

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2) The area and perimetre of a square are equal in number.
a) What is the length of its side?
b) What is the length of its diagonal?
c) What is the area of the square drawn on its diagonal?

a) 4a = a2 ⇒ a = 4
√

b) Length of the diagonal is 4 2
c) Area of the square drawn on the diagonal is (4√2)2 = 16 × 2 = 32

sq.unit

3) A bridge of length 600m is built across a river making 45◦ angle with the direction of flow.
a) Draw a rough diagram.
b) What is the width of the river?

a) Rough diagram is drawn below

b) Width of the river BC = 6√00 metre.

2

4) In traingle ABC ,∠A = 30◦, BC = 10cm

a) What is the length AB?
b) What is the length of the side AC?

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c) What is the length of the diagonal of the square drawn on AC?
d) What is the perimetre of the square?

a) In a 30 − 60 − 9o0pptorisaintegleto,ssi6d0e◦oipsp1o0s√ite3ctom30◦ is 10cm
Therefore side

b) Side opposite to 90◦ is 20cm
√

c) Length of diagonal of the square is 20 2cm
√√

d) Perimetre= 4 × 20 2 = 80 2cm

5) Consider an equilateral triangle of side 10cm
a) What is its altitude?
b) Draw a rough diagram of the square drawn on the altitude
c) What is the area of this square.
d) What is the length of its diagonal?

√
a) △ABD is a 30 − 60 − 90 triangle. AD is the altitude. AD = 5 3cm
b) Figure

√
c) Area = (5 3)2 = 25 × 3 = 75 sq.cm

√√ √
d) Length of the diagonal is 5 3 × 2 = 5 6cm

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2020-21 Academic year Works

Mathematics X
Trigonometry

45
Concepts

a) There are some special right triangles. The diagonal of a square makes two right triangles of angles

45◦, 45◦, 90◦
√

b) If the si√de opposite to 45◦ is 1 then the side opposite to 90◦ will be 2. The sides are in the ratio
1:1: 2

c) The altitude of an equilateral triangle makes two right triangles.The angles of these triangles are

30◦, 60◦, 90◦. to 30◦ is 1, the side opposite to 90◦ will be 2, side opposite tos 60◦ will be √
If the side opposite 3

Worksheet 45
1) In the parallelogram ABCD ,∠A = 60◦., AB = 12cm , AD = 10cm

a) What is the perpendicular distance from D to AB.
b) Find the area of the parallelogram

a) Look at the picture

∠D = 120◦,∠A = 180 − 120 = 60◦
△AED is a 30◦ − 60◦ − 90◦√triangle .The side opposite to 90◦is 10cm
.Therefore AE = 5, DE = 5 3cm .

√√
b) Area AB × DE = 12 × 5 31 = 60 3sq.cm

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2) In the rhombus ABCD,∠D = 150◦

a) What is the measure of ∠A?
b) What is the diatance between AB and CD
c) Find the area of the rhombus.

a) Look at the picture

∠A = 180 − 150 = 30◦
Draw a line DE perpendicular to AB
Triangle AED is a 30◦ − 60◦ − 90◦triangle
The side opposite to 90◦ is 8cm .The side opposite to 30◦ is 4cm
b) Area = AB × DE = 32sq.cm
3) In the figure O is the centre of the circle.∠ACB = 30◦

a) What is the measure of ∠AOB?
b) What kind of triangle is OAB?
c) If the radius of the circle is 12cm then what is the altitude of triangle OAB?
d) What is the area of triangle OAB?

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a) ∠AOB = 60◦

b) OA = OB, Angles opposite to the equal sides are equal. All angles are
60◦ .
This is an equilateral triangle

c) Look at the picture

Draw OD perpendicular to AB

Triangle ODAis a30◦ − 60◦ − 90◦triangle . 30◦ is 6 cm , OD = √
Side opposite to 90◦ is 12cm ,Side opposite to 6 3cm

d) Area 1 √√
2 × 12 × 6 3 = 36 3sq.cm

4) The diagonal of the rectangle ABC is 12cm , ∠BAC = 30◦

a) What is the length of the side AB?
b) What is the length of the side BC?
c) Calculate the area of the rectangle

a) △ABCis a 30◦ − 60◦, 90◦ triangle.

Side opposite to 90◦ is 12cm = √
Side opposite to 30◦ is 6cm AB 6 3cm

b) BC = 6cm
√

c) Area = AB × BC = 36 3sq.cm

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