Unit
11 Algebraic Fraction
11.1 Warm-up Activities
Discuss the following in your class and draw a conclusion.
• Give some examples of algebraic fraction.
• What is the L.C.M. of (x+2) and (x+3)?
• Simplify: 1 + 1
x–4 x+4
• Simplify: 1 + 1
x+2 x+3
• For what value of x is, 1 not defined?
x+2
• For what value of x is, 2x+1 not defined?
x+1
11.2 Simplification
The process of reducing an expression to the simplest form is known as simplification.
Generally, simplification of rational expressions can be done by taking the L.C.M. of the
denominators. The resulting terms should be reduced into the simplest form.
Let’s see the following examples.
I. Simplification of algebraic II. Simplification of algebraic fractions
fractions having a common denominator. having different denominators.
x² + y² + x² – y² x + 3 – x + 1
x – y x – y x + 4 x − 3
= x² + y² + x ² – y ² (x + 3)(x – 3) – (x + 1)(x + 4)
= (x + 4)(x – 3)
x–y
= 2x² = (x² − 9) − (x² + 5x + 4)
x–y (x + 4)(x – 3)
x2 − 9 − x2 − 5x − 4
= (x + 4)(x – 3)
= −5x − 13
(x + 4)(x – 3)
= −(5x + 13)
(x + 4)(x – 3)
Oasis School Mathematics-10 195
Worked Out Examples
Example: 1
Simplify : a–b – a+b + 4ab
a+b a–b a² + b²
Solution:
a – b – a + b + 4ab
a + b a – b a² + b²
(a² – 2ab + b²) – (a² + 2ab + b²) + 4ab
a² − b² a² + b²
=
= a–²4−abb² + a²4+abb²
= – 4ab(a² + b²) + 4ab(a² – b²)
(a² − b²)(a ² + b ²)
= – 4a3 b − 4ab3 + 4 a3 b – 4 ab3 = - 8ab3
a4 – b4 a4 - b4
Example: 2
2x − y 2x + y 16x3
Simplify: 4x² − 2xy + y² − 4x² + 2xy + y² + 16x4 + 4x²y² + y4
Solution:
= 4x² 2x − y y² − 4x² 2x + y y² + 16x4 16x3 + y4
− 2xy + + 2xy + + 4x²y²
= (2x − y)(4x² + 2xy + y²) – (2x + y)(4x² – 2xy + y²) + 16x³
(4x² − 2xy + y²)(4 x ² + 2 xy+ y ²) 2.4x².y² + (y²)²
(4x²)² + – 4 x ² y ²
= (4 [(2x)³ − y³] − [(2 x)³ + y ³] y ²) + (4 x² + 16x³ (2 x y)²
x² − 2xy + y²)(4 x ² + 2 xy+ y²)² −
= (4x² − 8x³ − y³ − 8x ³ − y³ y ²) + (4x² + 2xy + 16x³ ² + y ² − 2 x y)
2xy + y²)(4 x ² + 2 xy+ y²) (4 x
= (4x² − 2xy + −2y³ ² + 2 xy + y ²) + (4x² + 2xy + 16x³ x ² − 2 x y + y²)
y²)(4 x y²) (4
16x³ − 2 y ³
= (4x² − 2xy + y²)(4 x ² + 2 xy+ y ²)
2(8x³ − y ³)
= (4x² − 2xy + y²)(4 x ² + 2 xy+ y ²)
= 2(2x – y)(4x² + 2xy + y²) = 2(2x – y)
(4 x² − 2xy + y²)(4 x ² + 2 xy+ y ²) (4x² − 2xy + y²)
196 Oasis School Mathematics-10
Example: 3 (x – y)² − z² (y − z)² − x² (z− x)² − y ²
x² – (y+ z)² y² − (z + x)² z² – (x + y)²
Simplify: + +
Solution:
(x – y)² − z² + (y − z)² − x² + (z− x)² − y ²
x² – (y+ z)² y² − (z + x)² z² – (x + y)²
= (x – y + z)(x − y − z) + (y − z + x)(y − z − x) + (z− x + y)(z− x− y)
(x + y+ z)(x− y− z) (y + z + x)(y− z− x) (z + x + y)(z − x − y)
= x – y+ z + y − z + x + z −x+y
x + y+ z x + y + z x +y+z
= x–y+z+y−z+x+z−x+y
(x + y+ z)
(x + y + z)
= (x + y+ z) = 1
Example: 4
Simplify: 1 + 2a + a444+aaa7³x8+4–4−xa83ax848−a7x8
a+x a² + x²
Solution: 1 2a a444+aaa7³x8+4–4−xa83ax848−a7x8
+ a² + x²
= a x + +
1 2a 4a3 (a4 + x4 )
+ a² + x² (a4 + x4 )(a4 − x4
= a x + − )
1 2a 4a3
= a + x + a² + x² − a4 − x4
1 2a(a² – x²) – 4a3
= a + x + (a² + x²)(a ² – x ²)
= a 1 x + 2a³ – 2ax² − 4 a3
+ (a² + x²)(a ² – x ²)
= a 1 x + –2a(a ² + x ²) ²)
+ (a² + x²)(a ² – x
1 2a a − x − 2a
= a + x − (a + x)(a – x) = (a + x)(a− x)
−(a + x) = -1 = 1
= (a + x)(a− x) (a- x) x–a
Oasis School Mathematics-10 197
Example: 5
1 + a2 1 + 2a4
a2 (a2–x2) (x2+a2) (x8–a8)
Solution:
= a2 1 + a2 1 + 2a4
(x2–a2) (x2+a2) (x8–a8)
1 1 1 2a4
a2 a2–x2 a2+x2 (x8–a8)
[ ]= + +
[ ]
= 1 a2+x2+a2–x2 + 2a4
a2 (a2–x2) (a2+x2) (x8–a8)
1 2a2 2a4
a2 a4 – x4 (x8–a8)
[ ]= +
= 2 + 2a4
a4 – x4 (x4–a4) (x4+a4)
= 2(a4+x4)–2a4
(a4–x4) (a4+x4)
= 2a4 + 2x4–2a4
(a4–x4) (a4+x4)
= 2x4
a8 – x8
Exercise 11.1
1. (a) On what condition is the rational fraction p(x) not defined?
q(x)
(b) For what value of x, is 1 not defined?
2x–1
(c) What value of x makes the expression 5x+1 undefined?
2x–6
p(x)
(d) On what condition is q(x)–r(x) not defined?
2. Simplify:
(a) x2–xy ÷ x2(x–y) (b) x2–xy ÷ x2(x–y)
x2+xy x3+x2y x2+xy x3+x2y
(c) x2–xy+y2 ÷ x2+xy+y2 (d) x2 + y2
x–y x+y y(x–y) x(y-x)
(e) x2+y2 + x2 - y2
xy y(x+y) x(x+y)
3. Simplify the following:
(a) a+b − a−b + 2ab (b) 1 − x − 1+ x + 1 4x
a−b a+b a2 − b2 1 + x 1− x − x²
(c) x + y − x − y + 2(x² + y²) (d) a + b − a − b − 4ab
x − y x + y x² − y² a − b a + b a ² + b²
198 Oasis School Mathematics-10
22 1
4. Simplify: (a) (x– 2)(x− 3) + (x− 1)(3 − x) + (1 − x)(2 − x)
2(a − 3) (a − 1) a−2
(a– 4)(a− 5) (3 − a)(a− 4) (5 − a)(a− 3)
(b) + +
13 2
(c) (a − 3)(a+ 2) + (a+ 2)(4 − a) + (a − 3)(a − 4)
1 11
(d) (x − y)(z − x) + (y− x)(y− z) + (z− x)(z− y)
5. Simplify: (a) 112
x ² − 5x+ 6 + x ² − 3x+ 2 − x² − 4x + 3
(b) x² x −1 2 + x² x −2 6 + x² x−5
− 3x + − 5x+ − 8x + 15
(c) x² 1 + 21 + x ² 1 35 + x² − 1 + 15
− 10x − 12 x+ 8x
x2+ 2xy–8y2 x2–4y2 x
(d) x2–4y2 – x2–2xy + x+2y
6. Simplify: (a) a² a−b b² + a² a+b b² + a4 + 2b³ + b4
− ab + + ab+ a²b²
(b) 2a + b b² + 2a − b b² − 16a4 + 2b³ + b4
4 a ² + 2ab + 4 a ² − 2 ab+ 4a²b²
(c) 3a + 1 + 3a − 1 − 81a4 2
9 a ² + 3a + 1 9a² − 3a+1 + 9a² + 1
(d) x² x+3 9 + x² x−3 9 − x4 54 81
+ 3x + − 3x + 9x²
+ +
(e) 1 a+2 − 1 a−2 − 1 + 2a² a4
+ a + a² − a + a² a² +
7. Simplify: ((aa)) a² − (b − c)² + b² − (a− c)² + c² − (a − b)²
(a+ c)² − b² (a + b)² – c² (b + c)² − a²
((bb)) 4(44(a1c1²++−a2()3a+b)4²−(−141−9cba))²² ++ 2(9(21ba+1²+a−23()b4−)c²1−–−121aa64 )c²² + 16c² − (2a − 3b)²
(3b + 4c)² − 4a²
8. Simplify: (a) 2 – 1 + 1
(a+1) (a+2) (a+2)2 (a+1)2 (a+2)2
(b) 1 + 1 – 1
2(1–a2) 2(1+a2) 1-a4
(c) a + a – a3
2(a2–b2) 2(a2+b2) a4–b4
Oasis School Mathematics-10 199
9. Simplify: a² + ac − a−c − a 2c
a²c– c³ c(a + c) ² − c²
10. Simplify1: −(aa) − a2114+a+−1aa +− 1 4+aa² − 1 +8aa4 xx(b+−)aa −x4xx 2a–+−xa2aa − 4ax ² − 8a³x
1 +a x² + a x4 + a4
x 1(c) +xx2–2+ 4x1y22y + x² 2x + x4 4x³ 1 +(da) −b 1ba++22aa–ab−²bb2+2abbb++² 2aab −++ 4aa³ b+ 4 a³ b
+ 4y² + 16y4 a−²a++bb² a+4 +b b4 a4 + b4
− 2y
1 + x1y–1−1 yxxy22 − 1 − 1
1 +(exy) −
1 + y² x² y²
x² y² x²
11. Simplify : (a) 1+ x + 1− x + 4x4x+ ²x4x4+3 8x3 4 (b) x + y + x − y + 4xx²43x+xy4y–+yy4²x4 −y3x84 x−3 y 4
1− x 1+ x 11+ –x 1− x x − y x + y y
12. Simplify : (a) 1 + 2 + x4xx4484+–+14+ 8 (b) a 1 b + a 2a + a444a+a7a8+³b–44ba−83ba488−a7b8
+ x² + x8 −1 + ² + b²
x 1 1
(1a)+2 a3a+a2 –11 3a a 2 x – 31y 3y xy
13. Simplify : –a −1 1 + − a² + + a³ (xb)+ y −x2x– −y2y − y² − x² + x3 + y³
1 1
Note:
Simplification of algebraic fractions containing more than 3 terms is not in the syllabus prescribed
by CDC. These questions are given in miscellaneous exercise for further practise.
Answer
1. Consult your teacher 2. Consult your teacher
13.. C(ao)nsau62lt−abybo2ur t(eba)c0her(c) 22((xx.+-yCy)o)ns(udlt) ya8o4au-bbr34tea4c.her(a) 1 (b) 5
(x -2) (x -3) (a-3)(a-4)(a-5)
3. ((ac)) (aa6+2 −a2bb)(2a9-3()b(a)-04) (cd)) 2(x((xx-+-yy)2y()z)-x()d) a84a-bb345. 4(.a) (a0) ((xb-)2()1x(x-32-x3)(-)x7-3()b()c()a(-x3-)3(a)3(5-x4-)7(a)-5()d) x+2y
x
6. ((ca)) (aa2+2+(2aa)+b(+ab9-)b3)2(a(-b4)) 4(ad22)-(22(xaab--yb+))2(bz2-x()c) 92a(23-3a5a-.1+)1(a(d)) 0x22 -((3xbx-)3+)(9x-32(xe))(-x71-+3)a42(+c)a(4x-3)3(x-7) (d) x+2y
x
677.. (a((a)a))1a1126-2(+aab(8a5a)+b+1(bb)b)218x6.(8ba-(a)7ax8)4a(a22(+c-(122)1)aax2b-,8b+-8)2b(x5b276)y(0c8,) 9((c2ad)(2)3-a3a8a48a-a–1b-+7b)2b148 (d(9e).)xxa22x4-+3(23x-ycyx-23+4 )9(f)(e14)-xx12+4 a(4g2+) am24 m4 -1
71.0. ((ah))1106-aa8(5i) (ab24)+a1x2b682a-a7x(8j) 4(4c+)aa2x28 -82x5876.y(a8 ) 2((d(11)+-xa8x)8a)-7bb8(b)(e2()x(+xxx-42yy-y)y)24 9(.f)(a14)-xx12-41 (g(b) )m2bm41--a1
1110..(a((h)a))20(((111–+-x(ax)i)))(11a(+b24+)aa32b2()x(2x+-y(yjb)))) 4(4+1xa2-a2.y2()a(xx)338+x.1–y(1a3 ))(2b(1()111+-.bx1–x)(aa)) x(b12)3+.2(1x((a+x)-yy(b))(1)-2a9)x.1(21(a+)a(x3c)1)-1((bx()2b-)9(3x)b2-(y1x-x)a2x(-x3136+)y3)
10. ((ad)) 16a 2x (b) (x1-2y.)((xxa3)3+1 y3()b)111. 1(a3). (xa2)+0,1 (b()b)02x(2c) (c1)4(.x2a-+393c)2(xx12 -51.6x3)
((1x-a2 -)a(12+)(a43x)2 -a2 )
Project Work
Co(ldle) c(tx2-5a26d)a(i42fxxfe2-rae2n)t 1t2y.p(eas) 1of(bq) u1es1t3io. n(as)0o,f(ba)lg0eb(cr)aic14f.raa+c3tcion1s5.. x3Use the tool of
simplification for each type and present it in your classroom.
200 Oasis School Mathematics-10
Unit Equational
Problems
12
12.1 Warm-up Activities
Discuss the following in your class and draw a conclusion.
In two equations x+y=5 and x–y = 1 what values of 'x' and 'y' satisfy both these equations?
x = 1, y = 2 or x = 2, y = 3, or x = 3, y = 2.
• The present age of a boy is x years. What will be his age after 5 years? What was his age
before 5 years?
• Are 2x+y=5 and x–y=2 simultaneous equations?
• How to solve the equations 2x+y = 5 and x–y = –2?
12.2 Simultaneous Equations
Let’s take two equations x + y = 5 and 2x – y = 4.
Here variables 'x' and 'y' are related to each other.
We can obtain the values of x and y, which satisfy both equations. These equations are a
system of simultaneous equations.
In simultaneous equations, variables are related to each other. This relation of the variables
can be expressed in the form of an equation. If there are two or more such equations then we
can obtain the values of the variables by solving them.
Word Problems of Simultaneous Equations
When the relation between the variables is given in a statement, we have to convert that relation
in terms of an equation, assuming the variables as 'x' and 'y'. If two such equations are obtained
from the given condition, we can solve these two equations to get the values of 'x' and 'y'.
Worked Out Examples
Example: 1
The sum of two numbers is 52. If their difference is 16, find the numbers.
Solution:
Let the larger number be x and the smaller number be y.
From the given condition,
x + y = 52 ............. (i)
Oasis School Mathematics-10 201
Again, from the given condition, Alternative method
x – y = 16 ............. (ii) From the given condition,
x+y = 52
Now, adding equations (i) and (ii)
2x = 68 y = 52–x ..................(i)
x = 68 and x – y = 16
x = 2 From (i)
34
x–(52–x) = 16
Substituting the value of 'x' in (i) or, x–52+x=16
or, 2x = 16+52
x + y = 52
or, 2x = 68
or, y = 52 – 34 ∴ x = 68 = 34
2
or, y = 18 Substituting the value of x
∴ Required numbers are 34 and 18. in (i), y = 52–x
Example: 2 = 52–34 = 18
∴ Required numbers are 34 and 18.
Two numbers are such that the sum of the greater number and two times the smaller is 18.
If the smaller number is subtracted from the two times of the greater number, the result is
11, find the numbers.
Solution:
Let the greater number be x and the smaller number be y
From the given condition, x + 2y = 18 ............. (i)
Again, from the given condition,
2x – y = 11 ............. (ii)
Multiplying equation (ii) by 2 and adding these equations
x + 2y = 18 Alternative method
4x – 2y = 22 From the given condition,
40 x + 2y = 18
Adding, 5x = 40 x = 18–2y ..........(i)
5 Again, 2x-y=11 ...........(ii)
or, x = 8 2(18–2y)-y = 11
= 36–4y-y=11
or, x 18 36–5y=11
= 18 –5y = 11-36
Substituting the value of x in (i) = 18 – 8 -5y = -25
= y = 5.
x + 2y
or, 8 + 2y
or, 2y
or, 2y = 10 From (i)
or, y = 10 x = 18–2×5
y = 2 x = 18–10
or, 5 x=8
∴ The greater number = 8 ∴ Required numbers are 8 and 5.
and the smaller number = 5
202 Oasis School Mathematics-10
Example: 3
The cost of 5 kg apples and 6 kg oranges is Rs. 280 and the cost of 3 kg apples and 5 kg
oranges is Rs. 210. Find the cost per kg of apples and oranges.
Solution:
Let the cost per kg of apples be Rs. x and the cost per kg of oranges be Rs. y.
From the given condition,
5x + 6y = 280 ............ (i) Alternative method
Again, from the given condition From the given condition
3x + 5y = 210 ............ (ii) 5x + 6y = 280
Multiply equation (i) by 3 and (ii) by 5, then subtract (ii) from (i), 6y = 280 – 5x
15x + 18y = 840 y = 280–5x ...................(i)
6
15x + 25y = 1050
Again 3x + 5y = 210
Subtracting, (–) (–) (–)
280–5x
or, –7y = –210 or, 3x + 5 ( 6 ) = 210
y =
or, y = 210 or, 18x+1400–25x = 210
7 6
30 or, –7x + 1400 = 1260
Substituting the value of y in equation (i) or, 7x + 1400 – 1260
5x + 6y = 280 or, 7x = 140
or, 5x + 6 × 30 = 280 or, x= 140 = 20
7
or, 5x + 180 = 280 280–100
= 6 = 30
or, 5x = 100
or, ∴ Cost per kg apples = Rs. 20
100
x = 5 Cost per kg oranges = Rs. 30
or, x = 20
∴ Cost per kg apples = Rs. 20, cost per kg oranges = Rs. 30.
Example: 4
Two years ago, a father's age was nine times the son's age, 3 years later it will be five times
only. Find the present ages of the father and the son.
Solution:
Let, the present ages of the father and the son be 'x' and 'y' years respectively.
From the given condition,
x – 2 = 9(y – 2)
or, x – 2 = 9y – 18
or, x – 9y = – 18 + 2
or, x – 9y = – 16 ............. (i)
Again, from the given condition,
x + 3 = 5 (y + 3)
or, x + 3 = 5y + 15
Oasis School Mathematics-10 203
or, x – 5y = 15 – 3
or, x – 5y = 12 ............. (ii)
Now, subtracting equation (ii), from (i)
x – 9y = – 16
x – 5y = 12
(–) (+) (–)
– 4y = –28
or, 4y = 28
or, y = 28
4
or, y = 7
Substituting the value of y in (i)
x – 9y = – 16
or, x – 9 × 7 = – 16
or, x – 63 = – 16
or, x – 63 = – 16
or, x = –16 + 63
or, x = 47
∴ Father's age = 47 years, son's age = 7 years.
Example: 5
The present age of the father exceeds twice the age of the son by two years. When the
son will become as old as his father now, the sum of their ages will be 96 years. Find
their present ages.
Solution:
Let the present ages of the father and the son be 'x' years and 'y' years respectively.
From the given condition, x = 2y + 2
A gain, or, x – 2y = 2 ............ (i)
Son becomes as old as his father now after (x-y) years.
From the given condition ( x+x–y) + (y+x–y) = 96
2x–y+x = 96
3x–y = 96...............(ii)
Multiplying equation (ii) by 2 and subtracting from (i)
x – 2y = 2
6x – 2y = 192
(–) (+) (–)
–5x = –190
or, 5x = 190
or, x = 190
5
or, x = 38
204 Oasis School Mathematics-10
Substituting the value of x in (i)
x – 2y = 2
or, 38 – 2y = 2
or, 38 – 2y = 2
or, –2y = 2 – 38
or, – 2y = – 36
or, 2y = 36
or, y = 36
2
or, y = 18 years
∴ Father's age = 38 yrs, Son's age = 18 years.
Example: 6
The sum of the digits of a two digit number is 9. When 45 is subtracted from the number,
the digits will be reversed. Find the number.
Solution:
Let the digit in the tens place = x
The digit in the ones place = y Try to solve the given equation
by substitution method also.
Then the number = (10x + y)
From the given condition,
x + y = 9 ..................(i)
Again, from the given condition,
10x + y – 45 = 10y + x
or, 10x + y – 10y – x = 45
or, 9x – 9y = 45
or, 9(x – y) = 45
or, x – y = 45
9
or, x – y = 5 ................ (ii)
Now, adding equation (i) and (ii)
x+y = 9
x–y = 5
2x = 14
or, x = 14 =7
2
Substituting the value of 'x' in (i),
x + y = 9
9
or, 7 + y =
or, y = 9 – 7
or, y = 2
Therefore, the required number = 10x + y = 10 × 7 + 2 = 72
Oasis School Mathematics-10 205
Example: 7
Two numbers are in the ratio of 2:3. If 5 is added to both of them, their new ratio is 5:7.
Find the numbers.
Solution:
Let the two numbers be 'x' and 'y'.
From the given condition, x = 2
y 3
or,
3x = 2y
or, 3x – 2y = 0 ............ (i)
Again, from the given condition,
x+5 = 5
y+5 7
or, 7x + 35 = 5y + 25
or, 7x – 5y = 25 – 35
or, 7x – 5y = –10 .............. (ii)
Multiplying equation (i) by 5 and (ii) by 2 and subtracting (ii) from (i),
15x – 10y = 0 Alternative method
14x – 10y = –20
(–) (+) (+)
Let two numbers be 2x and 3x
x = 20 From the given condition,
Substituting the value of x in (i), 2x+5 = 5
3x+5 7
3x – 2y = 0
or, 14x+35 = 15x + 25
3 × 20 – 2y = 0 or, 15x–14x = 35–25
or, 2y = 60 or, x = 10
or, y = 60 ∴ Required numbers are
2
or, y = 30 2x = 2×10 = 20
∴ Required numbers are 20 and 30. and 3x = 3×10 = 30
Example: 8
If 1 is subtracted offrotmhetfhreacntiuomn,eirtabtoercoomf tehse12f.rFacintidonth, eitobreigcoinmael sfr31ac. tIifo2n.is subtracted from
the denominator
Solution:
Let the numerator of the fraction be x and the denominator y. Then the original fraction will
x
be y .
From the given condition, x–1 = 1
y 3
or, 3x – 3 = y
or, 3x – y = 3 ............ (i)
206 Oasis School Mathematics-10
Again, from the given condition, yx–2 = 1
2
or, 2x = y – 2
or, 2x – y = – 2 .............. (ii)
Subtracting equation (ii) from (i)
3x – y = 3
2x – y = –2
(–) (+) = (+)
x = 5
Substituting the value of x in (i)
3x – y = 3
or, 3 × 5 – y = 3
or, 15 – y = 3
or, – y = 3 – 15
or, – y = –12
or, y = 12
∴ The original fraction
x = 5
y 12
Example: 9
Bus 'A' starts from Gorkha to Pokhara at 7 a.m. at uniform speed. One hour later, bus 'B' starts
from Pokhara to Gorkha at uniform speed, which is 4 km/hr more than A's speed. Both of
them meet at 10 a.m. If the distance from Gorkha to Pokhara is 108 km, find the speed of A
and B.
Solution:
Let, The speed of bus A = x km/hr
The speed of bus B = y km/hr
Here, A starts the journey at 7 am and B starts the journey at 8 a.m.
Since, they meet at 10 a.m. = 3 hrs
Time period of A's journey
Time period of B's journey = 2 hrs.
Distance covered by A in 3 hrs = 3x km.
Distance covered by B in 2 hrs = 2y km.
From the given condition,
3x + 2y = 108 ............. (i)
Again, from the given condition,
y = x+4
or, x – y = –4 ............. (ii)
Multiplying equation (ii) by 2 and adding (i) and (ii),
3x + 2y = 108
2x – 2y = –8
Oasis School Mathematics-10 207
5x = 100
or, x = 100
or, 5
x = 20 km/hr
Substituting the value of 'x' in (ii), we get,
x – y = – 4
or, 20 – y = – 4
or, – y = –24
or, y = 24 km/hr
∴ Speed of bus A = 20 km/hr
Speed of bus B = 24 km/hr.
Exercise 12.1
1. (a) The present age of a man is x years. What will be his age after 10 years?
(b) Now my age is x years. What was my age 7 years before?
(c) The present age of two friends is x years and y years. What was their age z years
ago?
(d) Ansu and Aadhya have ‘x’ and ‘y’ marbles. If Ansu gave 10 marbles to Aadhya,
find the number of marbles each has.
(e) A number of two digits has ‘x’ in tens’ place and ‘y’ in ones’ place. What are the
number and the number formed by reversing its digit?
(f) The sum of a number and its reciprocal is 147, convert this statement into equation.
(g) The present ages of two brothers are 20 years and 15 years. What will be their
ages after x years?
(h) The present ages of a father and his son are ‘x’ years and ‘y’ years respectively.
What was the sons age when the father was as old as his son now?
2. (a) Find the two numbers whose sum is 50 and the difference is 4.
(b) The sum of two numbers is 80. If the smaller number is 40 less than the greater
number, find the numbers.
(c) The sum of two numbers is 36. If the bigger number is 10 more than the smaller
number, find the numbers.
(d) The difference between the two numbers is 8. If 3 times the greater number is
equal to 5 times the smaller, find the numbers.
(e) Divide 40 into two parts such that the bigger part is 8 more than the smaller part.
3. (a) The sum of two numbers is 15. If 2 times the smaller number is subtracted from
3 times the bigger number, the result is 15. Find the numbers.
208 Oasis School Mathematics-10
(b) The bigger number is 20 more than the smaller number. If the difference
between 3 times the bigger number and 4 times the smaller number is 35, find
the numbers.
(c) Two numbers are such that the greater number exceeds two times the smaller
number by 2 and the smaller number exceeds one–fourth of the greater number
by one. Find the numbers.
(d) Divide 26 into two parts such that one fourth of the smaller number exceeds
one–seventh of the larger number by 1.
4. (a) The cost of 5 kg rice and one kg flour is Rs 100. If the cost of 5 kg rice is as much
as the cost of 3 kg flour, find the cost per kg of the rice and flour.
(b) The cost of 5 pens and 6 copies is Rs. 190. If the cost of 1 pen is Rs. 5 more than
the cost of 1 copy, find the cost of a pen and the cost of a copy.
(c) A and B have altogether 100 chocolates. If A gives 10 chocolates to B, they would
have equal number of chocolates. Find the number of chocolates each has.
(d) Saurav and Asim have 80 marbles altogether. If Saurav gives 10 marbles to Asim,
Asim has three times as many as Saurav. Find the number of marbles each has.
(e) If A gives 2 sweets from what he has to B, then they will have equal number of
sweets. If B gives 2 sweets from what he has to A, then A will have three times as
many sweets as B. Find the number of sweets each has.
5. (a) The sum of the present ages of the father and his son is 60 years. 10 years ago, the
father was three times as old as his son. Find their present ages.
(b) A father is 20 years older than his son. After 5 years, the father will be 3 times as
old as his son. Find their present ages.
(c) Five years ago, a father was six times as old as his son. Two years hence, the age
of the father will be 1 year more the three times the age of his son. Find their
present ages.
(d) Six years ago, a man was six times the age of his son. Four years hence, thrice his
age will be equal to eight times his son’s age. Find their present ages.
(e) A son was born when his father was 28 years old. Three years hence, the ratio of
the age of the father to his son will be 5:1. Find their present ages.
(f) A year hence, a father will be 5 times as old as his son. Two years ago, he was three
times as old as his son will be four years hence. Find their present ages.
(g) One year ago, a father was six times as old as his son. Three years ago the age of
the father was four times the age his son will be one year hence. Find their present
ages.
6. (a) The sum of the ages of a father and his son is 84 years. When the father was as
old as his son now, he was five times as old as his son. Find their present ages.
(b) The sum of the ages of a father and his son is 45 years. If both live on until the
son’s age becomes equal to the father’s age now, the sum of their ages will be 95
years. Find their present ages.
Oasis School Mathematics-10 209
(c) Himanka says to Zenith, “When I will be as old as you are now, the ratio of our ages
(d) will be 3:5.” If the sum of their present ages is 48 years, find their present ages.
7. (a)
(b) Rajashree says to Anasuya, “I was twice as old as you were when I was as old as
(c) you are now.” If the sum of their present ages is 50 years, find their present ages.
(d)
(e) The sum of the digits of a two digit number is 9. If 27 is subtracted from the
number, the positions of the digits will be reversed. Find the number.
(f)
(g) A certain number of two digits is 4 times the sum of its digits. If 9 is added to the
number, then the positions of the digits will be reversed. Find the number.
8. (a)
(b) A number of two digits exceeds four times the sum of the digits by 3. If 36 is
(c) added to the number, the digits will be reversed. Find the number.
(d)
9. (a) A number of two digits is 3 more than 4 times the sum of the digits. If 27 is added
to the number, the position of the digits will be reversed. Find the number.
(b)
(c) When a number formed by reversing the digits of two numbers is added with
the original number, the result will be 99. If the same number is subtracted from
(d) the original number, the result will be 27. Find the original number.
10. (a) A number consists of two digits whose sum is 9. If three times the number is equal
to eight times the number formed by interchanging the digits, find the number.
In a two-digit number, the digit in the place of ones is five more than the digit in
the place of tens. If 9 is subtracted from the three times the original number, the
digits will be reversed. Find the original number.
The ratio of two numbers is 2:3. When 4 is subtracted from both of them, the new
ratio is 3:5. Find the numbers.
The ages of two boys are in the ratio 5:7. Eight years ago their ages were in the
ratio 7:13. Find their present ages.
Five years ago, the ratio of the ages of two brothers was 3:5. 5 years hence, ratio
of their ages will be 5:7. Find the ratio of their present ages.
Eight years ago, the ratio of the ages of two brothers was 2:3. After 8 years, the
ratio of their ages will be 4:5. Find their present ages.
In a fraction, if 1 is subtracted from the numerator and 3 is added to the
denominator, it becomes 1 . When 1 is added to the numerator and 2 is subtracted
2
fIthrnoeamdfretanhcoetimdoeninn,aoiftmo1rii,nstahatdeodrr,eeisdtubtloet citsohme31ne.suFmi45n.edFraitnhtoder,othtrhiegeoirnreiasgluifnlrtaailcstfir21oa.ncIt.fio1nis. subtracted from
If 2 and 1 respectively are subtracted from the numerator and denominator of a
fbreaccotimonesit1b. eFcinodmtehse21o.rIifg2inaanldfr1acatrieona.dded to the numerator and denominator, it
If the numerator of a fraction is multiplied by 4 and the denominator is reduced by
2, the result is 2. If the numerator of the fraction is increased by 15 and 2 is subtracted
from the double of the denominator, the result is 9 . Find the fraction.
7
The length of a rectangle is 12 cm more than the breadth. If the perimeter of the
rectangle is 96 cm, find the length and breadth of the rectangle.
210 Oasis School Mathematics-10
(b) Four times the breadth of a room is equal to 3 times the length. If the breadth
had been 1 m. more and the length 1 m. less, the room would have been a
square. Find the dimension of the room.
(c) The perimeter of a rectangle is 80 cm. If the length had been reduced by
one-sixth and breadth had been increased by its one-fourth, it would have
been a square. Find the length and breadth of the rectangle.
11. (a) Two buses are coming from two villages situated just in the opposite direction.
The uniform speed of one bus is 10 km/hr. more than that of the other one,
and they have started their travel at the same time. If the distance between the
villages is 450 km and they meet after 5 hours, find their speed.
(b) A bus was started at 6 am from Biratnagar to Kathmandu at a speed of 30 km/
hr. Two hours later, another bus was also started at a speed of 60 km/hr. from
Biratnagar. At what time will the second bus overtake the first bus?
Answer
1. Consult your teacher
2. (a) 27, 23 (b) 20, 60 (c) 13, 23 (d) 12, 20 (e) 16, 24
(c) 8, 3 (d) 14, 12 (e) 10, 6
3. (a) 9, 6 (b) 45, 25 (c) 60, 40 (d) 30, 50
(c) 35 years 10 years (d) 36 years 11 years
4. (a) Rs.15, Rs.25 (b) Rs. 20, Rs. 15
5. (a) 40 yrs, 20 yrs (b) 25 years, 5 years
(e) 32 yrs. 4 yrs (f) 29 years. 5 years (g) 31 yrs. 6 yrs. 6. (a) 54 years 30 years (b) 35years, 10 years.
(c) 36 yrs. 12 yrs. (d) 30 years 20 years 7. (a) 63 (b) 12 (c) 59
(d) 47 (e) 63 (f) 72 (g) 27 8. (a) 16, 24 (b) 15 years 21 years (c) 2:3 (d) 24 years 32 years
(c) 24 m, 16 m
9. (a) 11 (b) 1 (c) 4 (d) 3
17 4 5 8
10. (a) 30 cm, 18 cm (b) 8m, 6m
11. (a) 50 km/hr. 40 km/hr. (b) 10 a.m.
12.3 Quadratic Equations
Discuss the following with your friend and draw a conclusion. Solve the given
problem: x2 – 4 = 0.
• What are the values of x? Do both the values satisfy the given equation or not?
• What are the factors of x2 – 3x + 2? Solve the equation x2 – 3x + 2 = 0.
• What are the solutions to the above equation? Do both values satisfy the given
equation?
• Solve the equation x2 – 4x + 2 = 0.
• How do you factorise x2 – 4x + 2 ?
• If it is not factorised, how do you get the values of x?
Oasis School Mathematics-10 211
Give some examples of quadratic equations.
Remember !
In the quadratic equation ax2+bx+c=0 , a ± 0
x = –b ± b2 – 4ac
2a
If the given equation is not factorised, use this formula to get the values of x.
Some problems like the age of two people, area of rectangles, formation of two-digit numbers
are related in such a way that their relation can be expressed in the form of an quadratic
equation. These quadratic equations can be solved to get the required values. Every quadratic
equation has two roots. Both roots may not satisfy the given equation. The root which doesn't
satisfy the given condition is to be rejected.
Worked Out Examples
Example: 1
If 4 is subtracted from half of the square of a natural number, the result is 14. Find the
number.
Solution:
Let the required number be 'x'.
From the given condition, 21x² – 4 =
14
14 + 4
or, 1 x² =
2 18
36
or, 12x² =
or, x² =
or, x² = (± 6)²
or, x = ± 6
Hence, the required natural number is 6.
Example: 2
The sum of two numbers is 12 and the sum of their squares is 80. Find the numbers.
Solution:
Let the required numbers be 'x' and 'y'.
From the given condition,
x + y = 12
or, y = 12 – x................. (i)
Again, from the given condition,
x² + y² = 80
or, x² + (12 – x)² = 80
212 Oasis School Mathematics-10
or, x² + 144 – 24x + x² = 80
0
or, 2x² – 24x + 144 – 80 = 0
0
or, 2x² – 24 x + 64 = 0
0
or, x² – 12x + 32 = 0
0
or, x² – 8x – 4x + 32 = 8
or, x (x – 8) – 4 (x – 8) = 12 – x
12 – 8
or, (x – 8) (x – 4) = 4
4
Either, x–8 = 12 – x or, x–4 = 0
12 – 4 or, x=4
or, x =
When, x = 8, from equation (i),
y =
or, y =
or, y =
Again, when, x =
y =
or, y =
y = 8
∴ The required numbers are 4 and 8 or 8 and 4
Example: 3
Divide 16 into two parts such that twice the square of the larger part exceeds the square of
the smaller by 164.
Solution:
Let the larger part = x
then the smaller part = (16 – x)
Now, from the given condition,
2x2 – (16 – x)2 = 164
or, 2x2 – (256 – 32x + x2) = 164
or, x2 + 32x – 420 = 0
or, x2 + 42x – 10x – 420 = 0
or, x (x + 42) – 10 (x + 42) = 0
or, (x + 42) (x – 10) = 0
Either, (x + 42) = 0, or, x = – 42 (neglected)
or, x = 10
or, x – 10 = 0,
∴ Larger part = 10
Smaller part = (16 – 10) = 6
Example: 4
The product of two consecutive odd numbers is 143. Find the numbers.
Solution:
Let the two consecutive odd numbers be x and x + 2.
Oasis School Mathematics-10 213
From the given condition,
x(x + 2) = 143
x² + 2x = 143
x² + 2x – 143 = 0
x² + 13x – 11x – 143 = 0
or, x (x + 13) – 11 (x + 13) = 0
(x + 13) (x – 11) = 0
Either, x + 13 = 0
or, x = – 13 (neglected)
or, x – 11 = 0
or, x = 11
∴ Two numbers are x = 11 and x + 2 = 11 + 2 = 13
Example: 5
The present ages of the elder and younger brothers are 13 years and 7 years respectively. In
how many years will the product of their ages be 280?
Solution:
Let after 't' years the product of their ages be 280.
From the given condition,
(13 + t) (7 + t) = 280
or, 91 + 13t + 7 t + t² = 280
or, t² + 20 t + 91 = 280
or, t² + 20 t + 91 – 280 = 0
or, t² + 20t – 189 = 0
or, t² + 27 t – 7t – 189 = 0
or, t(t + 27) – 7( t + 27) = 0
or, (t + 27) (t – 7) = 0
Either, t + 27 = 0 or, t = – 27 ( not possible)
or, t – 7 = 0 or, t = 7
Hence, t = 7
∴ After 7 years the product of their ages will be 280.
Example: 6
The product of the ages of a father and his son is 180 years. When the son becomes as old as
his father now, the sum of their ages will be 84 years. Find their present ages.
Solution:
Let the present age of the father and his son be x years and y years respectively.
From the given condition, xy = 180 .............(i)
214 Oasis School Mathematics-10
The son will be as old as his father now after (x – y) years.
Then, x + (x - y) + y + (x – y) = 84
3x – y = 84
y = 3x – 84 ..........(ii)
From (i) and (ii)
xy = 180
or, x(3x – 84) = 180
or, 3x2 – 84x = 180
or, 3x2 – 84x –180 = 0
or, 3(x2 – 28x – 60) = 0
or, x2 – 28x – 60 = 0
or, x2 – (30 – 2) x – 60 = 0
or, x2 – 30x + 2x – 60 = 0
or, x(x – 30) + 2(x – 30) = 0
or, (x – 30) (x + 2) = 0
Either x – 30 = 0 ⇒ x = 30
or, x + 2 = 0 ⇒ x = –2 (not possible)
From (i) xy = 180
or, 30y = 180
or, y = 6
∴ Father’s age = 30 years, Son’s age = 6 years
Example: 7
The product of the digits of a two-digit positive number is 21. If 36 is added to the number,
the digits will interchange their places. Find the number.
Solution:
Let x be the digit in the place of tens and y be the digit in place of ones. Then the number is (10x + y).
If the digits interchange their places, then the new number is (10y + x).
From the condition,
xy = 21 ..................(i)
Also, 10x + y + 36 = 10y + x
or, 10x + y – 10y – x = –36
or, 9x – 9y = –36
or, 9 (x – y) = –36
or, x – y = –4
or, y = x+4 ..............(ii)
From (i) and (ii)
or, xy = 21
or, x(x + 4) = 21
or, x2 + 4x = 21
or, x2 + 4x – 21 = 0
Oasis School Mathematics-10 215
or, x2 + 7x – 3x – 21 = 0 or, x = –7 (not possible)
or, x(x + 7) – 3(x + 7) = 0 or, x = 3
or, 0
Either (x + 7) (x – 3) = 0
or, x + 7 = 0
From (i) x – 3 =
xy = 21
3xy = 21
y = 7
∴The required number = 10x + y = 37.
Example: 8
The perimeter of a rectangle is 72 m and the area is 320 m². Find the length and breadth of
the rectangle.
Solution:
Let the length and breadth of the rectangle be x m and y m respectively.
Then, perimeter of the rectangle = 2(x + y)
∴ 2(x + y) = 72
or, x + y = 36
or, y = ( 36 – x) ………….. (i)
Since, the area of rectangle is 320 m2
xy = 320
or, x(36 – x) = 320
or, 36x – x2 = 320
or, –x2 + 36x – 320 = 0
or, –(x2 – 36x + 320) = 0
or, x2 – 36x + 320 = 0
or, x2 – 20x – 16x + 320 = 0
or, x(x – 20) – 16(x – 20) = 0
or, (x – 20) (x – 16) = 0
Either x – 20 = 0 or x = 20
or, x – 16 = 0 or, x = 16
when x = 20
y = 36 – 20 = 16
when, x = 16
y = 36 – 16 = 20.
Taking the longer part as length,
length of the rectangle = 20 m and the breath of the rectangle = 16 m.
216 Oasis School Mathematics-10
Exercise 12.2
1. (a) If ‘x’ be a natural number, what are the three consecutive natural numbers that are
greater than x?
(b) ‘x’ is an odd number. What are the three consecutive odd numbers greater than x.
(c) ‘x’ is an even number. What are the three consecutive even numbers less than x?
(d) ‘x’ is a number which is a multiple of 5. What are the three consecutive number less
than x which are a multiple of 5.
(e) The present ages of a father and his son are ‘x’ and ‘y’ years respectively. How
many years before was the father as old as his son?
(f) The sum of a number and its reciprocal is 163, make the equation of this statement.
2. (a) If 6 is added to the square of a number, the result is 31. Find the numbers.
(b) If 3 is subtracted from the square of a number, the result is 6. Find the numbers.
(c) If 10 is added to two times the square of a positive number, the result is 60. Find the
number.
(d) If 5 times a number is subtracted from the square of a number, the result is 6. Find
the numbers.
(e) If a number is subtracted from its square, the result is 2. Find the numbers.
(f) If 4 times a natural number is equal to the square of a number, find the numbers.
(g) If 8 times a natural number is equal to the square of the number, find the number.
(h) The sum of a number and its reciprocal is 2 1 . Find the number.
2
(i) Find the number such that the sum of the number and 48 times its reciprocal is
equal to 19.
3. (a) Two natural numbers are in the ratio 3:5. If their product is 375, find the numbers.
(b) Two natural numbers are in the ratio 3:2. If the difference of their square is 45, find the
numbers.
(c) Two natural numbers are in the ratio 1:2 of the sum of their squares which is 125,
find the numbers.
4. (a) Find the two numbers whose sum is 5 and the product is 6.
(b) Divide 57 into two parts whose product is 350.
(c) Divide 16 into two parts such that twice the square of the larger part exceeds the
square of the smaller part by 164.
(d) The product of two consecutive positive integers is 56. Find the numbers.
(e) The product of two consecutive positive even numbers is 288. Find the numbers.
5. (a) If the difference between the squares of two consecutive natural numbers is 5, find
the numbers.
(b) Find the two consecutive positive even integers, the sum of whose squares is 340.
Oasis School Mathematics-10 217
6. (a) The present ages of a boy and his younger brother are 13 years and 7 years
respectively. How many years ago was the product of their ages 55 ?
(b) The present ages of a father and his son are 36 years and 12 years respectively. In
how many years will the product of their ages be 640?
(c) The difference of the present ages of two sisters is 5 years. After 4 years if the
product of their ages becomes 104, find their ages.
(d) One year ago, a man was 8 times as old as his son. Now his age is equal to the
square of his son's age. Find their present ages.
(e) The ratio of the present ages of two bothers is 2:1. Three years hence, the ratio of
the square of their ages will be 9:4. Find their present ages.
(f) The difference of the present ages of a sister and her younger brother is 6 years. The
product of their ages is equal to 4 times the sum of their ages. Find the present ages
of the sister and her brother.
(g) The product of the present ages of a mother and her daughter is 100. 10 years later,
the mother's age will be twice that of the daughter. Find their present ages.
(h) The ages of Rupsi and Rushma are in the ratio of 5:7. Eight years ago, their ages
were in the ratio of 7:13. Find their present ages.
(i) The product of the present ages of a father and his son is 96 years. When the son
becomes as old as his father is now, the sum of their ages will be 68 years. Find their
present ages.
(j) The age of a mother is 5 years less than twice the age of her daughter. When the
mother was as old a her daughter is now, the product of their ages was 125. Find
their present ages.
7. (a) A two-digit number is such that the product of its digits is 18. When 63 is subtracted
from the number, the digits interchange their places. Find the numbers.
(b) A number between 10 to 100 is four times the sum and three times the product of
its digits. Find the number.
(c) A two–digit number is such that the product of the digits is 16. When 54 is subtracted
from the number, the positions of the digits interchange. Find the number.
(d) A number of two-digits is equal to 4 times the sum of the digits. If the product of
the digits is 8, find the number.
(e) The product of the digits of a two-digit number is 18. The number formed by
interchanging the digits of the numbers will be 27 more than the original number.
Find the original number.
8. (a) The denominator of a fraction is 3 more than its numerator. The sum of the fraction
and its reciprocal is 2 9 . Find the fraction.
10
(b) The numerator of a fraction is 1 less than the denominator. If the sum of the fraction
and its reciprocal is 13 , find the fraction.
6
218 Oasis School Mathematics-10
9. (a) Aperson on tour has Rs. 360 for his daily expenses. If he extends his tour programme
by 4 days, he must cut down on his daily expenses by Rs. 3 per day. Find the
number of days of his tour programme.
(b) Some boys planned a picnic with a total budget of Rs. 4,200. But 5 boys cannot
attend the picnic and the cost of picnic increased by Rs. 70 for each boy. How many
boys joined the picnic?
10. (a) The lengths of the sides of a right-angled triangle are (x + 2) cm, 4x cm and (4x + 1)
cm. Find the length of the sides of the triangle.
(b) The hypotenuse of a right angled-triangle is 6 meters more than twice the shortest
side. If the third side is 2 meters less than the hypotenuse, find the sides of the
triangle.
(c) The hypotenuse of a right-angled triangle is 20 m. If the difference between the
lengths of the other sides is 4 meters, find the perimeter of the triangle.
11. (a) If the perimeter of a rectangular plot is 104 meters and its area is 640 m², find its
length and breadth.
(b) If the rectangular room has a perimeter 60 m and area 221 m², find the length and
breadth of the room.
12. (a) If I had walked 1 km per hour faster, it would have taken 100 minutes less to walk
a distance of 20 km. Find the speed of my walking.
(b) A train takes 3 hours less for a journey of 360 km if its speed is increased by
10 km/hr from its usual speed. What is the usual speed?
(c) A bus takes 3 hours less time than another bus to complete a journey of 360 km. If the
speed of the first bus is 20 km/hr more than that of the second bus, find their speed.
Answer
1. Consult your teacher 2. (a) 5,-5 (b) 3, -3 (c) 5 (d) 6, -1 (e) 2, -1 (f) 4 (g) 8
(h) 2, 1/2 (i) 16,3 3. (a) 15, 25 (b) 9, 6 (c) 5, 10 4. (a) 3, 2 (b) 50, 7 (c) 10, 6
(d) 7, 8 (e) 16, 18 5. (a) 2, 3 (b) 12, 14 6. (a) 2 years (b) 4 years
(c) 9 years, 4 years (d) 49 years, 7 years
(e) 6 years, 3 years (f) 12 years, 6 years (g) 20 years, 5 years (h) 15 years, 21 years
(i) 24 years, 4 years (j) 45 years, 25 years 7. (a) 92 (b) 12 (c) 82 (d) 24 (e) 36
8. (a) 2 (b) 2 9. (a) 20 days (b) 15 10. (a) 5 cm, 12 cm, 13 cm (b) 10 m, 24 m, and 26 m, (c) 48 m
5 3
11. (a) 32 m, 20 m (b) 17m, 13m 12. (a) 3 km/hr. (b) 30 km/hr. (c) 60 km/hr, 40 km/hr.
Project Work
• Collect 5 different problems of simultaneous equations and quadratic equation
used in our daily life. Present them in your class.
Oasis School Mathematics-10 219
Miscellaneous Exercise
H.C.F. and L.C.M
1. Find the H.C.F. of:
(a) x3 + 1 , x4 + 1 + 1 and x4 – x2 + 1 [Ans: x2 + 1 – 1]
x3 x4 x2
(b) ax2 + 2ax – 3x – 6, x3 – 2x2 + 8 – 4x and x2 – 4 [Ans: x + 2]
(c) y2 + 2y – 3, y3 + 3y2 – y – 3 and y3 + 4y2 + y – 6 [Ans: (y + 3) (y – 1)
(d) x2 + xy – 4x, x2 – y2 – 4x + 4y, x2 – 12x – y2 – 4y + 32 [Ans: x + y – 4]
(e) 81y4 – 9y2 + 6y – 1, 27y3 + 1, 9y3 – 3y2 + y [Ans: 9y2 – 3y + 1]
(f) x2–a2+4xy+2ab–b2+4y2, x3+2x2y+x2a–x2b [Ans: x+2y+a–b]
(g) a4–b4, (ax+by)2 + (bx–ay)2, a3 + ab2 [Ans: a2 + b2]
(h) x2–2x–8–y2–6y, x2+2xy+y2–16, x2+xy+4x [Ans: a2+b2]
(i) 3ax–bx–3ay+by, ax2–3bxy–axy+3by2, x2–3x–xy+3y [Ans: x–y]
2. Find the L.C.M. of:
(a) x4 + 2ax3 + a2x2 – b4, x4 + (2b2 – a2) x2 + b4 [Ans: (x2 + ax + b2) (x2 – ax + b2)]
(b) 3(7x3 – 10x2 – 7x + 10), 6(2x3 – x2 – 2x + 1) [Ans: 6(x2 – 1) (2x – 1) (7x – 10)]
(c) (1 – a2) (1 – b2) + 4ab, 1 – 2a + b – a2b + a2 [Ans: (1 – a + b + ab) (1 – a) (1 + a – b + ab)]
(d) x2–7x+12, x3–2x2–2x–3 [Ans: (x-3) (x-4) (x2+x+1]
(e) m2+3m–4, m3+2m2–2m+3 [Ans: (m–1) (m+3) (m+4) (m2-m+1]
(f) x3–3x2+3x–1, x3–x2–x+1, x4–2x2+2x–1 [Ans: (x-1)3 (x+1)]
(g) x3+2x2–9x–18, x3–2x2+8–4x, x2–9 [Ans: (x+2) (x-2)2 (x2–9)]
(h) a3 + 2a2 – 9a – 18, a3 – 2a2 + 8 – 4a, a2 – 4 [Ans: (a2–9) (a+2) (a–2)2]
3. Find the HCF and LCM of:
(a) x4 + y4 + (2y2–z2)x2, x4–y4 + 2x3z + x2z2 [Ans: HCF = (x2+xz+y2) LCM (x2+xz+y2)
(x2+zx–y2) (x2-zx+y2)]
(b) x4 + y4 + 2x2y2–4x2, x4–y4 + 4x3+4x2 [HCF = (x2+2x+y2), LCM = (x2+2x+y2)
(x2+2x–y2) (x2–2x+y2)
Surds
Solve:
(1) 5x – 9 = 1+ 5x – 3 [Ans: 5] (3) x+ a =4 2+ x– a [Ans: 2a or a ]
5x + 3 2 [Ans: 7] x– a x+ a 2
(2) 7x – 36 = 9 – 5 7x – 11 (4) x+ 2 + x– 2 = 6 [Ans: 4]
x– 2 x+ 2
6 + 7x 3
220 Oasis School Mathematics-10
Indices a 2 – b 2 a 2 + 1 a 2 – b 2 a 2 – 1
3 3 3 3 3 3
( ) ( ) ( ) ( ) 1. Simplify : (a) b3 b3
a–b × a 1 + b 1 × a+b × a 1 – b 1 [Ans: 1]
3 3 3 3
(b) z3 + 3z(y–z)2-y+ 3z2 + (y–z)3-y
(y–z)y (y–z)y-1
2. If a + b + c = m, prove that: x2a + x2a + xm–c + x2b + x2b + xm–a + x2c + x2c + xm–b = 1
xm–b xm–c xm–a
3. 2 –2 x3 + 3x = a2 – 1 .
a
If x2 + 2 = a 3 + a 3 , prove that :
4. If x –1 = 31/3 + 3-1/3, prove that 3x3– 9x2– 4 = 0
5. Prove that: a + 1 – a2/3 + 1 = a2/3 + 2
a1/3–1 a1/3+1 a1/3+1 a1/3–1
6. Solve:
(a) 3x+2 + 1 = 30 [Ans: ±1]
3x–2
[Ans: 3, 4]
(b) 4x + 128 – 6 × 2x+2 = 0 [Ans: 5,6]
(c) 6x–5 + 66–x = 7 [Ans: 2]
[Ans: 0, 1]
(d) 1 – 6 + 1 = 0 [Ans: 0, 2]
9x–1 3x+1 9
(e) 9 × 9x – 3x+2–3x+1 = –3
5
(f) 5x – 5x = 5 1
5 5
Algebraic fraction.
Simplify:
1. 1 – 1 + 1 + 2a4 [Ans: x82–a4a8]
2a3(a + x) 2a3(x – a) a2(x2 + a2) x8 – a8 [Ans: 0]
2. b–c – c+a + a+b
a2 – (b – c)2 b2 – (c + a)2 (a + b)2 – c2
3. 2x2 – 1 + 9y2 – 2x2 + 1 + 9y2 – 12xy [Ans: 0]
6xy 6xy 4x4 + 81y4
4. 1+ x 1 1 + 2x 1 – x x 1 + 4x3 [Ans: 8x7 ]
– x2 + + x4 + 1 x8 – 1
5. 2 1 x + 4 + 32 – 1024 [Ans: x 1 2 ]
+ 4 + x2 16 + x4 256 – x8 –
6. x 2 1 – 2 – 1 + 1 – 1 [Ans: 0]
+ x+2 (x + 1)2 (x + 1)2 (x + 2)2 (x + 2)2
7. ab – bc + ca [Ans: 1]
(c-b) (c-a) (a-b) (a-c) (b-a) (b-c) [Ans: x2+1)
x3 x² x 1
8. x − 1 − x + 1 − x − 1 + x + 1
Oasis School Mathematics-10 221
x4 x4 1 1 [Ans: 2x2]
9. x² + 1 + x² − 1 − x² + 1 − x² − 1
1 1 1 1
10. x − 5 − x − 3 + x + 5 − x + 3 Ans: 32x
(x2–9) (x2–16)
11. x 1 a − 2 a + x 1 a − 2 a Ans: 6a2x
− 2x + + 2x − (x2-a2) (4x2-a2)
111 ax bx cx Ans: 3
12. x + a + x + b + x + c + x³ + ax² + x³ + bx² + x³ + cx² x
13. 1 – 1 – 2a –4a3
1–a+a2 1+a+a2 1–a2+a4
[ ]Ans: (1-a2+a4) (1+a2+a4)
Verbal Problems
1. The difference of the present ages of a brother and his sister is 5 years. The product
of the ages is equal to six times the sum of their ages. Find their present ages.
[Ans: 10 years, 15 years]
2. The product of the present ages of two brothers is 160. Four years ago, the elder
brother was twice as old as his younger brother. Find their present ages.
[Ans: 16 years, 10 years]
3. ‘A’ said to ‘B’, “I was twice as old as you were when I was as old as you are.” If the sum of
their age is 50 years, find their present ages. [Ans: 30 years, 20 years]
4. The present ages of father and his son are x years and years respectively. When the
father was as old as his son is now, he was twice as old as his son.
(i) What is the age difference of the father and son?
(ii) What was the age of father and his son when the father was as old as his son?
(iii) Convert the above statement in the question into an equation.
(iv) If the sum of their present ages is 100 years, find their present ages.
[Ans: (i) (ii) and (iii) Consult your teacher, (iv) 60 years, 40 years]
5. A number of two digits is equal to four times the sum of its digits. If 18 is added to
the number, the digits are reversed. Find the number. [Ans: Rs. 24]
6. A number consists of two digits. If the number formed by reversing the digits is added
to it, the sum is 143, and if the same number is subtracted from it, the remainder is 9.
Find the number. [Ans: 76]
222 Oasis School Mathematics-10
7. A two-digit number is such that it is obtained by multiplying the sum of the digits by
8 and adding 3. It is also obtained by multiplying the difference of the digits by 13 and
subtracting 1. Find the number. [Ans: 51]
8. In a number of two digits, the ratio of the digit in the tens place and the digit in the
unit place is 3 : 1. If 3 is added to three times the sum of the digits, then the opposite
number is formed. Find the number. [Ans: 93]
9. A number consists of two digits. One of the digits is greater than the other by 5. When
the digits are reversed, the number becomes 3/8 of the original number, find the
number. [Ans: 72]
10. Dipesh said to Diwas, "If you give me half of your money, I shall have Rs. 100." Diwas
replied, " I shall have Rs. 100 if you give me only one third of your money." Find the
money each of them has. [Ans: Rs. 60, Rs. 80]
11. A rectangular field 20 m × 10 m is surrounded by a path of uniform width whose area
is 136m2, find the width of the path. [Ans: 2m]
12. A path of uniform width is running inside the rectangular field 22 m × 12 m. If the
area of the path is 64 m2, find the width of the path. [Ans: 1m]
13. The perimeter of a rectangular field is 32 m. Another rectangular field having same
perimeter has breadth 2m less than the first field and the area 12m2 more. Find the
length and breadth of first field. [Ans: 12m, 4m]
14. Some students of class X have reserved a bus with total budget of Rs. 1800 for their
tour. If 10 students don’t attain the tour, each would have to contribute Rs. 30 more.
(i) Find the number of students and the amount that each has to contribute.
(ii) What amount should each of them have to contribute if 6 more students join the tour.
[Ans: (i) 30, Rs. 60, (ii) Rs. 50]
15. To cover the distance of 600 km, a car takes 5 hours less time than a bus. If the speed of
car is 20 km/hr more than the speed of bus, find their speed.
[Ans: 40 km/hr, 60 km/hr]
Oasis School Mathematics-10 223
Full marks : 38
Attempt all the questions.
Group A [4 × 1 = 4]
1. (a) What is the H.C.F. of two algebraic expressions if there is no common factor?
(b) Find the value of 2 3 375 ÷ 3 3 192 .
1
(c) Solve: (10)2y-1 = 0.001 .
(d) " The age of a father was thrice the age of his son when he was as old as his son."
Convert this verbal form into an equation.
Group B [4×2=8]
4x × 2x−1 − 2x
2. (a) Simplify: 22x+1 × 2x−2 − 2x
(b) Simplify : 4 48x11y7 ÷ 4 3x3y−5
3. (a) Solve : x −3 = 2
x 5
(b) If a number is added to two times of its square, the result is 1. Find the number.
Group C [4×4=16]
4. Find the H.C.F. of : x3 + 1 , x2 + 3x + 2, x3 + 2x2+2x+1
5. Solve : 4x + 4-x = 441
6. Solve : x + 3 + x + 8 = 4x + 21
7. The product of the ages of a father and his son is 1,620. When the father was as old as his
son is now, he was five times as old as his son. Find their present ages.
Group D [2 × 5 =10]
8. Under which condition is the algebraic fraction 1 not defined? Prove that:
x-y
2x + 2x + 4x3 = 8x7
x2–y2 x2+y2 x4+y4 x8–y8
9. Find the factors of the two algebraic expressions x4+(2b2–a2) x2+b4 and x4+2ax3+a2x2–b4.
Is there any common factor between them? What is the product of the common factors
called? Multiply this product with the remaining factors of other expressions. What is
the result called?
224 Oasis School Mathematics-10
Geometry
44Estimated Teaching Hours
Contents
• Area of triangle and quadrilateral
• Construction
• Circle
Expected Learning Outcomes
At the end of this unit, students will be able to develop the following
competencies:
• To find the area of triangles and quadrilaterals
• To prove theorems related to area of triangle and parallelograms
• To solve the problems using theorems related to area of triangle and
parallelogram
• To construct a triangle and different types of quadrilateral equal in
area with given triangle or quadrilateral
• To verify the theorems on a circle experimentally
• To prove the theorems on a circle
• To solve problems using theorems of circle
• To solve the problems of tangents and circles using the properties of
a tangent
Materials Required
• set square, compass, scale, protractor, pencils, etc.
Oasis School Mathematics-10 225
Unit Area of Triangles
and Parallelograms
13
13.1 Warm-up Activities
Discuss the following in your class and draw a conclusion.
• What is the formula to calculate the area and perimeter of a rectangle?
• What is the formula to calculate the area and perimeter of a square?
• If 'd' is the diagonal of a square, what is its area in terms of 'd'?
• Is every rhombus a parallelogram?
• Is every rectangle a parallelogram?
• If d1 and d2 are the diagonals of a rhombus, what is its area?
13.2 Base and Altitude of Triangle and Parallelogram
Any side of a triangle can be considered the base. The perpendicular drawn from the vertex
to the base is the altitude. A
In ∆ABC, AD⊥BC, BE⊥AC and CF ⊥AB. F E
If AC is the base, BE is the altitude. O
DC
If AB is the base, CF is the altitude and if BC is the base, AD is the altitude.
B
B F
Similarly, any side of a parallelogram can be considered the base. The A C
perpendicular drawn from the vertex to the base is the altitude.
The perpendicular distance between two parallel sides is the altitude
of the parallelogram. D E
In parallelogram ABCD, AE⊥DC and AF ⊥ BC.
If DC is the base, AE is the altitude.
If BC is the base, AF is the altitude.
13.3 Area of Triangle
(i) When the base and altitude are given,
In this case the area of a triangle = 1 base × altitude.
2
AP
X
B DC Q R WY Z
226 Oasis School Mathematics-10
Area of ∆ABC = 1 BC × AD, A rea of ∆PQR = 1 QR × PQ Area of ∆XYZ = 1 YZ × XW
2 2 2
Note: In a right angled triangle, two perpendicular sides are taken as the base and altitude.
(ii) When three sides of a triangle are given: A
In this case, area of a triangle = s(s − a)(s − b)(s − c) cb
Where,
s = a + b + c = Semi perimeter of a triangle. B C
2 a
13.4 Area of Parallelogram AB
In the given parallelogram ABCD, if DC is the base, AE is the
altitude. Area of a parallelogram
= base × altitude D altitude C
= DC × AE
E
Base
Remember !
• Areas of two congruent figures are equal. A D
F
• Any side of a parallelogram can be taken as its base.
• A line passing through any vertex and perpendicular to the C
base of a parallelogram is called the altitude of a parallelogram.
In the given parallelogram ABCD, BE
• If BC is the base, AE is the altitude.
• If CD is the base, AF is the altitude.
Theorem 13.1 C
A diagonal of a parallelogram divides it into two triangles of equal area. D
Given : BD is the diagonal of the parallelogram ABCD.
To prove : Area of ∆ABD = Area of ∆DBC. AB
Proof:
S.N. Statements S.N. Reasons
1. In ∆ABD and ∆BDC, 1.
(i) AB = DC (S) (i) Being opposite sides of a parallelogram
(ii) ∠ABD = ∠BDC (A) (ii) Being alternate angles on AB//DC
(iii) BD = BD (S) (iii) Being a common side
2. ∆ABD ≅ ∆BDC 2. By Side-Angle-Side axiom
3. ∆ABD = ∆ BDC 3. Area of congruent triangles
Hence, the diagonal of a paralleogram divides it into two triangles of equal area.
Oasis School Mathematics-10 227
Alternative method
Given : ABCD is a parallelogram in which AC is the diagonal. A KD
C
To prove : ∆ABC = ∆ACD
Construction : Draw AH⊥BC, CK⊥AD.
Proof: BH
S.N. Statements S.N. Reasons
1. ∆ABC = 1 AH×BC 1. Area of triangle = 1 base × height.
2 2
2. ∆ACD = 1 AD×CK 2. Area of triangle = 1 base × height.
2 2
3. AH = CK 3. Distance between the same parallel lines.
4. ∆ABC = ∆ACD 4. From (1), (2) & (3).
Hence, a diagonal of a paralleogram divides it into two triangles of equal area.
Theorem 13.2
Parallelograms on the same base and between the same parallel lines are equal in area.
A ED F
BC
Given : Parallelogram ABCD and parallelogram EBCF are standing on the same base BC
and between the same parallel lines AF and BC.
To prove : Area of ||gm ABCD = Area of ||gm EBCF.
Proof :
S.N. Statements S.N. Reasons
1. In ∆ABE and ∆DCF, 1. (i) Being corresponding angles on parallel lines
(i) ∠AEB = ∠DFC (A) (EBІІFC)
(ii) ∠EAB = ∠CDF (A)
(iii) AB = DC (S) (ii) Being corresponding angles on parallel lines
(ABІІDC)
(iii) Opposite sides of a parallelogram are equal.
2. ∆ABE ≅ ∆DCF 2. By A.A.S. axiom
3. ∆ABE = ∆ DCF 3. Area of congruent triangles.
4. ∆ABE + Trap. EBCD 4. Adding common trapezium EBCD in (3).
= ∆ DCF+Trap EBCD
5. //gm ABCD = //gm EBCF 5. From (4), using whole part axiom.
Hence, parallelogams on the same base and between the same parallel lines are equal in area.
228 Oasis School Mathematics-10
Alternative method A F DE
Given : ABCD and BCEF are two parallelograms standing
on the same base BC and between the parallel lines
AE and BC. BG H C
To prove : Area of ||gm ABCD = Area of ||gm BCEF
Construction : Draw DH⊥BC and FG⊥BC
Proof: S.N. Reasons
S.N. Statements
1. ||gm ABCD = BC×DH 1. Area of parallelogram = base × height
2. ||gm BCEF = BC×FG 2. Area of parallelogram = base × height
3. DH = FG 3. Distance between the same parallel lines
4. ||gm ABCD = ||gm BCEF 4. From (1), (2) & (3)
Hence, a parallelograms on the same base and between the same parallel lines are equal in area.
Corollary:
(i) A parallelogram and a rectangle on the same base and between the same parallel lines
are equal in area.
(ii) Parallelograms on equal base and between the same parallel lines are equal in area.
(iii) A rhombus and a square on the same base and between the same parallel lines are equal
in area.
Theorem 13.2 FD E
Experimental Verification
Draw two parallelograms ABCD and FBCE having same base BC. A
Draw FH and DG perpendiculars to BC.
BH G C
Now, DG and FH are the altitudes of the parallelograms ABCD and FBCE. Fig (i)
A DF E
To verify: Area of ||gm gm ABCD = Area of ||gm gm FBCE. B GH C
Observation: Fig (ii)
Fig. BC DG FH Area of ||gm Area of Remarks
(i) ABCD=BC×DG ||gmFBCE=BC×FH
......cm .......cm .....cm .......cm2 .......cm2 Area of ||gm
ABCD = Area of
||gm FBCE
Oasis School Mathematics-10 229
Fig. BC DG FH Area of ||gm Area of Remarks
(ii) ABCD=BC×DG ||gmFBCE=BC×FH
......cm .......cm .....cm .......cm2 .......cm2 Area of ||gm AB-
CD=Area of ||gm
FBCE
Conclusion:
Hence, parallelograms on the same base and between the same parallel lines are equal in area.
Theorem 13.3
The area of a triangle is one half the area of a parallelogram standing on the same base and
between the same parallel lines.
Given : Triangle ABC and parallelogram DBCF both DH
stand on the same base BC and between the A F
C
parallel lines BC and AF.
To prove : Area of ∆ ABC = 1 Area of ||gm DBCF.
Construction : 2
Proof: Draw a line CH parallel to BA where H is a point
on AF. B
S.N. Statements S.N. Reasons
1. ABCH is a parallelogram 1. By construction
2. ∆ ABC = 1 ||gmABCH 2. Diagonal bisects the parallelogram.
2
3. Parallelograms on the same base BC and
3. ||gm ABCH = ||gm DBCF between the parallel lines BC and AF.
4. ∆ ABC = 1 ||gm DBCF 4. From (2) and (3).
2
Proved.
Hence, area of a triangle is half the area of a parallelogram standing on the same base and
between the same parallel lines.
Alternative method
Given : Parallelogram ABCD and triangle EBC A E D
both stand on the same base BC and lie K H C
between the parallel lines BC and AD.
To prove : ∆BEC = 1 ||gm ABCD.
2
B
Construction : Draw AK⊥BC and EH⊥BC.
Proof:
230 Oasis School Mathematics-10
S.N. Statements S.N. Reasons
1. Area of ||gmABCD = BC×AK 1. Area of parallelogram = base × height
2. Area of ∆BEC= 1 BC×EH 2. Area of triangle = 1 base × height
2 2
3. AK = EH 3. Distance between the same parallel lines
4. ∆EBC = 1 ||gm ABCD. 4. From (2) and (3).
2
Hence, the area of a triangle is one half the area of parallelogram standing an same base and
between same parallelograms.
Experimental Verification [Theorem 13.3] E
D
Draw a parallelogram EBCD and a triangle ABC having the same A
base BC and between the parallel lines AD and BC. BG C
Fig. (i) D
Draw AF and EG perpendiculars to the base BC or BC produced. F
Now, AF and EG are the altitudes of ∆ABC and ||gm EBCD EA
respectively.
To verify: Area of ∆ABC = 1 Area of||gm EBCD.
2
BG
FC
Observation: Fig. (ii)
Fig. BC AF EG Area of ||gm Area of ||gmFB- Remarks
ABCD=BC×AF CE=BC×FH
(i) ......cm .......cm .....cm .......cm2 .......cm2 Area of ∆ABC = 1
||gm EBCD 2
(ii) ......cm .......cm .....cm .......cm2 .......cm2 "
Conclusion: Hence, the area of a triangle is half the area of a ||gm on the same base and between
the same parallel lines.
Theorem 13.4 AED
Triangles on the same base and between the same parallel lines
are equal in area.
Given : Triangles ABC and DBC are standing on the B C
same base BC and lie between the parallel lines
BC and AD.
To prove : ∆ABC = ∆DBC.
Oasis School Mathematics-10 231
Construction : Draw CE||BA, where E is a point on AD.
Proof :
S.N. Statements S.N. Reasons
1. ABCE is a parallelogram 1. By construction (ABllCE) and given
2. ∆ABC = 1 parallelogram ABCE (AEllBC).
2 2. Diagonal bisects the parallelogram
3. ∆DBC = 1 parallelogram ABCE 3. Triangle and parallelogram on same base
2 BC and between the parallel lines BC and
AD
4. ∆ABC = ∆DBC
4. From (2) and (3).
Conclusion: Hence, triangles on the same base and between the same parallel lines are equal
in area.
Alternative method:
Given : Triangles ABC and DBC both lie on the AD
same base BC and between the parallel lines
AD and BC.
To prove : Area of ∆ABC = Area of ∆DBC
Construction: From A and D, draw AM and DN perpendiculars B MNC
to BC.
Proof :
S.N. Statements S.N. Reasons
1. Area of ∆ABC = 1 BC × AM 1. Area of a triangle = 1 base × height.
2 2
2. Area of ∆DBC = 1 BC × DN 2. Area of a triangle = 1 base × height.
2 2
3. AM = DN 3. Distance between two parallel lines.
4. Area of ∆ABC = Area of ∆DBC 4. From statements (1), (2) and (3).
Experimental Verification [Theorem 13.4] Hence proved.
AD
Draw two triangles ABC and DBC having the same base BC and BE CF
between the parallel lines BC and AD. A D
Draw AE and DF perpendiculars to BC or BC produced. Now AE
and DF are the altitudes of ∆ABC and ∆DBC respectively.
To verify: Area of ∆ABC = Area of ∆DBC.
B EFC
232 Oasis School Mathematics-10
Observation:
Figure BC AE DF Area of ∆ABC= Area of ∆DBC= Remarks
(i) ......cm .......cm .....cm 1 1
(ii) ......cm .......cm .....cm 2 BC × AE 2 BC × DF Area of ∆ABC=
Area of ∆DBC
.......cm2 .......cm2 Area of ∆ABC=
Area of ∆DBC
.......cm2 .......cm2
Conclusion: Hence, triangles on the same base and between the same parallel lines are equal
in area.
AD
Some important results:
1. Area of trapezium = 1 height × sum of the B C
2 H B
A
parallel sides.
Area of trap ABCD = 1 AH×(AD + BC).
2
2. Area of rhombus = 1 product of the diagonals. D O
2 C
1
Area of rhombus ABCD = 2 AC × BD D
h2
= 1 d1 × d2 A d
2
A
h1
BD
3. Area of quadrilateral = 1 diagonal (sum of altitude from B
2
C
opposite vertices to the diagonal)
Area of quadrilateral ABCD = 1 d ×(h1 + h2)
2
4. Area of kite = 1 product of the diagonals
2
= 1 AC × BD
2
Points to Remember ! C
• Median bisects a triangle. A
In the given figure, AD is the median.
∴ ∆ABD = ∆ADC
• Diagonal bisects a parallelogram. B DC
• Parallelograms on the same base and between the same parallel lines are equal in area.
• Area of a triangle is half the area of a parallelogram standing on the same base and
between the same parallel lines.
• Triangles on the same base and between the same parallel lines are equal in area.
Oasis School Mathematics-10 233
Worked Out Examples
Example: 1
In the given figure, ABCD is a square whose perimeter is 40 cm. Find the area of ∆EBC.
Solution: A DE
Given, Perimeter of square = 40 cm
4a = 40 cm [a = a side of a square]
a = 10 cm
We have, Area of a square = a2
= (10cm)2
= 100 cm2. B C
Area of ∆EBC = 1 area of square ABCD
2
[Area of a triangle is half of the area of a square standing on the same base and between the same parallel lines].
or, Area of ∆EBC = 1 × 100 cm2 = 50 cm2
2
Example: 2
In the given figure, ABCD is a parallelogram. DF = FE and area of parallelogram ABCD = 60
cm2. Find the area of ∆EDC. E
Solution:
Given, area of parallelogram ABCD = 60 cm2
Area of ∆FDC = 1 area of ||gm ABCD
2
1 AF
or, Area of ∆FDC = 2 × 60 cm2 = 30 cm2 B
Again, ∆EDC = 2∆FDC [Median bisects the triangle]
= 2×30 cm2 = 60 cm2
Hence, area of ∆EDC = 60 cm2. DC
Example: 3
In a quadrilateral ABCD, it is given that M is the mid-point of AC. Prove that quadrilateralral
ABMD = quadrilateral DMBC. DC
Solution:
Given : ABCD is a quadrilateral. M is the mid-point of AC. M
To prove : Quadrilateral ABMD = Quadrilateral DMBC AB
Proof :
S.N. Statements S.N. Reasons
1. ∆ADM = ∆DMC 1. Median bisects the triangle
2. ∆AMB = ∆BMC 2. Same as (1)
3. ∆ADM + ∆AMB = ∆DMC + ∆BMC 3. Adding (1) and (2)
4. Quadrilateral ABMD = Quadrilateral 4. From (3), whole part axiom.
DMBC
Hence, proved.
234 Oasis School Mathematics-10
Example: 4 D C
If O is a point inside the parallelogram ABCD, prove that O Y
∆AOB + ∆COD = ∆AOD + ∆BOC. B
X
Given : ABCD is a parallelogram. O is a point interior of it and
OC, OA, OB, OD are joined. A
To prove : ∆AOB + ∆COD = ∆AOD + ∆BOC
Construction : Draw XOY ||AB ||DC.
Proof:
S.N. Statements S.N. Reasons
1. XABY is a parallelogram 1. Being AX//BY and XOY // AB.
2. ∆OAB = 1 ||gm XABY 2. They stand on the same base AB
2 and between the parallel lines AB
and XY.
3. DXYC is a parallelogram 3. Being DX //CY at XOY // AB.
4. ∆DOC = 1 ||gm DXYC 4. Same as (2)
2
5. ∆OAB+∆DOC = 1 ||gm XABY+ 1 ||gm- 5. Adding (2) and (4)
DXYC 2 2
6. ||gm ABCD = ||gm XABY + ||gm DXYC 6. Whole part axiom
7. ∆OAB + ∆DOC = 1 ||gm ABCD 7. From (3) and (4)
2
8. Similarly ∆∆DDOOCA=+∆D∆COOAB+=∆12C|O|Bgm ABCD 8. Same as above by drawing line
9. ∆OAB + 9. parallel to DA and BC through O.
From (7) and (8).
Hence, proved.
Example: 5
In the given figure, P and Q are the mid-points of AB and A
AC respectively. Prove that.
(i) ∆POB = ∆QOC (ii) Quad. APOQ = ∆BOC P Q
Given : In ABC, P and Q are the mid-points of AB and AC
respectively.
To prove : ∆POB = Quad. APOQ, (ii) Quad. APOQ = ∆BOC O
Construction : Join PQ
Proof: BC
S.N. Statements S.N. Reasons
1. PQ || BC 1. The line joining the mid points of two
2. ∆PBC = ∆QBC sides of a triangle is parallel to the third
side.
2. Triangles on the same base and between
the same parallel lines
Oasis School Mathematics-10 235
S.N. Statements S.N. Reasons
3. ∆POB + ∆OBC = ∆QOC + ∆OBC 3. Whole part axioms
4. ∆POB = ∆QOC 4. From (3)
5. ∆BAQ = ∆BQC 5. Median bisects the triangle
6. ∆BAQ–∆POB = ∆BQC–∆QOC 6. Subtracting (4) from (5)
7. Quad. APOQ = ∆BOC. 7. From (6)
Example: 6 Hence, proved.
In the adjoining figure, ABCD is a parallelogram whose diagonals AC D QC
O
and BD intersect at O, a line segment through O meets AB at P and DC PB
at Q. Prove that the area of the quadrilateral APQD = 1 parallelogram
2
ABCD.
Given : ABCD is a parallelogram, diagonals AC and BD intersect
To prove : at OP and Q, and P and Q are points on sides AB and DC A
Proof :
respectively.
Area of quad. APQD = 1 (Area of ||gm ABCD)
2
S.N. Statements S.N. Reasons
1. In ∆APO and ∆QOC 1. (i) Being alternate angles on
(i) ∠OAP = ∠QCO (A) AB ||DC
(ii) AO = OC (S)
(iii) ∠AOP = ∠QOC (A) (ii) Diagonals of a parallelogram
bisect each other
(iii) Vertically opposite angles
2. ∆APO ≅ ∆OQC 2. By A.S.A. axioms
3. ∆APO = ∆OQC 3. Area of congruent triangles
4. ∆DAC = Quad. DAOQ + ∆QOC 4. Whole part axiom
5. Quad. APQD = Quad. DAOQ + ∆AOP 5. Whole part axiom
6. ∆DAC = Quad. APQD 6. From (3), (4) and (5)
7. ∆DAC = 1 ||gm ABCD 7. They stand on the same base DA
2 and between DA || CB
8. Quad. APQD = 1 ||gm ABCD 8. From (6) and (7)
2
Hence, proved.
236 Oasis School Mathematics-10
Example: 7 CS
In the adjoining figure, PABC and PQRS are two parallelograms of equal P OR
area. Prove that QC // BR. Q
Given : In the given figure, ||gm PABC = ||gm PQRS
To prove : QC // BR AB
Construction : Join QB and CR.
Proof :
S.N. Statements S.N. Reasons
1. ∆QBC = 1 ||gmPABC 1. Area of a triangle is half of the area of
2 parallelogram standing on the same base and
between the same parallel lines
2. ∆CQR = 1 ||gm PQRS 2. As statement (1)
2
3. ∆QBC = ∆CQR 3. From (1) and (2)
4. CQ || BR 4. From statements 1, being area of two triangles on
the same base is equal
Hence,, proved.
Exercise 13.1
1. (a) What is the relation between the area of two parallelograms standing on the same
base and between the same parallel lines?
(b) What is the relation between the area of a triangle and a parallelogram standing
on the same base and between the same parallel lines?
(c) If a parallelogram and a rectangle are on the same base and between the same
parallel lines, what is their relation?
(d) What is the relation of two triangles standing on the same base and between the
same parallel lines?
(e) If a triangle and a parallelogram are standing between the same parallel lines and
the base of a triangle is twice the base of a parallelogram, what is their relation?
2. (a) In the given figure, name two parallelograms A FD E
which are equal in area. Why?
(b) In the given figure, what is the relation between P B C
R T S
∆PQR and parallelogram TQRS? P
Q R
(c) In the given figure, what is PS called?
What is the relation between
∆PQS and ∆PRS? Why? Q
S
Oasis School Mathematics-10 237
(d) In the given figure, which two triangles are equal A D
in area? Why?
(e) In the given figure, write two pairs of triangles B C
which are equal in area. A D
3. (a) In the figure alongside, ABCD is a parallelogram. B C
P is a point on CD. If ∆DPA = 15 cm2, ∆BPC = 20 cm2, A D
find the area of ∆APB.
B P
(b) In the given figure, ABCD is a parallelogram, AP C
∆APD = 7 cm2 and
∆BCP = 5 cm2. Find the area of parallelogram ABCD. D B
(c) In the given figure, PQRS is a parallelogram in which T is P C
Q
the mid-point of PS. If the area of ∆PTQ is 6 cm2, find the area T
of quad. TQRS. R
D
S
(d) In the adjoining figure, if the area of the parallelogram A P
ABCD = 88 cm2, find the area of ∆BQC. Q
B C
E
A D
(e) In the given figure, ABCD is a parallelogram and EBC is a
triangle. If the area of ∆EBC is 7 sq. cm, find the area of the F
parallelogram ABCD. C
B
A DE
C
(f) ABCD is a parallelogram. A
If ∆ABF = 21 cm2, find the area of ∆EBC.
B D
(g) In the given figure, ABCD is a parallelogram and AH⊥BC. If 8cm
BC = 8 cm, AH = 8 cm, find the area of ∆AED. BH C
E
8cm
238 Oasis School Mathematics-10
(h) In the given figure, ABCD is a rhombus whose diagonals A D
AC and BD are 6 cm and 8 cm respectively. Find the area E
of ∆EBC.
B C
A
E D
(i) In the given figure, ABCD is a square with its diagonal
AC = 8 2 cm. Find the area of ∆EBC. C
D
B
C
A BF
(j) In the given figure, ABCD is a square and DEC is a triangle. If E C
the area of triangle DEC = 50 cm2, find the length of diagonal C
AC. B
A
(k) In the given figure, ABCD is a square whose perimeter is
40 cm and AB is produced to the point F. If E is the middle
point of CD, find the area of ∆EFC.
D E
D
(l) In the adjoining figure, AF//DC, ED//FC and ABCD is a
square. If AC = 5 2 cm, find the area of the parallelogram DEFC.
(m) In the adjoining figure, PS = 5 cm and SM = 8 cm. Calculate the A EB F
area of ∆PQN. P Q
M
NS Q R
A P
(n) Find the area of the quadrilateral PQRS given in the adjoining figure in R
which 3RB = 2PA = QS = 6 cm. B
S D
A 16cm
(o) In the given figure, ABCD is a trapezium. If AB = 10 cm, 10cm
BC = 22 cm, AD = 16 cm, AD||BC and DC⊥BC, calculate the
area of ∆ADC. B 22cm C
A
(p) In ∆ABC, D is the mid-point of BC. If AC = 10 cm and
DE = 6 cm, find the area of ∆ABC. E
BD C
Oasis School Mathematics-10 239
(q) In the given figure, PT = TQ and TS||QR. If the area of P T Q
∆TSP = 15cm2, find the area of
(i) ∆TQS (ii) ∆ TQR (iii) ∆PSQ S R
A D
(r) In the given figure, AE||DB and EB = BC. If area of E
∆ABD = 20cm2, find the area of (i) ∆DBC (ii) Quad. ABCD. B C
D C
4. In the adjoining figure, ABCD is a trapezium in which AB||DC. O
Prove that: ∆AOD = ∆BOC.
AB
A
5. In the given figure, D and E are the points on AB and AC of ∆ABC
respectively. If DE||BC, D O E
prove that: B C
(i) ∆ACD = ∆ABE
(ii) ∆OCE = ∆DOB. A B
6. In the adjoining figure, O is any point inside a parallelogram O
D
ABCD.
A
Prove that: RQ
(i) ∆OAB + ∆OCD = 1 ||gm ABCD C
2
(ii) ∆OAC + ∆OBD = 1 ||gm ABCD
2
7. In the given figure, R is the mid point of AB and PQCB is P
a parallelogram. Prove that area of ∆ABC = area BC
of parallelogram PQCB.
AD
8. In a quadrilateral ABCD, a line through D, parallel to AC,
meets BC produced at P.
Prove that: ∆ABP = quad. ABCD. B CP
PQ
9. In the given figure, PQRS is a trapezium and PQ||MN||SR. Prove that M N
R
area of ∆PSN = area of ∆QRM. S
C
A
E
10. In the adjoining figure, ACDE is a parallelogram.
F is the mid-point of BC. Prove that
∆ABC = Parallelogram ACDE B
F
D
240 Oasis School Mathematics-10
11. ABC is a triangle in which D is the mid–point of BC and E is the mid- A
point of AD. Prove that ∆ABC = 4 ∆ABE. E
BD C
A C
C
12. The vertex A of ∆ABC is joined to a point D on the side BC. The
Q
mid-point of AD is E. E
D
Prove that: ∆BEC = 1 ∆ABC.
2
B
13. D is the mid-point of side BC of ∆ABC and E is the mid-point of A
BD. If O is the mid–point of AE, O
Prove that: 8 ∆BOE = ∆ABC.
B ED
T
14. In the given figure, PQRS is a parallelogram, T is any point on the P
line RP produced. Prove that ∆TPS = ∆TQP.
SR
15. In a parallelogram ABCD, O is any point on the diagonal AC, D C
O
show that (i) ∆AOB = ∆AOD, (ii) ∆COB = ∆DOC.
AB
16. P, Q, R and S are respectively the mid–points of the DRC
sides AB, BC, CD and DA of the parallelogram ABCD. S Q
1 AP B
Show that the area of parallelogram PQRS = 2 area B
A
of ||gmABCD.
17. In the given figure AD//BE, AB//DC, prove that area of ∆AFC = Area FC
of ∆DFE. E
D
P S
18. In the figure, PQRS is a parallelogram. PM and SR are produced to
meet at N. N and Q are joined. QM R
Prove that area of ∆SMR = area of ∆QMN.
N
AD
19. In the adjoining figure, it is given that AD//BC and BD//CE. E
Prove that area of ∆ABC = area of ∆BDE. BC
Oasis School Mathematics-10 241
A D
N
20. In the given figure, ABCD is a parallelogram. If MN||BD, then, C
prove that area of ∆BAM = area of ∆AND.
B M
A
21. In the given ∆ABC, D is the mid-point of BC. E is the mid–point of F E
AD. F is the mid–point of AB and G is any point on BD. Prove that:
∆ABC = 8 ∆EFG.
B GD C
22. In the given diagram, ABCD and PQRD are two parallelograms. A PB
Prove that: //gm ABCD = //gm PQRD. D Q
C
R
Answer
1. Consult your teacher 2. Consult your teacher 3. (a) 12 6cm2 (b) 72 cm2 (c) 15 cm2 (d) 16 cm2
(e) 40.5 cm2 (f) 48 cm2 (g) 22 cm2 (h) 26 cm2 (i) 55 cm2
4. (a) 35 cm2 (b) 24 cm2 (c) 18 cm2 (d) 22 cm2 (e) 14 cm2 (f) 21 cm2 (g) 32 cm2 (h) 12 cm2 (i) 32 cm2
(j) 10 2cm (k) 25 cm2 (l) 25 cm2 (m) 20 cm2 (n) 15 cm2 (o) 64 cm2 (p) 60 cm2 (q) (i) 15 cm2
(ii) 15 cm2 (iii) 30 cm2 (r) (i) 20 cm2 (ii) 40 cm2
Project Work
On a graph paper, draw two given figures on the same base and between the same
parallel lines. From the graph, find the length of their base and height. Find their area
and draw a conclusion.
(i) two parallelograms (ii) a parallelogram and a rectangle
(iii) a parallelogram and a triangle (iv) two triangles
242 Oasis School Mathematics-10
Unit
14 Construction
14.1 Warm-up Activities
Discuss the following in your class and draw a conclusion.
• What are the features of a parallelogram?
• Is it possible to construct a parallelogram with two of its adjacent sides and angle
between them given?
• Is it possible to construct a parallelogram if the length of its two diagonals and angle
between them are given?
AD E
BC
• From the above figure name four triangles which are equal in area with ABC. Find out
the reason why they are equal in area.
• Is parallelogram ABCD=parallelogram BCED? why?
14.2 Construction of a parallelogram equal in area with given
parallelogram
Concept:
Parallelograms on the same base and between the same parallel lines are equal in area.
Example: 1
Construct a parallelogram whose adjacent sides are 4 cm and 6 cm and contained angle is
45°. Construct another parallelogram equal in area whose one angle is 60°.
PA QD Rough Sketch
4cm
PA Q D
B 6cm
4cm
600
450
C B 6cm C
Oasis School Mathematics-10 243
Steps:
(i) Draw a line segment BC = 6 cm.
(ii) Construct an angle of 45° at B, such that ∠ABC = 45°.
(iii) From B cut by an arc of 4cm to get A.
(iv) From A, cut by an arc of 6 cm and from C, cut by an arc of 4cm to get point D.
(v) Draw an angle of 60° at B such that ∠PBC = 60°.
(vi) Take an arc equal to PB and from C cut at Q.
(vii) Join PB and QC.
Hence, PBCQ is the required parallelogram.
Analysis
How is the area of the parallelogram ABCD equal to the area of the parallelogram PBCQ?
• The base of both a parallelograms is BC.
• Both lie between the same parallel lines PD and BC.
• Parallelograms on the same base and between the same parallel lines are equal in area.
Example: 2
Construct a parallelogram ABCD in which AB = 3 cm, AD = 4 cm, BD = 3.1 cm. Construct
another parallelogram equal in area whose one side is 3.5 cm.
B P CQ Rough Sketch
3.5cm 3.1cm
3cm B P 4cm C Q
3cm
3.1cm
A 4cm D A 4cm D
Steps:
(i) Draw a line segment AD = 4 cm.
(ii) From A and D cut by an arc of 3 cm and 3.4 cm respectively to get point B.
(iii) From D and B cut by an arc 3 cm and 4 cm to get C.
(iv) Now, ABCD is a parallelogram.
(v) Take an arc of 3.5 cm and cut from the points A and D to get points P and Q.
PADQ is the a required parallelogram
Exercise 14.1
1. a) Construct a parallelogram ABCD where AB = 6.3 cm, BC = 5.6 cm and
∠B = 600. Also construct another parallelogram equal in area to the parallelogram
ABCD, whose one side is 6.5 cm.
244 Oasis School Mathematics-10
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