Vedanta Excel in Mathematcs Book -10 Final (2078)

Indices

Example 2: Solve 3x + 2 × 5x – 1 = 405
Solution:

Here, 3x + 2 × 5x – 1 = 405

or, 3x × 3 2 + 5x × 5–1 = 405

or, 3x × 5x × 9 = 405
5

or, (3 × 5)x = 225

or, 15x = 152

? x =2

Example 3: Solve 2x + 3 + 1 – 9 = 0
2x
Solution:
2x + 3 + 1 – 9 = 0
Here, 2x

or, 2x × 23 + 1 – 9 = 0
2x

Let 2x = a 8a + 1 – 9 = 0

Then, a

or, 8a2 + 1 – 9a = 0
a

or, 8a2 – 9a + 1 = 0

or, 8a2 – 8a – a + 1 = 0

or, 8a (a – 1) – 1 (a – 1) = 0

or, (a – 1) (8a – 1) = 0

Either, a – 1 = 0 or, 8a – 1 = 0

i.e. a = 1 i.e. a = 1 = 1 = 2–3
8 23
i.e. 2x = 2° i.e.
? x =0 ? 2x = 2–3
So, x = 0, – 3 x = –3

Example 4: If a = bc, b = ca and c = ab, prove that abc = 1.

Solution:

Here, a = bc, b = ca and c = ab Alternative process
1 Here, a = bc,

Since c = ab, i.e. a = cb a = (ca)c [ b = ca]
a = cac
Now, a = bc [Given]

1 1

or, cb = bc [Putting a = cb]

1 a = (ab)ac [ c = ab]

or, cb = (ca)c [Putting b = ca] a = aabc
? abc = 1 Proved.
1

or, cb = cac

or, 1 = ac
b

or, abc = 1 proved.

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Indices

Example 5: If ap.aq = (ap) , prove that p(q – 2) + q (p – 2) = 0.
Solution:

Here, ap.aq = (ap)

or, ap + q = apq

or, p + q = pq

or, 2(p + q) = 2pq (multiplying both sides by 2)

or, pq + pq – 2p – 2q = 0

or, pq – 2p + pq – 2q = 0

or, p (q – 2) + q (p – 2) = 0 Proved.

Example 6: If xa = yb = zc and y2 = xz, show that 2 = 1 + 1 .
bac
Solution:

Here, xa = yb, b Alternative process
Also, zc = xa, Let, xa = yb = zc = k
Now, i.e. x = ya
or,
or, a b ab 11 1
or,
or, i.e. z = x c = ya c = yc

y2 = xz Then, x = ka, y = kb , z = k c

bb Now, y2 = xz

y2 = ya . y c 1 11

bb or, k2 b = ka . k c

y2 = ya c or, 2 = 1 + 1
c
kb ka
b b
2 = a + c or, 2 = 1 + 1 Proved.
b a c
2 =b 1 + 1
a c

or, 2 = 1 + 1 Proved.
b a c

Example 7: If p x = q y = r z and xyz = 1, prove that p + q + r = 0.
Solution:
Here, p x = q y = r z = k (suppose)

Then, p x = k

1

or, xP = k

or, x = kp

Similarly, y = kq and z = kr

Now, xyz = 1

or, kp.kq.kr. = 1

or, kp + q + r = k0

or, p + q + r = 0 Proved

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Indices

EXERCISE 10.2

General section b) 9x – 1 = 3x + 1 c) 42x – 1 = 2x + 1
1. Solve. 1 f) 103y – 3 =0.0101
0.04 i) 3x + 1 – 3x = 54
a) 2x – 4 = 4x – 6
l) 4x – 1 = ( 2)x
d) 32x + 1 = 92x –1 e) 25x + 3 =
x
g) 3 × 81x = 9x + 4 h) 2x + 1 – 2x = 8
j) 3x + 2 + 3x + 1 = 113 o) (0.5)2 = 0.25
k) 4x + 1 = 1
8x r) 5x + 5x + 1 + 5x + 2 = 155

m) ( 2)3x – 1 = ( 4)x – 2 n) ( 9)x – 3 = ( 3)x + 2

p) 3x – 2 + 3x = 10 q) 32x + 3 – 2.9x + 1 = 1
9 3

2. Solve.

a) 2x + 3 × 3x + 4 = 18 b) 2x – 3 × 3x – 4 = 3–1 c) 23x – 5 ax – 2 = 2x – 2 a1 – x
d) 5x – 3 × 32x – 8 = 225 e) 2x – 5 × 5x – 4 = 5 7
Creative section f) 72x + 1 × 52x – 1 = 5

3. Solve.

a) 22x + 3.2x – 4 = 0 b) 22x – 6.2x + 1 + 32 = 0 c) 4.3x + 1 – 9x = 27

d) 2a – 2 + 23 – a = 3 e) 3.2x + 1 – 4x = 8 f) 5.4x + 1 – 16x = 64
26 h) 5x + 5–x = 25215
g) 51 – x + 5x – 1 = 5 i) 5x + 1 + 52 – x = 126

j) 2x + 16 = 10 k) 3x +3 + 1 – 28 = 0 l) 7x + 343 = 56
2x 3x 7x
1 1 1 = 414 1 1
m) 4x + 4x = 16 16 n) 2x + 2x o) 3x + 3x = 9 9

p) 3x + 2 + 1 = 30
3x – 2

4. a) If ax = b, by = c and cz = a, prove that xyz = 1.

b) If x = yz, y = zx and z = xy, prove that xyz = 1.

c) If ax = by and b = a2, show that x – 2y = 0.

d) If a = 10x, b = 10y and aybx = 100, show that xy = 1.

e) If xa.xb = (xa)b, prove that a + b = ab – 2.
b a
3 1 1
f) If xa = yb = zc and y3 = xz, show that b = a + c

g) If ap = bq = cr and b2 = ac, prove that q = 2pr
p+r
1 1 2x.
h) If 2x = 3y = 12z, show that z = y +

i) If x a = y b = z c and abc = 1, prove that x + y + z = 0.

5. a) In how many years does Rs 2,000 amount to Rs 2,420 at 10% p.a. compound
interest?

b) In how many years does Rs 8,000 amount to Rs 9,261 at 5% p.a. compound interest?

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Unit Surds

11

11.1 Surds - Review

0, 1, 2, –3, –5, 2 , 4 , etc. are the examples of rational numbers. A rational number
can be 3 9
p
expressed in q form where q z 0. When a rational number is expressed in

decimal, whether the decimal part may be terminating or none-terminating recurring

decimal. For example,

1 = 0.5, 2 = 0.4, 3 = 0.75, 5 = 0.625 , etc. are terminating decimals.
2 5 4 8

1 = 0.333…, 5 = 0.833…, 4 = 0.5714285714..., etc. are none-terminating recurring
3 6 7

decimals.

On the other hand, there is another set of numbers which are not rational and cannot
p
be expressed in q form. These numbers are called irrational numbers. When an

irrational number is expressed in decimal, the decimal part is neither terminating

nor none–terminating recurring. 2, 3 , 3 6 , 4 8 , S, etc. are the example of irrational
numbers. Such irrational numbers are called surds.

In n a , ‘n’ is called the order of the surd and ‘a’ is called the radicand.
Here, n a is called the nth order surd, where n is the natural number and a is the
rational number greater than zero (0).

11.2 Laws of surds

There are some verified rules which are used in the operations of surds. Such rules
are called the laws of surds. The table given below shows some of the laws of surds.

Laws of surds Examples

(i) n a = 1 So, n a n 1 35 33

n, =a 3 2 = 23, = 53 = 5

(ii) n ab = n a . n b 3×5= 3× 5

(iii) n a= na 4 4= 44
b nb 5 45

(iv) n a + n a = 2 n a 3 5 + 3 5 =23 5

m na n ma mn a 3 64 = 4 = 2 = 2×3 64 = 6 64

(v) = =

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Surds

11.3 Like and unlike surds

2 3 , 5 3 , 3 , etc. are the like surds. They have the same order and the equal
radicand.
3 5 , 3 3 5 , 7 3 5 , etc. also have the same order and the equal radicand. So, they are
also the like surds.

The surds with the same order and the equal radicands are known as like surds.
On the other hand, 2 3 , 2 , 3 2 , 7 4 6 , etc. are unlike surds. Unlike surds do not
have the same order and the equal radicands.

11.4 Simplification of surds

(i) Addition and Subtraction of Surds

We can add or subtract the rational factors of the surds only in the case of like
surds. The process is exactly the same as in the case of addition and subtraction
of the algebraic terms with the same base and powers.

For example,

5 2 + 4 2 = (5 + 4) 2 = 9 2 As like 5x + 4x = 9x
7 3 4 – 3 3 4 = (7 – 3) 3 4 = 4 3 4 As like 7x – 3x = 4x

(ii) Multiplication and division of surds

We can multiply or divide the rational factors as well as the radicands only in the
case of the surds with the same order. For example,

3 6×5 7 =3×5 6×7 = 15 42
435 × 234 = 4 × 235 × 4 = 8 3 20

8 15 y 2 5 = (8 y 2) 15 ÷ 5 = 4 3

Worked-out examples

Example 1: Simplify (i) 50 + 18 – 8 2

(ii) 3 128 + 3 3 54 – 2 3 250

Solution:

(i) 50 + 18 – 8 2 = + 9 × 2 – 8 2

=5 2+3 2–8 2

=8 2–8 2 =0

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Surds

(ii) 3 128 + 3 3 54 – 2 3 250 = 3 64 × 2 + 3 3 27 × 2 – 2 3 125 × 2

= 4 3 2 + 3.3 3 2 – 2.5 3 2

= 13 3 2 – 10 3 2

= 332

Example 2: Simplify (i) 5 × 15 × 3 (ii) 3 128 – 3 16
Solution: 23 2

(i) 5 × 15 × 3 = 5 × 15 × 3

= 5×5×3×3

= 5 × 3 = 15

(ii) 3 128 – 3 16 = 3 64 × 2 – 3 8 × 2
23 2 23 2

= 43 2 – 23 2
23 2

= 23 2 = 1
23 2

Example 3: Simplify (3 5 – 4 2 ) (2 5 + 3 2 )
Solution:

Here, (3 5 – 4 2 ) (2 5 + 3 2 ) = 3 × 2 5. 5 + 3.3 5. 2 – 4 × 2 2. 5 – 4 × 3 2. 2
= 6 × 5 + 9 10 – 8 10 – 12 × 2
= 30 + 10 – 24
= 6 + 10

Example 4: Simplify 5x – 9 .
Solution: 3 + 5x

Here, 5x – 9 = ( 5x)2 – 32 = ( 5x + 3) ( 5x – 3) = 5x – 3
3 + 5x 3 + 5x
3 + 5x

Example 5: Simplify 7 3 16 – 2 3 54 .
5 3 32 + 2 3 108

Solution:

Here, 7 3 16 – 2 3 54 = 7 3 8 × 2 – 2 3 27 × 2
5 3 32 + 2 3 108 5 3 8 × 4 + 2 3 27 × 4

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Surds

= 7 × 232 – 2 × 332
5 ×2 3 4 + 2 × 3 3 4

= 14 3 2 – 6 3 2
10 3 4 + 6 3 4

= 832 = 1 3 2 = 1 31
16 3 4 2 4 2 2

EXERCISE 11.1

General section

1. Simplify: b) 5 3 + 6 3 – 3 c) 5 – 4 5 + 6 5
a) 2 + 5 2 + 2 2 e) 3 7 + 4 3 7 – 9 3 7 f) 2 4 6 – 4 6 – 3 4 6

d) 8 3 4 – 3 4 – 3 3 4

2. Simplify: b) 6 × 3 × 2 2 c) 3 9 × 3 3 × 3 3 2
a) 2 × 3 × 5

d) 5 8 ÷ 2 2 e) 5 3 108 ÷ 3 3 2 f) 4 360 ÷ 3 20

3. Simplify:

a) 32 + 8 – 72 b) 27 + 75 – 8 3

c) 4 45 – 3 20 +8 5 d) 12 – 75 + 48
e) 3 16 + 3 54 – 3 250 f) 5 3 81 – 2 3 24 + 3 375
g) 4 4 405 – 3 4 80 – 2 4 5 h) 3 2 + 4 2500 – 4 64 + 6 8

4. Simplify:

a) ( 3 + 2) ( 3 – 2) b) ( 5 – 3) ( 5 + 3)

c) (2 5 + 3 2) (2 5 – 3 2) d) ( 2 + 3)2

e) ( 5 – 3)2 f) ( x + a – x – a)2

g) (2 2 – 3) (3 2 + 3) h) (3 5 – 4 2) (2 5 + 2 3)

5. Simplify: b) x2 – 9 c) 3 25 – x2
x–3 3x + 5
a) a2 – b2
a–b e) 3x – 16 f) 49 – 5x
4 + 3x 7 – 5x
d) x – 4
x +2

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Surds

6. Simplify. b) 4 3 54 – 2 3 250 c) 2 75 + 4 108 – 3 48
6 3 128 33 3
a) 24 + 54
10 6 e) 50 + 18 f) 5 3 81 – 2 3 24
7 8 – 128 2 3 48 + 3 3 162
d) 3 3 81 – 3 3 24 + 2 3 375
13 3 192

11.5 Rationalisation

Let’s take a surd 3 .

3 is an irrational number. When it is multiplied by 3 , the product is 3 and it is
a rational number. The process of changing a surd into a rational number is called
rationalisation. Look at the following examples:

2× 2 = ( 2)2 = 2 o 2 is the rationalising factor of 2.

2 3× 3 = 2( 3)2 = 6 o 3 is the rationalising factor of 2 3 .

( x + a) ( x + a) = ( x + a)2= x + a o x + a is the rationalising factor of x + a.

Thus, if the product of two surds is a rational number, each of them is called the
rationalising factor of the other.

11.6 Conjugate

Let’s take a binomial surd, 5 + 3 .
The rationalising factor of 5 + 3 is 5 – 3 .
Here, 5 – 3 is called the conjugate of 5 + 3 or vice versa.
Thus, a binomial surd can be rationalised multiplying by its conjugate.

Worked-out examples

Example 1: Rationalise the denominators of: (i) 2 (ii) 3 6
3 25
Solution:

(i) Multiplying the numerator and denominator by 3 ,

2 =2 × 3 = 23
33 3 3

(ii) Multiplying the numerator and denominator by 5,

3 6 =3 6 × 5 = 3 30 = 3 30
2 5 2 5 5 2 ×5 10

Example 2: Rationalise the denominator and simplify: 3 5+ 3 .
5– 3
Solution:

Multiplying the numerator and denominator by 5 + 3 .

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Surds

3 5+ 3 = 3 5+ 3 × 5+ 3
5– 3 5– 3 5+ 3

= 3 × 5 + 3 15 + 15 + 3
( 5)2 – ( 3)2

= 15 + 4 15 + 3 = 18 +4 15 = 2 (9 +2 15) =9+2 15
5–3 2 2

Example 3: Rationalise the denominator of: x+2– x–2 .
x+2+ x–2
Solution:

Here, x + 2 – x – 2 = x + 2 – x – 2 × x + 2 – x – 2
x+2+ x–2 x+2+ x–2 x+2– x–2

= ( x + 2 – x – 2 )2
( x + 2)2 – ( x – 2 )2

=( x + 2)2 – 2 x + 2. x – 2 ) + ( x – 2 )2
x+2–x+2

= x + 2 – 2 x2 – 4 + x – 2
4

= 2x – 2 x2 – 4
4

= 2(x – x2 – 4) = x – x2 – 4
42

Example 4: Rationalise the denominator of: 3 5.
3+ 2–
Solution:

Here, 3 = 3 × ( 3 + 2) + 5

3+ 2– 5 ( 3 + 2) – 5 ( 3 + 2) + 5

= 3 + 6 + 15
( 3 + 2)2 – ( 5)2

= 3 + 6 + 15
( 3)2 + 2. 3. 2 + ( 2)2 – 5

= 3 + 6 + 15
3 + 2 6+ 2 – 5

= 3+ 6+ 15 × 6
26 6

=3 6+6+ 90
2×6

= 6+3 6+ 9 × 10
12

= 6+3 6+3 10 = 3(2 + 6+ 10) = 2 + 6+ 10
12 12 4

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Surds

Example 5: Simplify 5– 3 – 5+ 3 .
5+ 3 5– 3
Solution:
( 5 – 3)2 – ( 5+ 3)2
Here, 5– 3 – 5+ 3 = ( 5 + 3) ( 5– 3)
5+ 3 5– 3

= ( 5)2 – 2. 5. 3+( 3)2 – ( 5)2 – 2. 5. 3–( 3)2
( 5)2 – ( 3)2

= –4 15
5–3

= –4 15 = –2 15
2

Note: In the above example, the denominator of each term is conjugate to each other. In
such case, instead of rationalising the denominators, we can take their L.C.M. and
then simplify them.

Example 6: Simplify 32 6 + 6 3– 43 2 .
3+ 2+ 6+
Solution:

Here, 3 2 + 6 – 43
3+ 6 2+ 3 6+ 2

= ( 3 2 ( 3 – 6) 6) +( 6 ( 2 – 3) 3) –( 4 3 ( 6 – 2)
3 + 6) ( 3 – 2 + 3)( 2 – 6 + 2) ( 6 – 2)

= 3 2 ( 3 – 6) + 6 ( 2 – 3) – 4 3 ( 6 – 2)
( 3)2 – ( 6)2 ( 2)2– ( 3)2 ( 6)2 – ( 2)2

= 3 2 ( 3 – 6) + 6 ( 2 – 3) – 4 3 ( 6 – 2)
–3 –1 4

=– 2× 3+ 2× 6– 6× 2+ 6× 3– 3× 6+ 3× 2

= – 6 + 12 – 12 + 18 – 18 + 6 = 0

Example 7: Simplify a2 + 1 + a2 – 1 + a2 + 1 – a2 – 1
a2 + 1 – a2 – 1 a2 + 1 + a2 – 1
Solution:

Here, a2 + 1 + a2 – 1 + a2 + 1 – a2 – 1
a2 + 1 – a2 – 1 a2 + 1 + a2 – 1

= ( a2 + 1 + a2 – 1 )2 + ( a2 + 1 – a2 – 1 )2
( a2 + 1 – a2 – 1 ) ( a2 + 1 + a2 – 1 )

= ( a2 + 1)2 +2 a2 + 1. a2 – 1 ) + ( a2 – 1 )2 + ( a2 + 1)2 – 2 a2 + 1. a2 – 1 ) + ( a2 – 1 )2
( a2 + 1)2 – ( a2 – 1 )2

= a2 + 1 + 2 a4 – 1 + a2 – 1 + a2 + 1 – 2 a4 – 1 + a2 – 1 = 4a2 = 2a2
a2 + 1 – a2 + 1 2

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Surds

EXERCISE 11.2

General section

1. Rationalise the denominators and simplify:

a) 1 b) 1 c) 3 d) 4
2+1 3–2 5– 2 2 3– 2

e) 5 f) 2 2 g) 5 3 h) 4 5
3 7+2 3 6+ 3 2 3– 2 2 3+ 5

2. Rationalise the denominators and simplify:

a) 2 + 1 b) 3 – 1 c) 3 – 2 d) 5 + 3
2–1 3+1 3+ 2 5– 3

e) 2 3 – 3 2 f) 3 5 – 3 g) a + b – a – b h) x + 1 – x – 1
2 3+3 2 5– 3
a+b+ a–b x+1+ x–1

i) a + 2 – a – 2 j) a + 5 + a – 5 k) 2 l) 2
3+ 2+1
a+2+ a–2 a+5– a–5 2+ 3– 5

3. Simplify:

a) 3 5 – 1 b) 7 + 2 3 c) 3 + 5
5 3 5 2

d) 7 – 3 e) 2 + 3 f) ) 7 + 5
3 4 5 2 3 27

Creative Section - A b) 5 + 3 5 – 5 – 3 5

4. Simplify: 5+2 5–2
a) 3 + 1 + 3 – 1
3–1 3+1

c) 3 + 2 + 3 – 2 d) x + a – x – a
x– a x+ a
3– 2 3+ 2

e) x + x2 – 1 – x – x2 – 1 f) a – a2 – 1 + a + a2 – 1
x – x2 – 1 x + x2 – 1 a + a2 – 1 a – a2 – 1

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Surds

5. Simplify: b) 72 – 48 – 45 + 2 98
a) 3 20 + 4 + 5 + 3 50 128
5 5–3
d) 3 2 – 4 3 + 6
c) 2 10 – 2 5 – 10
3+1 6+2 2+1 6+ 3 6+ 2 3+ 2

e) 7 3 + 25 + 32 f) 5 2 – 8 5 + 3 10
10 + 3 6+ 5 15 + 3 3 5( 2 + 1) 10 + 2 2+ 5

Creative Section - B

6. Simplify:

a) x + 1 + x – 1 + x + 1 – x – 1 b) a + b + a – b + a + b – a – b

x+ 1 – x– 1 x+ 1 + x– 1 a+ b – a– b a+ b + a– b

c) 2x + 3 + 2x – 3 + 2x + 3 – 2x – 3 d) x2 + 2 + x2 – 2 + x2 + 2 – x2 – 2

2x + 3 – 2x – 3 2x + 3 + 2x – 3 x2 + 2 – x2 – 2 x2 + 2 + x2 – 2

11.7 Simple surd equations

Let's consider an equation x = 5.
Here, the unknown variable is a surd. Such an equation is known as the surd
equation.

To solve a surd equation, we should remove the radical from the variable. For this, we
should give nth power to both sides of the equation to remove nth order of radical.

∴ If n x = a, then

nx n

= an

n

i.e. xn = an
i.e. x = an.

Worked-out examples

Example 1: Solve x + 5 = 3.
Solution:

Here, x + 5 = 3
Squaring both sides of the equation, we get,

( x + 5)2 = 32
or, x + 5 = 9
or, x = 9 – 5
or, x = 4
Now, substituting x = 4 in the original equation, we get,

4 + 5 =3
or, 9 =3
or, 3 = 3 which is true.
So, the required value of x is 4.

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Surds

Example 2: Solve x + 4 = x – 2.
Solution:

Here, x + 4 = x – 2

Squaring both sides of the equation, we get,

or, ( x + 4)2 = (x – 2)2
or,
or, x+4 = x2 – 4x + 4
Either, x2 – 5x =0
or, x (x – 5) =0
=0
x = 0, i.e. x = 5
x–5

Substituting x = 0 in the original equation, we get,

0+4 =0–2

or, 4 = – 2
or, 2 = – 2 which is false.

Substituting x = 5 in the given equation, we get,

5+4 =5–2

or, 9 = 3
or, 3 = 3 which is true.
So, the required value of x is 5.

Example 3: Solve x – 9 – x = 3

Solution:

Here, x – 9 – x = 3
or, x – 9 = 3 + x

Squaring both sides of the equation, we get,

( x – 9 )2 = (3 + x )2
or, x – 9 = 32 + 2.3. x + ( x )2
or, x – 9 = 9 + 6 x + x
or, 6 x = – 18

x =–3

Again, squaring both side,

( x )2 = (– 3)2
or, x = 9

Now, substituting x = 9 in the original equation, we get,

9–9 – 9 =3
or, 0 – 3 = 3
or, – 3 = 3 which is false.
So, the equation does not have any solution.

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Surds

Example 4: Solve 3x + 1 – x – 1 = 2
Solution:
Here, 3x + 1 – x – 1 = 2
or, 3x + 1 = 2 + x – 1
Squaring both sides of the equation, we get,

( 3x + 1)2 = (2 + x – 1)2
or, 3x + 1 = 22 + 2.2. x – 1 + ( x – 1)2
or, 3x + 1 = 4 + 4 x – 1 + x – 1
or, 2x – 2 = 4 x – 1

x–1 =2 x–1
Again, squaring both sides,

(x – 1)2 = (2 x – 1 )2

or, x2 – 2x + 1 = 4 (x – 1)

or, x2 – 2x + 1 – 4x + 4 = 0
or, x2 – 6x + 5 = 0
or, x2 – 5x – x + 5 = 0
or, x(x – 5) – 1(x – 5 = 0
or, (x – 5) (x – 1) = 0
Either, x – 5 = 0, i.e., x = 5

or, x – 1 = 0, i.e., x = 1

When, x = 5, 3 × 5 + 1 – 5 – 1 = 2, i.e. 2 = 2 which is true.
When, x = 1, 3 × 1 + 1 – 1 – 1 = 2, i.e. 2 = 2 which is true.
So, the requested values of x are 5 and 1.

Example 5: Solve 3 x – 4 = 15 + 3 x
x +2 x + 40
Solution:

Here (3 x – 4) ( x + 40) = ( x + 2) (15 + 3 x )

or, 3x +120 x – 4 x – 160 = 15 x + 3x + 30 + 6 x

or, 95 x = 190

or, x = 2
or, ( x )2 = 22

x =4

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Surds

Now, substituting x = 4 in the given equation, we get

3 4 – 4 = 15 + 3 4
4 +2 4 + 40

or, 6–4 = 21
4 42

or, 1 = 1 which is true
2 2

So, the required value of x is 4.

Example 6: Solve x–1 =4+ x –1
Solution: x +1 2

Here, x–1 =4+ x –1
x +1 2

or, ( x )2 – 12 = 8+ x –1
x +1 2

or, ( x + 1) ( x – 1) 7+ x
x +1 =2

or, 2 x – 2 = 7 + x

or, 2 x – x = 7 + 2

or, x = 9

or, ( x )2 = 92

or, x = 81

Now, substituting x = 81 in the given equation, we get,

81 – 1 =4+ 81 – 1
81 + 1 2

or, 80 =4+4
10

or, 8 = 8 which is true

Hence, the required value of x is 81.

Example 7: Solve x + 5 + 6 = 3
x+5– 6

Solution:

Here, x+5+ 6 =3
x+5– 6

By using the rule of componendo and dividendo, we get

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Surds

x+5+ 6+ x+5– 6 3+1 If a = c then
x+5+ 6– x+5+ 6 = 3–1 b d

or, 2 x + 5 = 4 a+b = c+d
26 2 a–b c–d

is called the rule of

x+5 componendo and
or, = 2
dividendo.
6
Squaring both sides, we get,

x+5 2
6
= 22

or, x + 5 = 4
6

or, x = 19

Now, substituting x = 19 in the given equation, we get
19 + 5 + 6 = 3
19 + 5 – 6

or, 24 + 6 = 3
24 – 6

or, 2 6 + 6 = 3
2 6– 6

or, 3 6 = 3
6

or, 3 = 3 which is true.

Hence, the required value of x is 19.

Example 8: Solve x+ 2 + x– 2 =6
x– 2 x+ 2

Solution:

Here, x+ 2 + x– 2 =6
x– 2 x+ 2

or, ( x + 2)2 + ( x – 2)2 = 6
( x – 2) ( x + 2)

or, ( x )2 +2. x . 2 + ( 2)2 + ( x )2 – 2. x . 2 + ( 2)2 = 6
( x )2 – ( 2)2

Vedanta Excel in Mathematics - Book 10 164 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Surds

or, x + 2 + x + 2 = 6
x–2

or, 2x + 4 = 6x – 12

or, 4x = 16
or, x = 16
Now, Substituting x = 4 in the given equation, we get,

4+ 2 + 4– 2 =6
4– 2 4+ 2

or, 2 + 2 + 2 – 2 = 6
2– 2 2+ 2

or, 22 + 2 . 2 . 2 + ( 2)2 + 22 – 2. 2. 2 + ( 2)2 = 6
(2 – 2) (2 + 2)

or, 4 + 2 + 4 + 2 = 6
4–2

or, 12 = 6
2

or, 6 = 6 which is true.
So, the required value of x is 4.

EXERCISE 11.3

General section

1. Solve:

a) x + 1 = 5 b) 2x – 5 = 7 c) 4x – 3 = 3
f) 3 x + 5 – 1 = 2
d) 2x + 3 – 3 = 0 e) 3x – 2 – 2 = 3 i) 4 a – 3 = 5a + 7
g) 4 2x – 1 – 2 = 1 h) x + 5 = 2x – 3 l) 4 6x + 5 = 4 x – 20
j) 3 4x + 1 = 3 x + 13 k) 3 7x + 11 = 3 3x + 5
c) x + 8 – 2 = x
2. Solve: f) 2x – 4x2 – 15 = 3
i) x2 + 5 – 1 = x
a) x – 7 = x – 1 b) x + 3 = x + 1
d) x – 4 + x = 2 e) x = 6 – x – 24 c) x – 25 = 9
x –5
g) x + x2 – 20 = 10 h) 9x2 – 20 = 3x – 2

3. Solve: b) x –9 = 1
x–1 x +3

a) x + 1 = 1

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d) 5x – 4 5x + 2 e) 5y – 4 = 2 – 5y – 3 f) x–3 = 2
5x + 2 = 2 – 2 5y – 2 2 x 5

g) x–4 = 3 h) y + 5 = 3 i) x + 7 = 3
x 7 y– 5 x– 7

Creative Section - A b) 2x +9 = 13 – x
4. Solve: d) 3x – 7x +2 = 2
f) 2x2 + x – 3 = x – 1
a) 2x + 7 = x + 2 h) 4 – x + x + 9 = 5
c) 3x + 4 + x = 12 j) x + 5 + x + 12 = 2x + 41
e) 2x + 1 = 4x2 + 3x + 6

g) 3x + 1 – x – 1 = 2
i) 4x – 3 + 2x + 3 = 6

5. Solve: 105
b) x + x – 15 = x – 15
a) x+ 5+x = 15
5+x

c) 2 x – 4x – 3 = 1 d) 5 x –3 = 3+5 x
4x – 3 x +2 x +5

e) x + x + 13 = 91 f) x + x – 1 – x = 1
Creative Section - B x + 13

6. Solve:

a) x–1 x –1 b) x – 4 = 2 + x –2
x +1 =4+ 2 2+ x 2

y – 25 =4+ y –5 d) 5y – 4 = 2 + 5y – 3
c) 5 + y 5 5y + 2 2

e) 3x – 4 – 3x – 2 = 2 f) 7x – 36 =9–5 7x – 11
2 + 3x 2 6 + 7x 3

7. Solve:

a) x + 4 + 2 = 2 b) x + 6 + 3 = 3
x+4 – 2 x+6 – 3

c) x+2 – x–2 = 1 d) x + 4 + x – 4 = 2
x+2 + x–2 2 x+4 – x–4

e) x + 5 + x – 5 = 4 f) x + a + x – a = 6
x– 5 x+ 5 x– a x+ a

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Unit Equations

12

12.1 Simultaneous equations - review

Let’s take a linear equation with two variables, y = x + 2.
The equation has as many pairs of solutions as we wish to find.
The table given below shows a few pairs of solutions.

x 0 1 2 3 5 –1 –2 –3

y 2 3 4 5 7 1 0 –1

From the above table (0, 2), (1, 3), (3, 4), (3, 5), … (–3, –1), ... are a few pairs of
solutions that satisfy the equation y = x + 2.

Now, take another equation, y = 3x – 2.

It has also many pairs of solutions. For example, (0, -2), (1, 1), (2, 4), (3, 7), ... and so
on. Each pair of solution satisfies the equation y = 3x – 2.

Here, (2, 4) is the common pair of solution that satisfies both the equations
y = x + 2 and y = 3x – 2 simultaneously. Such pair of equations that have only one
pair of solution which satisfies both the equations simultaneously at a time are called
simultaneous equations.

12.2 Application of simultaneous equations

Simultaneous equations are broadly used to find two unknown quantities. Considering
two unknown quantities which are asked to find in a word problem by any two
variables such as x and y, we can make a pair of simultaneous equations under the
two given conditions. By solving the equations, we can find the unknown quantities.

Worked-out examples

Example 1: Solve x + 2y = 8 and x + y = 5.

Solution:

x + 2y = 8 ……… (i)
x + y = 5 ……….(ii)

Subtracting equation (ii) from (i),

x + 2y = 8 The coefficient of x is the same in both equations and so
rx r y = r 5 we subtract one equation from the other to eliminate x.

y=3

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Substituting the value of y in equation (ii), we get,
x+3 =5

or, x = 5 – 3 = 2
? x = 2 and y = 3

Example 2: The sum of two numbers is 45. If the greater number is 9 more than the
Solution: smaller one, find the numbers.

Let the greater number be x and the smaller one be y.

From the first given condition, The sum of the number is 45.
x + y = 45 ? x + y = 45

or, x = 45 – y ………….. (i)

From the second given condition, The greater number is 9 more than smaller one.
x = y + 9 …………… (ii) ?x= y+9

Substituting the value of x from equation (i) in (ii), we get,
45 – y = y + 9

or, – y – y = 9 – 45
or, – 2y = – 36
or, y = 18

Now, substituting the value of y in equation (i), we get,
x = 45 – 18 = 27

So, the required greater number is 27 and the smaller one is 18.

Example 3: The total cost of 3 kg of apples and 5 kg of oranges is Rs 245. If the cost of 5
kg of apples is the same as the cost of 8 kg of oranges, find the rate of cost
of these fruits.

Solution:

Let the rate of cost of apples be Rs x per kg.
And, the rate of cost of oranges be Rs y per kg.

From the first given condition,

3x + 5y = 245 ………… (i)

From the second given condition,

5x = 8y

or, x= 8 y ……………….. (ii)
5

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Equations

Substituting the value of x from equation (ii) in equation (i), we get,

3 × 8y + 5y = 245
5
24y + 25y
or, 5 = 245

or, 49y = 1,225

or, y = 1,225 = 25
49
Now, substituting the value of y in equation (ii), we get,

x= 8 × 25 = 40
5
So, the required rate of cost of apples is Rs 40 per kg and that of oranges is Rs 25 per kg.

Example 4: In a fraction, the numerator is 1 less than the denominator. If 1 is added to
12.
the numerator and 5 to the denominator, the fraction become Find the
original fraction.

Solution:

Let the numerator and the denominator of the original fraction be x and y respectively. Then,

the required fraction is x .
y
From the first given condition,

x = y – 1 .................. (i)

From the second given condition,

x+1 = 1
y+5 2

or, 2x + 2 = y + 5

or, 2x – y = 3 ............... (ii)

Now, substituting the value of x from equation (i) in equation (ii), we get,

2(y – 1) – y = 3

or, 2y – 2 – y = 3

or, y = 5

Again, putting the value of y in equation (i) we get,

x=y–1=5–1=4

So, the required fraction is 4 .
5

Example 5: The monthly income of Ram and Hari are in the ratio of 3 : 2 and their

expenses are in the ratio of 5 : 3. If each of them saves Rs 1000 in a month,

find their monthly incomes.

Solutions:

Let the monthly income of Ram be Rs x.
And the monthly income of Hari be Rs y.

? The monthly expense of Ram = Rs (x – 1,000)

The monthly expense of Hari = Rs (y – 1,000)

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From the first given condition,
x 3
y = 2

or, thexs=eco32nyd ……………………..….. (i)
From given condition,

x – 1,000 = 5
y – 1,000 3

or, 3x – 3,000 = 5y – 5,000

or, 3x = 5y – 2,000 ………. (ii)

Substituting the value of x from equation (i) in equation (ii), we get,
3y
3 × 2 = 5y – 2,000

or, 9y = 10y – 4,000

or, y = 4,000

Now, substituting the value of y in equation (i), we get.

x = 3 × 4000 = 6,000
2
Hence, the monthly income of Ram is Rs 6,000 and that of Hari is Rs 4,000.

Example 6: Two years ago, father was 6 times as old as his son was. Three years hence
Solution: he will be 5 years older than 3 times the age of his son. Find their present
ages.

Let the present age of the father be x years.
And the present age of his son be y years.

From the first given condition, Two years ago, the age of the father was (x – 2)
x – 2 = 6 (y – 2) years and that of son was (y – 2) years.
Father was 6 times as old as his son means
or, x = 6y – 12 + 2 x – 2 = 6 (y – 2)

or, x = 6y – 10 ……… (i) 3 years hence, father will be (x + 3) years
From the second given condition, and the son will be (y + 3) years.
5 years older than 3 times the age of the son
x + 3 = 5 + 3 (y + 3) means x + 3 = 5 + 3 (y + 3)
or, x = 5 + 3y + 9 – 3

or, x = 3y + 11 ……………. (ii)

Substituting the value of x from equation (i) in equation (ii), we get,

6y – 10 = 3y + 11

or, 6y – 3y = 11 + 10

or, 3y = 21

or, y = 7
Now, substituting the value of y in equation (i), we get,

x = 6 × 7 – 10

or, x = 32

Hence, the present age of the father is 32 years and that of the son is 7 years.

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Equations

Example 7: 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.

Solution:
Let the present age of the father be x years and the present age of the son be y years.

From the first condition, From the second condition,

x + 1 = 5 (y + 1) x – 2 = 3 (y + 4)

or, x = 5y + 5 – 1 or, x = 3y + 12 + 2

or, x = 5y + 4 …………. (i) or, x = 3y + 14 ………….. (ii)

Substituting the value of x from equation (i) in equation (ii), we get,

5y + 4 = 3y + 14

or, 5y – 3y = 14 – 4

or, 2y = 10

or, y = 5

Now, substituting the value of y in equation (i), we get, x = 5 × 5 + 4 = 29

Hence, the present age of the father is 29 years and that of the son is 5 years.

Example 8: The ratio of the present age of A and B is 2 : 7. Five years hence, the ratio
of their ages will be 3 : 8. Find their present ages.

Solution:

Let the present age of A is x years. Alternative method

And the present age of B is y years.

From the first given condition, Let the present age of A be 2x years and
that of B be 7x years.
x = 2
y 7 From the given condition,

or, 7x = 2y 2x + 5 = 3
7y + 5 8
or, x = 2y ……….… (i) or, 21x + 15 = 16x + 40
7
From the second given condition, or, 21x – 16x = 40 – 15

x+5 = 3 or, 5x = 25
y+ 5 8
or, x = 5
or, 8x + 40 = 3y + 15
? Age of A = 2x = 2× 5 years = 10 years

or, 8x = 3y + 15 – 40 Age of B = 7x = 7 × 5 years = 35 years.

or, 8x = 3y – 25 ………… (ii)

Substituting the value of x from equation (i) in equation (ii), we get,
2y
8 × 7 = 3y – 25

or, 16y = 21y – 175

or, y = 35

Now, substituting the value of y in equation (i), we get,

x = 2 × 35 = 10
7

Hence, the present age of A is 10 years and that of B is 35 years.

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Example 9: The sum of the digits of a two–digit number is 8. If 18 is subtracted from the
Solution: number, the places of the digits are reversed. Find the number.

Let the digit at tens place of the number be x
and the digit at ones place of the number be y.

? The number so formed is 10x + y. Let's consider a number 26.
In 26, 2 is in tens place and 6 is in ones place.
From the first given condition, ? 26 = 10 × 2 + 6
x+y=8 Similarly, if x is in tens place and y in ones place, the
x = 8 – y …………… (i) number is 10x + y.

From the second given condition, The original number is 10x + y. When the
10x + y – 18 = 10y + x places of the digits are reversed, the new
number so formed is 10y + x.
or, 10x + y – 10y – x = 18

or, 9x – 9y = 18

or, 9 (x – y) = 18

or, x – y = 2 …………. (ii)
Substituting the value of x from equation (i) in equation (ii), we get,

8–y –y =2
or, – 2y = 2 – 8
or, y = 3
Now, substituting the value of y in equation (i), we get, x = 8 – 3 = 5
Hence, the required number is 53.

Example 10 : A two digit number is 3 times the sum of its digits. The sum of the number

formed by reversing its digits and 9 is equal to 3 times the original number.
Find the number.

Solution:

Let the digit at tens place of the number be x and the digit at ones place of the number be y.
Then, the number so formed is 10x + y.
Again, the number formed by reversing the digits is 10y + x.

From the first given condition, From the second given condition,
10x + y = 3(x + y) (10y + x) + 9 = 3(10x + y)

or, 10x + y = 3x + 3y or, 10y + x + 9 = 30x + 3y
or, 7y – 29x = – 9 ...................... (ii)
or, y = 7x .......................... (i)
2

Substituting the value of y from equation (i) in equation (ii), we get,

7 × 7x – 29x =–9
2
or, 49x – 58x = –18

or, –9x = – 18

or, x = 2 7x 7 × 2
2 2
Now, putting the value of x in equation (i), we get, y = = =7

Hence, the required number is 27.

Vedanta Excel in Mathematics - Book 10 172 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Example 11: Upendra started his motorbike journey from Damak to Lahan at 7 a.m. with
an average speed of 40 km/hr. An hour later Tashi also started his journey
from Damak to Lahan with a uniform speed of 60 km/hr. At what time
would they meet each other?

Solutions:

Suppose Upendra meets Tashi after x hours.
But, Tashi meets Upendra after (x – 1) hour.

Now, the distance travelled by Upendra in x hours = 40x km.
Also, the distance travelled by Tashi in (x – 1) hours = 60(x – 1) km.

When they travelled the equal distance, they meet each other.
? 40x = 60(x – 1)
or, 2x = 3x – 3
or, x = 3
? They meet after 3 hours.
Since Upendra started his journey at 7 a.m., they meet at 10 a.m.

Example 12: Two buses were coming from two villages situated just in the opposite
direction. The uniform speed of one bus is 10 km/hr more than that of
another one and they had started their travelling in the same time. If the
distance between the villages is 450 km and they meet after 5 hours, find
their speeds.

Solutions:

Let the speed of the faster bus be x km/hr 450 km
and the speed of slower bus be y km/hr.

From the first given condition, Distance travelled by faster bus = 5x
x – y = 10 ………..….. (i) Distance travelled by slower bus = 5y

From the second given condition, When they meet, they travelled 450 km.

5x + 5y = 450 travelling from opposite direction.
or, 5(x + y) = 450 So, 5x + 5y = 450

or, x + y = 90 ……………... (ii)

Adding equations (i) and (ii),

x – y + x + y = 10 + 90

or, 2x = 100

or, x = 50

Now, substituting the value of x in equation (ii), we get,

50 + y = 90

or, y = 40

So, the speed of the faster bus is 50 km/hr and the slower one is 40 km/hr.

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Example 13: A few number of students are made to stand in rows. If 5 more students
were kept in each row, there would be 3 rows less. If 5 less students were
kept in each row, there would be 5 rows more. Find the number of students.

Solutions:

Let, the number of rows be x and the number of students in each row be y.

Then, the total number of students = xy

From the first condition, From the second condition,

(x – 3) (y + 5) = xy (x + 5) (y – 5) = xy

or, xy + 5x – 3y – 15 = xy or, xy – 5x + 5y – 25 = xy

or, 5x – 3y = 15 .......... (i) or, –5xy + 5y = 25 .......... (ii)

Adding equation (i) and (ii), we get,

5x – 3y – 5x + 5y = 15 + 25

or, 2y = 40

or, y = 20

Now, putting the value of y in equation (i), we get,

5x – 3 × 20 = 15

or, x = 15

Then, the number of students = 15 × 20 = 300

Hence, there are 300 students.

EXERCISE 12.1

General section

1. Solve each pair of the following simultaneous equation:

a) y = x – 2 b) x = 8 – 2y c) y = 2x – 3
x+y=4 x+y=6 2y + x = 4

d) x + y = 42 e) x – y = 30 f) x – 20 = 5 (y – 20)
x – 6 = 5 (y – 6) x + 5 = 2(y + 5) x – 2y = 10

g) x + y = 5 h) x + y = 10 i) 10x + y = 7 (x + y)

10x + y – 9 = 10y + x 10x + y + 18 = 10y + x 10x + y – 36 = 10y + x

2. a) The sum of two numbers is 60 and their difference is 10. Find the number.

b) A number is twice the other number. If their sum is 30, find the numbers.

c) The sum of two numbers is 59. If the smaller number is less than the bigger one by
7, find the number.

d) The difference of two numbers is 12. If the greater number is three times the smaller
one, find the numbers.

e) If three times the sum of two, numbers is 42 and five times their difference is 20,
find the numbers.

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f) The total cost of a watch and a radio is Rs 500. If the watch is cheaper than the
radio by Rs 150, find their cost.

g) The sum of the age of a father and his son is 38 years. If the father is 22 years older
than his son, find their present ages.

h) 1 of the age of a mother is the age of her daughter. If the difference of their ages is
6
35 years, find the age of the mother.

i) Father is three times as old as his son. If the difference of their ages is 24 years, find
their present ages.

j) The perimeter of a rectangular field is 120 m. If the field is 10 m longer than its
breadth, find the length and the breadth of the field.

Creative section - A

3. a) The total cost of 4 kg of apples and 6 of oranges is Rs 570. If the cost of 3 kg of
apples is the same as the cost of 5 kg of oranges, find the rate of cost of apples and
oranges.

b) The cost of tickets of a comedian show of 'Gaijatra' is Rs 150 for an adult and Rs 50
for a child. If a family paid Rs 550 for 5 tickets, how many tickets were purchased
in each category?

c) A man buys a few number of books at Rs 72 each and a few number of pens at
Rs 25 each. If he buys 11 articles and pays Rs 510 altogether, find the number of
each article bought by him.

4. a) oaIfdri4dgeiindsatlsoufrtbhatcertadicoetnen?domfrionmatotrheofnthuemoerriagtionraloffraacftrioanct,iiotns ,vaitlsuevableuceombeecso12me. sW31ha. tIfis5thies

b) The numerator of a fraction is 5 less than its denominator. If 1 is added to each, its

value becomes 1 . Find the original fraction.
2
c) 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
d) The ratio of the monthly income and expenditure of Sunayana is 5 : 3. If she
increases her income by Rs 5,000 and decreases the expenditure by Rs 3,000, the
new ratio becomes 5 : 2. Find her income and expenditure.

e) The monthly income of Rita and Bishwant are in the ratio of 4 : 3 and their expenses
are in the ratio 3:2. If each of them saves Rs 2000 in a month, find their monthly
income.

5. a) If twice the son's age in years is added to the father's age, the sum is 70. But twice
the father's age is added to the son's age, the sum is 95, find the present ages of the
father and his son.

b) The sum of the ages of the father and son is 44 years. After 8 years, the age of the
father will be twice the age of the son. Find their present ages.

c) Three years ago the sum of the ages of a father and his son was 48 years and three
years hence father's age will be three times that of his son's age. Find the present
ages of the father and his son.

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d) Eight years age, the daughter's age was thrice the son's age. Now the daughter's age
is 4 years more than the son's age. Find their present ages.

e) 2 years ago, father's age was nine times the son's age but 3 years later it will be 5
times only. Find the present ages of the father and the son.

f) 5 years ago a man's age was 5 times the age of his daughter's. 3 years hence, twice
his age will be equal to 6 times his daughter's age. Find their present ages.

g) A year hence a father will be 5 times as old as his son. Two years ago the father was
3 times as old his son will be 4 years hence. Find their present ages.

h) 14 years ago the age of the mother was 4 times the age of her daughter. The present
age of the mother is 2 times the age of her daughter will be 4 years hence. What are
their present ages?

i) 3 years later a mother will be 4 times as old as her son. 3 years ago, the mother's age
was two times her son's age will be 8 years hence. What are their present ages?

j) The present age of the father is three times the age of his son. If the age of the son
after 10 years is equal to the age of the father before 20 years, find the present ages
of the father and the son.

k) The sum of the ages of a father and his son is 40 years. If they both live on till the
son becomes as old as the father is now, the sum of their ages will be 96 years. Find
their present ages.

l) The sum of present ages of a father and his son is 80 years. When the father's age
was equal to the present age of the son, the sum of their ages was 40 years. Find
their present ages.

m) The ages of two girls are in the ratio of 5 : 7. Eight years ago their ages were in the
ratio 7 : 13. Find their present ages.

n) Three years ago, the ratio of the ages of A and B was 4 : 3. Three years hence, the
ratio of their ages will be 11 : 9. Find the present ages of A and B.

6. a) A number consists of two digits. The sum of its digits is 16. if 18 is subtracted from
the number, the digits interchange their place. Find the number.

b) The sum of the digits of a two-digit number is 10. When 18 is added to the number,
its digits are reversed. Find the number.

c) The digit at tens place of a two digit number is two times the digit at ones place.
When 27 is subtracted from the number, its digit are reversed. Find the number.

d) The digit at ones place of a two digit number is three times the digit at tens place.
When 54 is added to the number, its digits are reversed. Find the number.

e) A number consisting of two digits is three times the sum of its digits. If 45 is added
to the number, the digits will be interchanged. Find the number.

f) A certain number of two digits is seven times the sum of the digits. If 36 is subtracted
from the number, the digits will be reversed. Find the number.

g) The sum of the digits in a two-digit number is 11. The number formed by
interchanging the digits of the number will be 45 more than the original number.
Find the original number.

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h) A number of two-digits exceeds four times the sum of its digits by 3. If 36 is added
to the number, the digits are reversed. Find the number.

i) A number consists of two digits. If the number formed by reversing its 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.

j) 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.

k) A two digit number is 4 times the sum of its digits. The sum of the number formed
by reversing its digits and 9 is equal to 2 times the original number. Find the
number.

l) Five times the sum of the digits of a two digit number is 9 less than the number
formed by reversing its digits. If four times the value of the digit at ones place is
equal to half of the place value of the digit at tens place, find the number.

Creative section - B
7. a) In a city, the taxi charges consists of two types of a charges: a fixed charge together

with the charge for the distance covered. If a person travels 10 km, he pays Rs 180
and for travelling 12 km, he pays Rs 210. Find the fixed charges and the rate of
charge per km.
b) A lending library has a fixed charge for the first four days and an additional charge
for each day thereafter. Dorje paid Rs 50 for a book kept for 9 days, while Kajal paid
Rs 35 for the book she kept for 6 days. Find the fixed charge and the charge for each
extra day.
8. a) If Prakash gives one of the marbles from what he possesses to Kamala, then they
will have equal number of marbles. If Kamala gives one of the marbles from what
she possesses to Prakash, Prakash will have double number of marbles. Find the
number of marbles possessed by them initially.
b) When Gopal gives Rs 10 from his money to Laxmi her money becomes double than
that of the remaining money of Gopal. When Laxmi gives Rs 10 to Gopal his money
is still Rs 10 less than the remaining money of Laxmi. Find their original sum of
money.
9. a) Janak started his bicycle journey from Kohalpur to Dhangadhi at 6:00 a.m. with an
average speed of 20 km/hr. Two hours later Ganesh also started his journey from
Kohalpur to Dhangadhi with an average speed of 30 km/hr. At what time would
they meet each other if both of them maintain non-stop journey.
b) Two buses were coming from two different places situated just in the opposite
direction. The average speed of one bus is 5 km/hr more than that of another one
and they had started their journey in the same time. If the distance between the
places is 500 km and they meet after 4 hours, find their speed.
10. a) When the length of a rectangular field is reduced by 5 m and breadth is increased
by 3 m, its area gets reduced by 9 sq. m. If the length is increased by 3 m and
breadth by 2 m the area increases by 67 sq. m. Find the length and breadth of the
room.
b) A few number of students are made to stand in a certain number of rows equally.
If 3 more students were kept in each row, there would be 2 rows less. If 3 less
students were kept in each row, there would be 3 rows more. Find the number of
students.

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12.3 Quadratic equation - review

Let's consider an equation x + 5 = 0.

In this equation, the highest power of the variable x is 1. So, it is called a first degree
equation of one variable. A first degree equation is also called a linear equation.

On the other hand, let's consider another equation x2 + 9x + 20 = 0.

In this equation, the highest power of the variable x is 2. So, it is called a second
degree equation. A second degree equation of one variable is called a quadratic
equation.

Thus, a quadratic equation is a second degree polynomial equation in a single
variable. The standard form of a quadratic equation is ax2 + bx + c = 0, where a, b,
c  R and a ≠ 0.

A quadratic equation of the form ax2 + c = 0, where the middle term with the variable
containing power 1 is missed, is known as pure quadratic equation. x2 – 9 = 0,
4x2 – 25 = 0, etc. are pure quadratic equations. The standard form of quadratic
equation which is ax2 + bx + c = 0 is called the adfected quadratic equation.
x2 + 5x + 6 = 0, 2x2 + 3x – 2 = 0, etc. are the example of adfected quadratic equations.

12.4 Application of quadratic equations

A certain type of verbal problems can be solved by the method given by quadratic
equation. For this purpose, we should consider the unknown quantity as a variable
such as x, y, z, a, b, c, etc. Then, the verbal problem is expressed in a quadratic
equation form. By solving the quadratic equation, we find the values of the variable.
In some cases, only one solution of the quadratic equation may be meaningful and
it will be the required value. Any value which does not satisfy the condition of the
problem should be rejected.

Worked-out examples

Example 1: Solve 2x2 + 3x – 2 = 0.

Solution:

Here, 2x2 + 3x – 2 = 0

or, 2x2 + 4x – x – 2 = 0

or, 2x (x + 2) – 1 (x + 2) = 0

or, (x + 2) (2x – 1) = 0

Either, x + 2 = 0, i.e. x = – 2

or, 2x – 1 = 0, i.e. x = 1
2
1
?x = –2 and 2

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Example 2: If the sum of two numbers is 9 and their product is 18, find the numbers.

Solution:
Let one of the numbers be x.
Then, the other number is (9 – x)
According to the given condition,

x (9 – x) = 18
or, 9x – x2 = 18
or, x2 – 9x + 18 = 0
or, x2 – 6x – 3x + 18 = 0
or, x (x – 6) – 3 (x – 6) = 0
or, (x – 6) (x –3) = 0

Either, x–6 =0

i.e. x = 6

or, x – 3 = 0

i.e. x = 3

? When x = 6, then 9 – x = 9 – 6 = 3

When x = 3, then 9 – x = 9 – 3 = 6

Hence, the required numbers are 6 and 3.

Example 3: A two digit number is such that the product of the digits is 8. When 18 is
added to the number the digits interchange their places. Find the number.

Solution:

Let the digit at tens place be x and ones place be y.

Then, the number so formed = 10x + y

When the digits interchange their places, the number so formed = 10y + x.

Now, according to the first condition, xy = 8 ................... (i)

According to the second condition,

(10x + y) + 18 = 10y + x

or, 10x + y – 10y – x = – 18

or, 9x – 9y = – 18

or, x – y = –2

or, x = y – 2 ................... (ii)

Substituting the value of x from equation (ii) in equation (i), we get,

(y – 2) y = 8

or, y2 – 2y – 8 = 0

or, y2 – 4y + 2y – 8 = 0

or, y (y – 4) + 2 (y – 4) = 0

or, (y – 4) (y + 2) = 0

Either, y – 4 = 0, i.e. y = 4

or, y + 2 = 0, i.e. y = –2

When y = 4, then x × 4 = 8, i.e. x = 2

When y = – 2, then x × (–2) = 8 which is false.

Hence, the required number is 24.

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Example 4: The product of two consecutive even numbers is 168. Find the numbers.

Solution:

Let the required consecutive even numbers be x and x + 2 respectively.
According to the given condition,

x (x + 2) = 168

or, x2 + 2x – 168 = 0

or, x2 + 14x – 12x – 168 = 0

or, x (x + 14) – 12 (x + 14) = 0

or, (x + 14) (x – 12) = 0

Either, x + 14 = 0, i.e., x = – 14 (rejected)

or, x – 12 = 0, i.e., x = 12

So, the first even number = 12

The next consecutive even number = x + 2 = 12 + 2 = 14

Example 5: The difference of the ages of two brothers is 4 years and the product of
their ages is 221. Determine the ages of two brothers.

Solution:

Let the age of the younger brother be x years.
Then, the age of the older brother will be (x + 4) years.
According to the problem,

x u (x + 4) = 221
or, x2 + 4x – 221 = 0
or, x2 + 17x – 13x – 221 = 0
or, x (x + 17) – 13 (x + 17) = 0
or, (x + 17) (x – 13) = 0

Either, x + 17 = 0
i.e. x = – 17

But age cannot be negative. So, it is rejected.

or, x – 13 = 0

i.e. x = 13

Also, x + 4 = 13 + 4 = 17

Hence, the required ages of the brothers are 13 years and 17 years.

Example 6: The present ages of a father and his son are 35 years and 12 years
Solution: respectively. Find how many years ago the product of their ages was 210.

Let the required number of years is x years.
? The age of the father before x years = (35 – x) years

The age of the son before x years = (12 – x) years.

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According to the problem, = 210
(35 – x) (12 – x) = 210
=0
or, 420 – 35x – 12x + x2 =0
or, x2 – 47x + 210 =0
or, x2 – 42x – 5x + 210 =0
or, x (x – 42) – 5 (x – 42)
or, (x – 42) (x – 5)

Either, x – 42 = 0
i.e. x = 42
But, the value does not satisfy the condition. So, it is rejected.
or, x – 5 = 0
i.e. x = 5

So, 5 years ago, the product of their ages was 210.

Example 7: 6 years ago a man’s age was 6 times the age of his son. The product of the
present ages of the father and his son is 396. What are their present ages?

Solution:

Let the present age of the father be x years.
The present age of the son be y years.
6 years ago, the age of the father = (x – 6) years
6 years ago, the age of the son = (y – 6) years

According to the first condition,
(x – 6) = 6(y – 6)

or, x = 6y – 30 ………………….. (i)

According to the second condition,
x × y = 396 ……………………….. (ii)

Substituting the value of x from equation (i) in equation (ii), we get,

(6y – 30) × y = 396

6y2 – 30y – 396 = 0
or, 6(y2 – 5y – 66) = 0
or, y2 – 11y + 6y – 66 = 0
or, y(y – 11) + 6(y – 11) = 0

or, (y – 11) (y + 6) = 0
Either, y – 11 = 0, i.e. y = 11

Or, y + 6 = 0, i.e. y = -6, but age cannot be negative, so, rejected.

Now, substituting the value of y in equation (i), we get,

x = 6 × 11 – 30 = 36
So, the present ages of the father and his son are 36 years and 11 years respectively.

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Example 8: A room is 2 m longer than its breadth. If the area of the floor of the room is
120 m2, find its length and breadth.

Solution:

Let the breadth of the room be x m.
Then, the length of the room will be (x + 2) m

According to the problem, = 120
Area of the room = 120
=0
or, x u (x + 2) =0
or, x2 + 2x – 120 =0
or, x2 + 12x – 10x – 120 =0
or, x (x + 12) – 10 (x + 12)
or, (x + 12) (x – 10)

Either, x + 12 = 0
i.e. x = – 12

But the breadth of the room cannot be negative. So, it is rejected.
or, x – 10 = 10
i.e. x = 10

Also, x + 2 = 10 + 2 = 12

So, the length and the breadth of the room are 12 m and 10 m respectively.

Example 9: A rectangular meadow has an area of 28 sq. metres and perimeter 22 m.
Find the length and breadth of the meadow.

Solution:

Here,the perimeter of the meadow = 22 m
? 2 (l + b) = 22
or, l + b = 11
or, l = 11 – b ………………………. (i)

According to the problem,
The area of the meadow = 28 m2
? l u b = 28 ………………………………………. (ii)
Substituting the value of l from equation (i) in equation (ii),

(11 – b) b = 28

or, 11b – b2 = 28

or, b2 – 11b + 28 = 0

or, b2 – 7b – 4b + 28 = 0

or, b (b – 7) – 4 (b – 7) = 0

or, (b – 7) (b – 4) = 0
Either, (b – 7) = 0
i.e. b = 7
or, b – 4 = 0

i.e. b = 4

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Equations

? When, b = 7m, then l = (11 – 7) m = 4 m
or, when, b = 4 m, then l = (11 – 4) m = 7 m

So, the required length and breadth of the meadow are 7 m and 4 m or 4 m, and 7 m
respectively.

Example 10: The area of a rectangular room is 45 sq.m. If the length had been 3 m less
and the breadth 1 m more, it would have been a square. Find the length
and the breadth of the room.

Solution:

Let the length and the breadth of the rectangular room be l m and b m respectively.

From the first condition,

l × b = 45 ............... (i)

From the second condition,

l–3=b+1 (since the room becomes a square)

or, l = b + 4 ................... (ii)

Substituting the value of l from equation (ii) in equation (i) we get,

(b + 4) × b = 45

or, b2 + 4b – 45 = 0

or, b2 + 9b – 5b – 45 = 0

or, b (b + 9) – 5 (b + 9) = 0

or, (b + 9) (b – 5) = 0

Either, (b + 9) = 0, i.e. b = – 9 which is impossible and rejected.

or, (b – 5) = 0, i.e. b = 5

When b = 5, then l × b = 45, i.e. l = 9

So, the required length and breadth of the room are 9 m and 5 m respectively.

Example 11: The sides of a right-angled triangle containing the right-angle are less than
its hypotenuse by 2 cm and 4 cm respectively. Find the lengths of the sides
of the triangle.

Solution:

Let the hypotenuse, perpendicular and base of the right angled triangle be h, p, and b
respectively.

From the given condition,

p = h – 2 and b = h – 4
By using pythagoras theorem in the right angled triangle,

h2 = p2 + b2
or, h2 = (h – 2)2 + (h – 4)2
or, h2 = h2 – 4h + 4 + h2 – 8h + 16

or, h2 – 12h + 20 = 0

or, h2 – 10h – 2h + 20 = 0

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or, h (h – 10) – 2 (h – 10) = 0

or, (h – 10) (h – 2) = 0

Either, h – 10 = 0, i.e. h = 10 cm

or, h – 2 = 0, i.e. h = 2 cm which is impossible.

Now, when h = 10 cm, p = (10 – 2) cm = 8 cm

Also, when h = 10 cm, b = (10 – 4) cm = 6 cm

So, the required lengths of the sides of the triangle are 10 cm, 8 cm, and 6 cm respectively.

Example 12: Last week Mrs. Yadav bought some vegetables for Rs 150. This week the
rate of cost of vegetables increases by Rs 5 per kg and she can buy 1 kg less
vegetables for the same amount of money. By what percent is the rate of
cost of vegetables increased?

Solution:

Let the rate of cost of vegetables in last week be Rs x per kg.

Then, the quantity of vegetables purchased in last week = 150 kg
x

In this week, the increased rate of cost of vegetables = Rs (x + 5) per kg

Then, the quantity of vegetables purchased in this week = 150 kg
x+5
According to the question,

150 – 150 =1
x x+5

or, x2 + 5x – 750 = 0

or, (x + 30) (x – 25) = 0

or, x = –30 or 25

But, rate of cost cannot be negative.

So, the rate of cost of vegetables in last week was Rs 25 per kg.

Again, percentage of increased amount of rate of cost = Rs 5 × 100%
25

= 20%

Hence, the rate of cost of vegetables is increased by 20% in this week.

EXERCISE 12.2

General section

1. Solve: b) x2 – 5 = 0 c) 9x2 = 5
a) x2 – 9 = 0 5 5 4

d) x2 – 3x + 2 = 0 e) x2 – 5x + 6 = 0 f) x2 + 9x + 18 = 0
g) x2 – 2x – 8 = 0
h) x2 – 3x – 28 = 0 i) 5x2 + 8x – 21 = 0

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2. a) When 5 is subtracted from the square of a number the difference will be 11. Find
the numbers.

b) If 7 is subtracted from the half of the square of a natural number, the result is 25.
Find the number.

c) If 5 is added to the square of a number and the sum will be 30, find the numbers.

d) If 22 is added to the one-third of the square of a natural number the sum will be 49.
Find the number.

e) Divide 11 into two parts so that their product will be 24.
f) If the double of a natural number is equal to the square of the numbers, find the

number.
Creative section - A
3. a) If the sum of two numbers is 10 and their product is 24, find the numbers.

b) The product of two positive numbers is 56. If their difference is 10, find the
numbers.

c) A number exceeds another number by 3 and their product is 40. Find the numbers.
d) If the product of two consecutive even numbers is 224, find the numbers.
e) The product of two consecutive odd numbers is 143. Find the numbers.
f) The sum of two numbers is 21 and the sum of their squares is 261. Find the

numbers.
g) When a natural number is decreased by 20, the result is 96 times the reciprocal of

the number. Find the number.
h) If the sum of the square of two consecutive natural even numbers is 52, find the

numbers.
i) Find a number which is when added to its square the sum will be 72.
4. a) In a two-digit number, the product of the digits is 18 and the sum is 9. Find the

number.
b) In a two-digit number, the digit at tens place is 1 more than the digit at unit place.

If the product of the digits is 20, find the number.
c) A two-digit number is such that the product of the digits is 24. When 45 is added

to the number, the digits interchange their places. Find the number.
d) A number consists of two-digits whose product is 18. If 63 is subtracted from the

number, the digits are interchanged. Find the number.
e) A two-digit number is four times the sum and three times the product of its digits.

Find the number.
f) A number of two-digits is equal to four times the sum of the digits. If the product

of the digits is 8, find the number.
g) The product of digits in a two-digit number is 18. The number formed by

interchanging the digits of the number will be 27 more than the original number.
Find the original number.
h) In a number of two digits, the square of the sum of its digits is 100. If 2 is subtracted
from three times of the number, the digits are reversed. Find the number.
i) The square of the sum of the digits of a two-digit number is 64. If 3 is added to four
times the number, the digits are reversed. Find the number.

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5. a) The difference of the present ages of two sisters is 5 years. If the product of their
ages is 204, find the ages of the two sisters.

b) The sum of the present ages of a brother and his younger sister is 16 years. If the
product of their ages is 63, find their ages.

c) The sum of the present ages of elder and younger brothers is 46 years and the
product of their ages is 504. Find their present ages.

d) Sunayana is 7 years younger than Bishwant. If the product of their ages is 144, find
their present ages.

e) The difference of the ages of two brothers is 4 years. 5 years ago, if the product of
their ages was 96, find their present ages.

f) The difference of the ages of two sisters is 5 years. After 4 years, if the product of
their ages becomes 104, find their present ages.

g) The difference of the ages of a sister and her younger brother is 6 years. The
numerical value of the product of their ages is equal 4 times the sum of their ages.
Find the present ages of the sister and her brother.

h) The daughter's age is 4 years more than the son's age. 2 years ago, the product of
their ages was 32. Find the present ages.

i) The product of the present ages of two sisters is 150. 5 years ago, the elder sister
was twice as old as her younger sister. Find their present ages.

j) One year hence, a father's age will be 5 times the age of his son. The product of the
present ages of the father and his son is 145. Find their present ages.

6. a) The present ages of two brothers are 15 years and 22 years respectively. After how
many years will the product of their ages be 408?

b) The present ages of a mother and her daughter are 30 years and 14 years respectively.
Find how many years ago the product of their ages was 192.,

7. a) The length of a room is 2 m longer than its breadth. If the area of the room is
63 sq.m, find the length and the breadth of the room.

b) The breadth of a rectangular garden is 4 m shorter than its length and the area of
the garden is 192 sq.m. Find the perimeter of the garden.

c) A rectangular piece of land is 40 m long and 25 m wide. A path of uniform width
and 426 m2 area surrounds the land from outside. Find the width of the path.

d) If the area of a rectangular courtyard is 63 sq.m and its perimeter is 32 m, find the
length and the breadth of the courtyard.

e) A rectangular meadow has an area of 247 sq.m and its perimeter is 64 m. Find the
length and the breadth of the meadow.

f) The height of a tank is 2 m and its length is 3 m more than its breadth. If 80 cu.m
of water can be stored in the tank, find the length and the breadth of the tank.

8. a) If the sides of a right-angled triangle are (x – 2) cm, x cm and (x + 2) cm, find the
length of each of its side.

b) The perimeter of a right-angled triangle is 12 cm and its hypotenuse is 5 cm. Find
the other two sides of the triangle.

c) The area of a right-angled triangle is 24 sq. cm. If the perpendicular of the triangle
is 2 cm shorter than its base, find two sides of the triangle.

Vedanta Excel in Mathematics - Book 10 186 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

d) The sides of a right-angled triangle containing the right-angle are less than its
hypotenuse by 5 cm and 10 cm respectively. Find the length of the sides of the
triangle.

Creative section - B
9. a) The area of a rectangular field is 720 sq. m. and perimeter is 108 m. By what percent

is the longer side of the field to be decreased to make it a square? Why?
b) The area of a rectangular conference hall is 300 m2. If the length were decreased by

5 m and the breadth increased by 5 m, the area would be unaltered. Find the length
of the room.
10. a) A piece of cloth costs Rs 800. If the piece was 5 m longer and the rate of cost of cloth
per metre was Rs 8 less, the cost of the piece would have remained unchanged.
How long is the piece and what is the rate of cost per meter?
b) Mrs. Magar bought a certain kilograms of vegetables for Rs 120 last week. This
week, the rate of cost of vegetables decreases by Rs 20 per kg and she can buy 1 kg
more vegetables for the same amount of money. By what percentage is the rate of
cost of vegetables decreased?
11. a) Some students planned a picnic. The budget for the food was Rs 2,500. As five of
them unable to join the picnic, the cost of the food for each students increased by
Rs 25. Find how many students went for the picnic?
b) A new year party bill for a number of people is Rs 4,800. If there were 4 people
more, the bill each person had to pay, would have reduce by Rs 200. Find the
number of people attending the party.

Objective Questions
Tick the correct alternatives.

1. If x is a multiple of y, what is the HCF of x and y?

(A) 1 (B) x (C) y (D) xy

2. The L.C.M. of x2y + xy2 and x3y – xy3 is

(A) xy (B) x (x2 – y2) (C) y (x2 – y2) (D) xy (x2 – y2)

3. If H and L be the HCF and LCM of two expressions A and B respectively, then the
relation of A, B, H and L is:

(A) A×H = B×L (B) A×L = B×H (C) A + B = H + L (D) A×B = H×L

4. What is the order of the surd 2 3 4 ?

(A) 2 (B) 3 (C) 4 (D) 12

5. What is the rationalizing factor 3 2 + 5?

(A) 2 (B) 2 – 5 (C) 3 2 – 5 (D) 3 2 + 5

6. If x + 5 = 5 – x , what is the value of xx?

(A) 1 (B) 4 (C) 16 (D) 256

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7. What is the simplest form of x + xy y y2 ?
xy – y2 –

(A) x + y (B) 1 (C) x–y (D) x+y
xy xy

8. What is the value of (1 – x m – n) – 1 + (1 – x n – m ) – 1?

(A) m (B) n (C) x (D) 1
(D) xa – 1
9. The simplified form of xa+3 – xa is ... (D) ± 4
xa+1 – xa

(A) x – 1 (B) x + 1 (C) x2 + x + 1

10. The value of x in the equation 2x + 1 = 881 is
2x

(A) ± 1 (B) ± 2 (C) ± 3

11. If 5 – 3 = 0.04, what is the value of 5 ?

(A) 2 (B) 3 (C) 4 (D) 5

12. If a = bx, b = cy and c = az, what is the value of xyz?

(A) -1 (B) 0 (C) 1 (D) 2

13. If a number is decreased by its 75% then the remainder is 75, the number is

(A) 200 (B) 300 (C) 400 (D) 500

14. If the reciprocal of a number is added to the number, it becomes 2. What is the number

(A) 0 (B) 1 (C) 2 (D) 3

15. If the product of ages of a boy 3 years ago and 3 years later is equal to the square of his
age a year ago, then his present age is

(A) 9 years (B) 6 years (C) 5 years (D) 4 years

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Unit Geometry - Area of Triangles
and Quadrilaterals
13

13.1 Relation between the area of triangles and quadrilaterals

In this chapter, we will discuss the relation of area of triangles and quadrilaterals
which are formed between parallel lines. The relation of area of triangles and
quadrilateral standing on the same base or on the equal base and between the same
parallel lines are verified by using the following various theorems.

Theorem 1

Diagonal of a parallelogram bisect the parallelogram.

or

The area of each of the triangles formed by drawing a diagonal of a parallelogram is half

of the area of the parallelogram. DC

Given: ABCD is a parallelogram in which AC is its diagonal.

To prove: ' ABC = 1 ABCD A
Proof 2
B

Statements Reasons
1. In ∆ ABC and ∆ ACD
1.

(i) AB = DC (S) (i) Sides of the parallelogram

(ii) BC = AD (S) (ii) Same as (i)

(iii) AC = AC (S) (iii) Common side

(iv) ? ∆ ABC # ∆ ACD (iv) S.S.S. axiom
2. Area of ∆ ABC = Area of ∆ ACD 2. Area of congruent triangles

3. ∆ ABC + ∆ ACD = ABCD 3. Whole part axiom

4. 2 ∆ ABC = ABCD 4. From statements (2) and (3)

5. Area of ∆ ABC = 1 area of ABCD 5. From statement (4)
2

Proved

Theorem 2
Parallelograms on the same base and between the same parallels are equal in area.

Experimental verification

Step 1: Three parallelograms ABCD with different sizes of bases and heights are drawn.

Another parallelogram BCEF on the same base BC and between the same parallels

BC and AE is drawn in each figure.

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Geometry– Area of Triangles and Quadrilateral

Step 2: DM A BC is drawn in each figure, where DM is the height (altitude) of the

parallelograms. A D F E A FD E

A FD E

B MC BM C B MC
(i) (ii) (iii)

Step 3: The base BC and the height DM are measured and the areas of ABCD and BCEF

are calculated. Result

Fig. Base Height Area of ABCD Area of BCEF ABCD = BCEF
(BC × DM)
BC DM (BC × DM)
(i)

(ii) ABCD = BCEF

(iii) ABCD = BCEF

Conclusion: Parallelograms on the same base and between the same parallels are equal in

area.

Theoretical proof A FD E

Given: Parallelograms ABCD and BCEF are on the same base BC
and between the same parallels BC and AE.

To prove: ABCD = BCEF BC

Proof

Statements Reasons
1. In ∆s ABF and DCE
1.

(i) ‘ AFB = ‘ DEC (A) (i) BF // CE and corresponding angles

(ii) ‘ BAF = ‘ CDE (A) (ii) BA // CD and corresponding angles

(iii) AB = DC (S) (iii) Opposite sides of a parallelogram

(iv) ∆ ABF # ∆ DCE (iv) A. A. S. axiom
2. ∆ ABF = ∆ DCE 2. Area of congruent triangles
3. ∆ ABF + trapezium BCDF 3. Adding the same trapezium BCDF to

= ∆ DCE + trapezium BCDF both sides
4. ABCD = BCEF
4. Whole part axiom

Corollary A DE Proved

Parallelograms on the equal bases and between the same H
parallels are equal in area.

In the figure, AH // BG and BC = FG B C FG
So, area of ABCD = area of EFGH

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Geometry – Area of Triangles and Quadrilateral

Theorem 3

The area of a triangle is equal to half of the area of a parallelogram standing on the same
base and between the same parallels.

Experimental verification
Step 1: Three different figures with a triangle BCD and a parallelogram ABCE on the same

base BC and between the same parallels AD and BC are drawn.

Step 2: AM A BC is drawn in each figure. AM is the height of ∆ BCD and ABCE.

A ED DA E AD E

MB C BM C BM C
(i) (ii)
(iii)

Step 3: The base BC and the height AM in each figure are measured. The areas of ∆ BCD and
ABCE are calculated.

Fig. Base BC Height AM Area of ∆ BCD Area of ABCE Result
(i) 1
(ii) ∆ BCD = 2 ABCE
(iii)
∆ BCD = 1 ABCE
2

∆ BCD = 1 ABCE
2

Conclusion: The area of triangle is equal to half of the area of a parallelogram on the same

base and between the same parallels. A FE D
Theoretical proof

Given: ∆ BCD and parallelogram ABCE are on the base BC

and between the same parallels AD and BC.

To prove: ∆ BCD = 1 ABCE B C
2
Construction: BF // CD is drawn. BF meets AD at F.

Proof

Statements Reasons

1. BCDF is a parallelogram 1. BC // FD and BF // CD

2. ∆ BCD = 1 BCDF 2. Diagonal bisects a parallelogram.
2

3. BCDF = ABCE 3. They are on the same base and between the same

parallels.

4. ∆ BCD = 1 ABCE 4. From statements (2) and (3)
2

Proved

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Geometry– Area of Triangles and Quadrilateral

Alternative process A ED
Construction: EF A BC is drawn.

EF is the height of the parallelogram ABCE and ∆ BCD.

Proof B FC

Statements Reasons
Area of a parallelogram is base × height
1. ABCE = BC × EF 1.
2.
2. ∆ BCD = 1 BC × EF Area of a triangle is 1 base × height
2 2

3. ∆ BCD = 1 BCEA 3. From statements (1) and (2)
2

Proved

Corollary

If a triangle and a parallelogram stand on equal bases and A DG

between the same parallels, the area of the triangle is equal to

one half of the area of the parallelogram. In the figure, AG // BF

and BC = EF

So, area of ' ABC = 1 area of DEFG. B CE F
2
Theorem 4

Triangles on the same base and between the same parallels are equal in area.

Experimental verification

Step 1: Three pairs of triangles ABC and BCD are drawn on the same base BC and between
the same parallels AD and BC.

Step 2: AM A BC and DN A BC are drawn.

AD A D AD

BM NC MB CN B CN
M
(i) (ii) (iii)

Step 3: The heights AM and DN and the base BC of each figure are measured. The areas of
∆ ABC and ∆ BCD are calculated.

Fig. Base Height Area of Base Height Area of Result
BC DN ∆ BCD
BC AM ∆ ABC ∆ ABC = ∆ BCD
(i) ∆ ABC = ∆ BCD
(ii) ∆ ABC = ∆ BCD
(iii)

Conclusion: Triangles on the same base and between the same parallels are equal in area.

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Geometry – Area of Triangles and Quadrilateral

Theoretical proof A E D

Given: ∆ ABC and ∆ BCD are on the same base BC and
between the same parallels BC and AD.

To prove: ∆ ABC = ∆ BCD

Construction: BE // CD is drawn. BE meets AD at E. BC

Proof

Statements Reasons

1. BCDE is a parallelogram 1. Opposite sides are parallel

2. ∆ ABC = 1 BCDE 2. They are on the same base and between the same
2 parallels.

3. ∆ BCD = 1 BCDE 3. Diagonal bisects a parallelogram.
2

4. ∆ ABC = ∆ BCD 4. From statements (2) and (3)

Alternative process A D Proved

Construction: AM A BC is drawn.

AM is the height of ∆ ABC and ∆ BCD.

MB C

Statements Reasons

1. Area of ∆ ABC = 1 BC × AM 1. Area of a triangle is 1 (base × height)
2 2

2. Area of ∆ BCD = 1 BC × AM 2. Area of a triangle is 1 (base × height)
2 2

3. ∆ ABC = ∆ BCD 3. From statements (1) and (2)

Corollary A Proved

Triangles on equal bases and between the same parallels are D
equal in area.
F
In the figure, AD // BF and BC = EF

So, area of ' ABC = area of ' DEF B CE

Worked-out examples

S R
T

Example 1: In the figure, the area of ' PQT is 28 cm2. Find the
area of ' PRS.
Solution:
(i) Area of P Q

PQRS = 2 area of ' PQT [' PQT and PQRS are standing on
= 2 × 28 cm2 the same base and between the same
= 56 cm2 parallels.]

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Geometry– Area of Triangles and Quadrilateral

(ii) Area of ' PRS = 1 Area of PQRS [Diagonal PR bisects PQRS]
2
1
= 2 × 56 cm2 = 28 cm2

So, the required area of ' PRS is 28 cm2. D E C
Example 2: In the given figure, E is the mid point of DC. If

the area of ABCD is 24 cm2, find the area of the

quadrilateral ABCE.

Solution: = 1 Area of ABCD = 1 × 24 cm2 = 12 cm2 A B
(i) Area of ' ABE 2 2

(ii) Area of ' BCE = Area of ' AED [Both the triangles are standing on the equal base

1 and between the same parallels.]
2
? Area of ' AED = × 12 cm2 = 6 cm2 [Area of (' BCE + ' AED) = 12 cm2]

(iii) Area of quadrilateral ABCE = Area of (' ABE + ' BCE)
= 12 cm2 + 6 cm2 = 18 cm2

So, the required area of the quadrilateral ABCE is 18 cm2.

Example 3: In the figure, ABCD is a rhombus in A DE
which the diagonals AC = 20 cm and
BD = 28 cm. If AD is produced to E, find the
area of ' BCE.

Solution: B C
S
(i) Area of rhombus ABCD = 1 × AC × BD T
2
1
= 2 × 20 × 28 cm2 = 280 cm2

(ii) Area of ' BCE = 1 area of ABCD
2
1
= 2 × 280 cm2 = 140 cm2

So, the required area of ' BCE is 140 cm2. P

Example 4: In the given figure, PQRS is a square with its Q
Solution: diagonal PR = 4 3 cm. If PS is produced to T, find
the area of ' TQR.

R

(i) Area of the square PQRS = 1 × PR2
2

= 1 × (4 3 cm)2 = 24 cm2
2 =
(ii) Area of ' TQR = 1 area of PQRS 1 × 24 cm2 = 12 cm2
2 2

So, the required area of ' TQR is 12cm2.

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Geometry – Area of Triangles and Quadrilateral

Example 5: In the given ∆ PQR, S and T are the mid–points of P
PQ and PR respectively. Prove that ∆ RPS = ∆ QPT.

Solution: In ∆ PQR, S is the mid–point of PQ and T is the S T
Given: R
mid–point of PR. O
To prove: ∆ RPS = ∆ QPT Q

Construction: S and T are joined.

Proof

Statements Reasons
ST joins the mid–points of two
1. ST // QR 1. sides of ∆ PQR
2. They are on the same base QR
2. ∆ QRT = ∆ QRS 3. and between QR // ST.
The same ∆ QRO is subtracted
3. ∆ QRT – ∆ QRO = ∆ QRS – ∆ QRO from both the sides of the
statement (2)
4. ∆ ORT = ∆ OQS 4. Remaining part of whole
The same quadrilateral PSOT is
5. ∆ ORT + quad. PSOT = ∆ OQS + quad.PSOT 5. added to both the sides of the
statement (4)
6. ∆ RPS = ∆ QPT 6. Whole part axiom

Example 6: In the given figure, O be any point on the diagonal D F Proved
BD of ABCD. If EF // AD and GH // CD, EF and GH
C
pass through O, prove that AEOH = CGOF. HO G

Solution: B

Given: ABCD is a parallelogram. O is any point on the diagonal A E
BD, EF // AD, GH // CD.

To prove: AEOH = CGOF.

Proof

Statements Reasons
1. EBGO is a parallelogram 1. Opposite sides are parallel.
2. ∆ EBO = ∆ BGO 2. Diagonal BO bisects EBGO.
3. DFOH is a parallelogram 3. Opposite sides are parallel.
4. ∆ HOD = ∆ OFD 4. Diagonal OD bisects DFOH.
5. ∆ ABD = ∆ BCD 5. Diagonal BD bisects ABCD.
6. ∆ ABD – ∆ EBO – ∆ HOD 6. The same triangles EBO and HOD are

= ∆ BCD – ∆ EBO – ∆ HOD subtracted from both the sides of (5).
7. ∆ ABD – ∆ EBO – ∆ HOD
7. From statements (2), (4) and (6)
= ∆ BCD – ∆ BGO – ∆ OFD
8. AEOH = CGOF 8. Remaining part of a whole

Proved

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Geometry– Area of Triangles and Quadrilateral

Example 7: In the given figure, PARM is a parallelogram. B is M R

any point on diagonal AM. Prove that 'PMB and X

' RMB are equal in area. B

Solution : PA

Given: PARM is a parallelogram. AM is its diagonal and B is any point on the diagonal.

To prove: Area of 'PMB = Area of 'RMB

Construction: P and R are joined. Diagonals AM and PR bisect each other at X.

Proof

Statements Reasons

1 X is the mid-point of diagonals AM and PR 1. From construction and given

2. 'BPX = 'BRX 2. BX is the median of ' BPR

3. 'PMX = 'RMX 3. MX is the median of ' PMR
4. 'BPX + 'PMX = 'BRX + 'RMX 4. Adding the statements (2) and (3)

5. 'PMB = 'RMB 5. Whole parts axiom
6. ? 'PMB and 'RMB are equal in area 6. From statement (5)

Example 8: In the given figure, D is the mid-point of side A Proved

BC of ' ABC, E is the mid-point of AD, F is the C
mid-point of AB and G is any point of BD. Prove that F E C

'ABC = 8'EFG. B GD
Solution :

Given: D, E and Fare the mid-points of BC, AD and AB respectively. A

G is any point of BD. FE

To prove: 'ABC = 8 'EFG

Construction: D and F are joined. B GD
Proof

Statements Reasons

1 EF//BD 1. E and F are the mid-points of the sides

AD and AB of ' ABD.

2. 'ABD = 1 'ABC 2. Median AD bisects ' ABC.
2

3. 'EFD = 'EFG 3. Both of them are standing on the same
base and between the same parallels.

4. 'ADF = 1 'ABD 4. Median DF bisects ' ABD.
2 5. Median FE bisects 'ADF

5. 'EFD = 1 'ADF
2

6. 'EFG = 1 × 1 'ABD 6. From statements (3), (4) and (5)
2 2

7. ' EFG = 21×12×12 ' ABC 7. From statements (1) and (6)

' EFG = 1 ' ABC. So, ' ABC = 8 ' EFG
8

Proved

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Geometry – Area of Triangles and Quadrilateral

EXERCISE 13.1

General section

1. Find the area of the shaded regions.

a) D E C b) S R c) 7.4 cm
D

12.5cm
10 cm

12 cm

A B PT Q A B
16.5 cm R T
18 cm 9.6 cm
S

2. a) In the figure, parallelogram PQSR and ' PQT are standing
on the same base and between, the same parallel lines. The
area of ' PQT is 50 cm2. Find the area of ' PRS.

P Q

AD

b) In the figure, the area of trapezium ABCD is 100 sq.cm

and the area of ' ADC is 40 sq.cm. Find the area of

' DEC. B EC

P W X
Z Y
c) In the figure, WXYZ is a rhombus in which XW
is produced to the point P. If WY = 10 cm and
XZ = 9 cm, find the area of 'PYZ.

P ST

d) In the given figure, PQRS is a square whose each side is
4 cm. Find the area of ' QRT.

Q R
D
A
E
e) In the figure, AE // BC, square ABCD and ' EBC are
standing on the same base BC and between the same C
parallels. If AC = 6 cm, find the area of ' EBC. B

B F

A

f) 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 mid
point of DC, find the area of ' EFC.

DE C

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Geometry– Area of Triangles and Quadrilateral

D C

g) In the adjoining figure, AF // DC, ED // FC and ABCD is a EB
square. If AC = 5 2 cm, find the area of the parallelogram E
DEFC.

A F
D
A

h) In the given figure, if ABCE is a rectangle, find the area 8 cm
of ' ADE.
C
B 10 cm

P Q
M
i) In the adjoining figure, PS = 5 cm and SM = 8 cm.
Calculate the area of ' PQN. SR

N

Q

j) PQRS is a quadrilateral in which PR = 10 cm, perpendiculars P

from S and Q PR are 3.4 cm and 4.6cm respectively. Calculate MN
the area of the quadrilateral.

RS
Q

k) Find the area of the quadrilateral PQRS R A P
given in the adjoining figure in which B
3RB = 2PA = QS = 12 cm.

l) Calculate the area of the following trapeziums. S

(i) A 18 cm B (ii) 17 cm Q (iii) E 12 cm B

P A

13 cm 26 cm 13 cm

E D 23 cm C S 7 cm R D C
28 cm

m) In the given figure, ABCD is a trapezium. If AB = 10 cm, 10 cm A 16 cm D
BC = 22 cm, AD = 16 cm, AD // BC and DC A BC, calculate the B 22 cm C
area of ' ADC.

n) The areas of two parallelograms are equal and their altitudes are 6 cm and 9
cm. If the base of the first parallelogram is 12 cm, find the base of the second
parallelogram.

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