CBSE CLASS – X
(30/1, 2, 3)
MATH SOLUTIONS
Series LRH
(ALL INDIA)
SECTION – C(Set-1, 2 and 3)
This solution as per the order of questions given in set – 1, additional questions that appeared in set – 2 and
Set – 3 are given at the end.
16.
Prove that
16.
Proof: root 3 is irrational
3 is an irrational number.
Suppose,
3 is rational. This means, we can write 3 as
⇒
p
(q ≠ 0)
q
3=
p
q
If p and q have common factors, then we can reduce it into the lowest terms, say
a
a
⇒ 3= .
b
b
(Here remember that ‘a’ and ‘b’ have no common factor other than one, because
a
is already in its
b
lowest terms).
Multiplying the equation by ’b’ we get
b
3 =a
Squaring both sides, we get
b2 × 3 = a2
This means three times b2 is equal to a2. In other words ‘3’ is a factor of a2.
Now, according to the theorem, if 3 divides a2 it will divide ‘a’ also.
If ‘3’ divides ‘a’, then ‘a’ should be a multiple of 3. i.e. a = 3 × some number
Say, a = 3c
We already have an equation 3b2 = a2
Substituting the value of ‘a’ in this equation, we get
3b2 = (3c)2 = 9c2
⇒ b2 = 3c2
2
This means ‘3’ divides ‘b ’. This again means ‘3’ divides ‘b’
Therefore ‘3’ divides both ‘a’ and ‘b’. This contradicts the fact that ‘a’ and ‘b’ have no common factor
other than one.
This contradiction is because of our wrong assumption that
So, we conclude that
17.
3 is rational.
3 is irrational.
Solve the following pair of linear equations for x and y:
b
a
x + y = a2 + b2
a
b
x + y = 2ab
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OR
The sum of the numerator and the denominator of a fraction is 4 more than twice the numerator. If 3 is
added to each of the numerator and denominator, their ratio becomes 2 : 3. Find the fraction.
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18.
In A.P., the sum of its first ten terms is – 80 and the sum of its next ten terms is – 280. Find the A.P.
19.
In figure 4, ABC is an isosceles triangle in which AB = AC. E is a point on the side CB produced, such
that FE ⊥ AC. If AD ⊥ CB, prove that
AB × EF = AD × EC
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20.
Prove the following:
(1 + cot A – cosec A) (1 + tan A + sec A) = 2
OR
Prove the following:
Sin A (1 + tan A) + cos A (1 + cot A) = sec A + cosec A
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21.
Construct a triangle ABC in which AB = 8 cm, BC = 10 cm and AC = 6 cm. Then construct another
4
of the corresponding sides of ∆ ABC.
triangle whose sides are
5
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22.
Point P divides the lien segment joining the points A(– 1, 3) and B(9, 8) such that
AP
k
= . If P lies on
PB
1
the line x – y + 2 = 0, find the value of k.
23.
If the points (p, q); (m, n) and (p – m, q – n) are collinear, show that pn = qm.
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24.
The rain-water collected on the roof of a building, of dimensions 22 m × 20 m, is drained into a
cylindrical vessel having base diameter 2 m and height 3.5 m. If the vessel is full up to the brim, find the
height of rain-water on the roof. [Use π =
22
]
7
OR
In figure 5, AB and CD are two perpendicular diameters of a circle with centre O. If OA = 7 cm, find the
area of the shaded region. [Use π =
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22
]
7
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25.
A bag contains cards which are numbered from 2 to 90. A card is drawn at random from the bag. Find
the probability that it bears
(i)
a two digit number,
(ii)
a number which is a perfect square.
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SET – 2
16.
Prove that
16.
Proof: root 5 is irrational
Suppose,
⇒
5 is an irrational number.
5 is rational. This means, we can write 5 as
5=
p
q
p
(q ≠ 0)
q
If p and q have common factors, then we can reduce it into the lowest terms, say
a
a
⇒ 5= .
b
b
(Here remember that ‘a’ and ‘b’ have no common factor other than one, because
a
is already in its
b
lowest terms).
Multiplying the equation by ’b’ we get
b
5 =a
Squaring both sides, we get
b2 × 5 = a2
This means five times b2 is equal to a2. In other words ‘5’ is a factor of a2.
Now, according to the theorem, if 5 divides a2 it will divide ‘a’ also.
If ‘5’ divides ‘a’, then ‘a’ should be a multiple of 5. i.e. a = 5 × some number
Say, a = 5c
We already have an equation 5b2 = a2
Substituting the value of ‘a’ in this equation, we get
5b2 = (5c)2 = 25c2
⇒ b2 = 5c2
This means ‘5’ divides ‘b2’. This again means ‘5’ divides ‘b’
Therefore ‘5’ divides both ‘a’ and ‘b’. This contradicts the fact that ‘a’ and ‘b’ have no common factor
other than one.
This contradiction is because of our wrong assumption that
So, we conclude that
21.
5 is rational.
5 is irrational.
Construct a triangle ABC in which BC = 8 cm, ∠ B = 60o and ∠ C = 45o. Then construct another triangle
whose sides are
3
of the corresponding sides of ∆ ABC.
4
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23.
Find the value of k, if the points A(7, – 2), B(5, 1) and C(3, 2k) are collinear.
SET – 3
2 is an irrational number.
16.
Prove that
16.
Proof: root 2 is irrational
Suppose, 2 is rational. This means, we can write 2 as
⇒
2=
p
q
p
(q ≠ 0)
q
If p and q have common factors, then we can reduce it into the lowest terms, say
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a
a
⇒ 2= .
b
b
Page : 10
(Here remember that ‘a’ and ‘b’ have no common factor other than one, because
a
is already in its
b
lowest terms).
Multiplying the equation by ’b’ we get
b 2 =a
Squaring both sides, we get
b2 × 2 = a2
This means two times b2 is equal to a2. In other words ‘2’ is a factor of a2.
Now, according to the theorem, if 2 divides a2 it will divide ‘a’ also.
If ‘2’ divides ‘a’, then ‘a’ should be a multiple of 2. i.e. a = 2 × some number
Say, a = 2c
We already have an equation 2b2 = a2
Substituting the value of ‘a’ in this equation, we get
2b2 = (2c)2 = 4c2
⇒ b2 = 2c2
This means ‘2’ divides ‘b2’. This again means ‘2’ divides ‘b’
Therefore ‘2’ divides both ‘a’ and ‘b’. This contradicts the fact that ‘a’ and ‘b’ have no common factor
other than one.
This contradiction is because of our wrong assumption that 2 is rational.
So, we conclude that 2 is irrational.
21.
Construct a triangle ABC in which BC = 9 cm, ∠ B = 60o and AB = 6 cm. Then construct another
triangle whose sides are
2
of the corresponding sides of ∆ ABC.
3
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23.
Find the value of k, if the points A(8, 1), B(3, – 4) and C(2, k) are collinear.
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