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RBSE 10th Maths Model Paper Set 9 with Answers in English

March 23, 2022 by Fazal Leave a Comment

Students must start practicing the questions from RBSE 10th Maths Model Papers Set 9 with Answers in English Medium provided here.

RBSE Class 10 Maths Model Paper Set 9 with Answers in English

Time: 2:45 Hours
Maximum Marks: 80

General Instructions:

  • All the questions are compulsory.
  • Write the answer of each question in the given answer book only.
  • For questions having more than one part the answers to their parts are to be written together in continuity.
  • Candidate must write first his / her Roll No. on the question paper compulsorily.
  • Question numbers 17 to 23 have internal choices.
  • The marks weightage of the questions are as follows:
Section Number of Questions Total Weightage Marks for each question
Section A 1 (i to xii), 2(i to vi), 3(i to xii) = 30 30 1
Section B 4 to 16 = 13 26 2
Section C 17 to 20 = 4 12 3
Section D 21 to 23 = 3 12 4

Section – A

Question 1.
Multiple Choice Questions :
(i) If HCF (a, 8) = 4, LCM (a, 8) = 24, then a is (1)
(a) 8
(b) 10
(c) 12
(d) 14
Answer:
(c) 12

(ii) The sum and product respectively of zeroes of the polynomial x2 – 4x + 3 are (1)
(a) 3, 3
(b) 4, 3
(c) – 4, 3
(d) \(\frac{4}{3}\), 1
Answer:
(b) 4, 3

(iii) If (6, k) is a solution of the equation 3x + y – 22 = 0, then the value of k is 1
(a) -4
(b) 4
(c) 3
(d) – 3
Answer:
(b) 4

RBSE 10th Maths Model Paper Set 9 with Answers in English

(iv) If one root of the equation ax2 + bx + c = 0 is three times of the other, then k is equal to …………………… (1)
(a) 3
(b) lg
(c) 16
(d) 3
Answer:
(b) lg

(v) The sum of first 20 odd numbers is (1)
(a) 400
(b) 200
(c) 500
(d) 800
Answer:
(d) 800

(vi) To divide a line segment AB in the ratio 2 : 5, first a ray AX is drawn so that ZBAX is an acute angle and then at equal distances points are marked on the ray AX such that the minimum number of these points is : (1)
(a) 2
(b) 5
(c) 4
(d) 7
Answer:
(d) 7

(vii) Which of the following point is equidistant from (3, 2) and (- 5, – 2) ? (1)
(a) (0, -2)
(b) (0, 1)
(c) (0, 2)
(d) (2, 0)
Answer:
(a) (0, -2)

(viii) 9 sec2θ – 9 tan2 θ is equal to : 1
(a) 9
(b) 1
(c) 0
(d) 8
Answer:
(a) 9

(ix) If sin θ + sin2 θ = 1, then cos2 θ + cos4 θ is equal to : 1
(a) 0
(b) – 1
(c) 1
(d) 2
Answer:
(c) 1

(x) The statistical data are of two types. These types are : (1)
(a) technical data and presentation data
(b) primary data , and secondary data
(c) primary data and personal data
(d) none of these
Answer:
(b) primary data , and secondary data

(xi) The number of times a particular item occurs in a class interval is called its : (1)
(a) mean
(b) frequency
(c) cumulative frequency
(d) none of these
Answer:
(c) cumulative frequency

RBSE 10th Maths Model Paper Set 9 with Answers in English

(xii) A card is drawn from a packet of 100 cards numbered 1 to 100. The probability of drawing a number which is perfect square is : (1)
(a) \(\frac{1}{10}\)
(b) \(\frac{1}{100}\)
(c) \(\frac{1}{50}\)
(d) \(\frac{9}{100}\)
Answer:
(a) \(\frac{1}{10}\)

Question 2.
Fill in the blanks :
(i) The value of k for which x – 5x = 10 and 3x – 3ky = – 5 have no solution, will be ______________ (1)
Answer:
5

(ii) If a + 1, 2a + 1, 4a – 1 are in AP, then value of a is ______________ . (1)
Answer:
2

(iii) Given a triangle with side AB = 8 cm. To get a line segment AB’ = \(\frac{3}{4}\) of AB. It is required to divide the line segment AB in the ratio ______________ (1)
Answer:
3 : 1

(iv) If the points (- 1, 3), (2, P) and (5, -1) are collinear, then the value of P will be ______________ (1)
Answer:
1

(v) The equation sin2x = \(\frac{a^{2}+b^{2}}{2 a b}\) is possible if ______________ (1)
Answer:
a = b

(vi) The points A (0, 0), B(3, √3) and C(3, -√3) are vertices of a triangle, then triangle will be ______________ (1)
Answer:
equilateral

RBSE 10th Maths Model Paper Set 9 with Answers in English

Question 3.
Very Short Answer Type Questions :
(i) HCF and LCM of two numbers are 9 and 360 respectively. If one number is 45, find the other number. (1)
Answer:
HCM = 9, LCM = 360 and one number = 45
∵ other number = \(\frac{\mathrm{LCM} \times \mathrm{HCF}}{\text { One number }}\)
∴ Required number = \(\frac{360 \times 9}{45}\) = 8 × 9
= 75

(ii) Write division algorithm for polynomials.
Answer:
If p (x) and g (x) are any two polynomials with g (x) ≠ 0, then p (x) = g (x) × q (x) + r (x); where r (x) ≠ 0 or degree of r(x) < degree of g (x).

(iii) Check whether (x – 3) is a factor of polynomial f(x) = x3 + x2 – 17x + 15 or not. (1)
Answer:
f (x) = x3 + x2 – 17x + 15
f(3) = (3)3 + (3)2 – 17 (3) + 15
= 27 + 9 – 51 + 15 = 0

(iv) For which values of P, will the lines represented by the following pair of linear equations be parallel 3x – y – 5 = 0 and 6x – 2y – P = 0. (1)
Answer:
The condition for parallel lines is
RBSE 10th Maths Model Paper Set 9 with Answers in English 1
So, P can have any real values other than 10.

(v) If x = 3 is one root of quadratic equation x2 – 2kx – 6 = 0, then find the value of k. (1)
Answer:
Since, x = 3 is one root of the quadratic equation x2 – 2kx – 6 = 0.
∴ (3)2 – 2k × 3 – 6 = 0
⇒ 9 – 6k – 6 = 0
⇒ – 6k + 3 = 0
⇒ k = \(\frac{3}{6}\) ⇒ k = \(\frac{1}{2}\)

(vi) Find the roots of the quadratic equation 6x2 – x – 2 = 0.
Answer:
The given equation
6x2 – x – 2 = 0
⇒ 6x2 – (4 – 3)x – 2 = 0
⇒ 6x2 – 4x + 3x – 2 = 0
⇒ 2x(3x – 2) + 1(3x – 2) = 0
⇒ (3x – 2) (2x + 1) = 0
⇒ 3x – 2 = 0 or 2x + 1 = 0
⇒ x = \(\frac{2}{3}\) or x = – \(\frac{1}{2}\)
Hence, x = \(\frac{2}{3}\) and x = – \(\frac{1}{2}\) are the roots of given quadratic equation.

RBSE 10th Maths Model Paper Set 9 with Answers in English

(vii) Define scale factor.
Answer:
Scale factor is the ratio of the sides of the triangle to be constructed with corresponding sides of the given triangle.

(viii) Find a relation between x and y such that the point (x, y) is equidistant from the points (3, 6) and (- 3, 4). (1)
Answer:
Given P(x, y) be the equidistant from the points A(3, 6) and B(- 3, 4), then
PA = PB
⇒ PA2 = PB2
⇒ (3 – x)2 + (6 – y)2 = (- 3 – x)2 + (4 – y)2
⇒ 9 + x2 – 6x + 36 + y2 – 12y
= 9 + x2 + 6x + 16 + y2 – 8y
⇒ x2 + y2 + 6x – 8y + 6x + 12y – x2 – y2 = 45 – 25
⇒ 12x + 4y = 20
⇒ 3x + y – 5 = 0
Hence, this is the required relation.

(ix) Evaluate: \(\frac{\sin 18^{\circ}}{\cos 72^{\circ}}\)
Answer:
\(\frac{\sin 18^{\circ}}{\cos 72^{\circ}}\) = \(\frac{\sin \left(90^{\circ}-72^{\circ}\right)}{\cos 72^{\circ}}\)
= \(\frac{\cos 72^{\circ}}{\cos 72^{\circ}}\) = 1

(x) If tan A = cot B, prove that A + B = 90°.
Answer:
We have,
tan A = cot B
⇒ tan A = tan (90° – B)
⇒ A = 90° – B
⇒ A + B = 90°.

(xi) The mean of the following data is 15. Find the missing number P.
12, 15, 32, P, 28, 12, 35
Answer:
According to Question, Sum of data
Mean = \(\frac{\text { Sum of data }}{\text { Total number }}\)
⇒ 21 = \(\frac{12+15+32+p+28+12+35}{7}\)
⇒ 21 × 7 = 134 + p
⇒ p = 147 – 134
⇒ p = 13

(xii) A die is thrown once, what is the probability of getting an even prime number? (1)
Answer:
When a die is thrown once, the number of possible outcomes is 6 [1, 2, 3, 4, 5, 6]. Let E be event of getting a even prime number.
The number of outcomes favourable to E (i.e. getting even prime no. i.e. 2) is 1.
P(E) = \(\frac{\text { No. of favourable outcomes }}{\text { No. of all possible outcomes }}=\frac{1}{6}\)

RBSE 10th Maths Model Paper Set 9 with Answers in English

Section – B

Question 4.
Given that √3 is an irrational number. Prove that (2 + √3) is an irrational number. 2
Answer:
Let us assume, that (2 + √3) is a rational, it can be expressed in the form of \(\frac{a}{b}\), where a and b are coprime positive integers and b ≠ 0.
∴ 2 + √3 = \(\frac{a}{b}\)
(∵ HCF of a and b is 1)
⇒ \(\frac{a}{b}\) = √2
⇒ \(\frac{a-2 b}{b}\) = √3
\(\frac{a-2 b}{b}\) = rational number
(∵ a and b are positive integers) Thus, from eqn. (i) is a rational number.
But this contradicts the fact that is an irrational number. So, our assumption that 2 + √3 is a rational number, is wrong.
Hence, (2 + √3) is an irrational number.

Question 5.
Find a quadratic polynomial whose zeroes are reciprocal of the zeroes of the polynomial f(x) = ax2 + bx + c, a ≠ 0, c ≠ 0. (2)
Answer:
f(x) = ax2 + bx + c (given)
Let α and β be the zeroes of f(x)
Then, α + β = –\(\frac{b}{a}\) And αβ = \(\frac{c}{a}\)
Zeroes of required polynomial are \(\frac{1}{\alpha}\) and \(\frac{1}{\beta}\).
∴ Sum of zeroes = \(\frac{1}{\alpha}+\frac{1}{\beta}\) = \(\frac{\alpha+\beta}{\alpha \beta}\) = \(\frac{-\frac{b}{a}}{\frac{c}{a}}\)
= \(\frac{-b}{a} \times \frac{a}{c}\) = \(\frac{-b}{c}\)
And product of zeroes
RBSE 10th Maths Model Paper Set 9 with Answers in English 2
Required polynomial = x2 – (sum of zeroes)x + product of zeroes
RBSE 10th Maths Model Paper Set 9 with Answers in English 3

Question 6.
Solve the equations 3x – 5y = 4 and 9x – 2y = 7 for x and y. (2)
Answer:
The given equation are:
3x – 5y = 4 ….. (i)
9x – 2y = 7 ……. (ii)
From equation (ii), we get
x = \(\frac{7+2 y}{9}\) …… (iii)
Substituting the value of x in equation
(i), we get
\(\frac{3(7+2 y)}{9}\) – 5y = 4
⇒ \(\frac{7+2 y}{3}\) – 5y = 4
⇒ 7 + 2y – 15y = 12
⇒ – 13y = 12 – 7 = 5
⇒ y = \(\frac{5}{-13}\) = \(\frac{-5}{13}\)
Putting the value ofy in equ. (iii), we get
RBSE 10th Maths Model Paper Set 9 with Answers in English 4

RBSE 10th Maths Model Paper Set 9 with Answers in English

Question 7.
Solve: \(\frac{1}{a+b+x}\) = \(\frac{1}{a}+\frac{1}{b}=\frac{1}{x}\); where a + b ≠ 0
Answer:
The given equation is:
RBSE 10th Maths Model Paper Set 9 with Answers in English 5
⇒- (a + b) × ab = (a + b)x (a + b + x)
⇒ (a + b)x (a + b + x) + ab (a + b) = 0
⇒ (a + b) [x (a + b + x) + ab] = 0
⇒ x (a + b + x) + ab = 0
⇒ ax + bx + x2 + ab = 0
⇒ x2 + ax + bx + ab = 0
⇒ x(x + a) + b (x + a) =0
⇒ (x + a) (x + b) = 0
⇒ (x + a) = 0 and (x + b) = 0
⇒ x = – a and x = – b

Question 8.
Find the 11th term of the AP : – 17, – 12, – 7,….
Answer:
The given sequence of AP is – 17, – 12, – 7, ……..
Here, a1 = – 17, a2 = – 12, a3 = – 7 d = a2 – a1 = – 12 – (-17) = – 12 + 17 = 5
We know that nth term of AP is
an = a + (n – 1)d
⇒ a11 = – 17 + (11 – 1) × 5
= – 17 + 50 = 33
Hence, 11th term of given AP = 33.

Question 9.
In a flower bed, there are 23 rose plants in the first row, 21 in the second, 19 in the third, and so on. There are 5 rose plants in the last row. How many rows are there in the flower bed? (2)
Answer:
The number of rose plants in the 1st, 2nd, 3rd, … rows are 23, 21, 19, …, 5
a2 – a1 = 21 – 23 = – 2
a3 – a2 = 19 – 21 = – 2
∵ a2 – a1 = a3 – a2
∴ This sequence forms an AP.
Here a = 23, d = 21 – 23 = – 2, an = 5
We know that nth term of an AP is
an = a + (n – 1) d
⇒ 5 = 23 + (n – 1) × (- 2)
⇒ 5 = 23 – 2n + 2
⇒ 2n = 25 – 5
⇒ n = \(\frac{20}{2}\) = 10
Hence, the number of rows are there in the flower bed = 10.

RBSE 10th Maths Model Paper Set 9 with Answers in English

Question 10.
Draw a circle of radius 3.5 cm. From a point P, 6 cm from its centre, draw two tangents to the circle. (2)
Answer:

  • Draw a circle with O as the centre and radius 3.5 cm
  • Mark a point P outside the circle such that OP = 6 cm.
  • Join OP and draw prependicular bisector of PO meeting PO at M.
  • Draw a circle with M as the centre and radius equal to PM = OM intersecting the given circle at points Q and R.
    RBSE 10th Maths Model Paper Set 9 with Answers in English 6
  • Join PQ and PR then PQ and PR are required tangents.

Question 11.
Draw a ∆ABC in which BC = 7 cm, ∠B = 45° and ∠A = 105°. Then construct another triangle whose sides are \(\frac{3}{4}\) times corresponding sides of ∆ABC. (2)
Answer:

  • Draw a line segment BC = 7 cm.
  • In ∆ABC, ∠B = 45°, 2A = 105°
    ∴ ∠C = 180° – (45° + 105°) = 30°
    At B draw and angle ∠B = 45° and draw ∠C = 30° intersecting each other at A to get ∆ABC
    RBSE 10th Maths Model Paper Set 9 with Answers in English 7
  • At B draw any ray BX making an acute angle with BC on the side opposite to the vertex A
  • Along BX mark 4 points B1(B2, B3 and B4 such that BB1 = B1B2 = B2B3 = B3B4
  • Join B4C
  • From B3 draw B<sub3C’ | | B4C meeting BC at C’
  • From C’ draw C’A’ | | AC meeting AB at A
    Then A’BC’ is required triangle, each of whose side is \(\frac{3}{4}\) of corresponding sides of ∆ABC.

Question 12.
Prove that 1 + \(\frac{\cot ^{2} \alpha}{1+{cosec} \alpha}\) = cosec α
Answer:
LHS = 1 + \(\frac{\cot ^{2} \alpha}{1+{cosec} \alpha}\)
= 1 + \(\frac{{cosec}^{2} \alpha-1}{{cosec} \alpha+1}\)
= \(1+\frac{({cosec} \alpha+1)({cosec} \alpha-1)}{({cosec} \alpha+1)}\)
= 1 + cosec α – 1
= cosec α = R.H.S.

Question 13.
If sec θ = x + \(\frac{1}{4 x}\), prove that sec θ + tan θ = 2x or \(\frac{1}{2 x}\)
Answer:
RBSE 10th Maths Model Paper Set 9 with Answers in English 8

RBSE 10th Maths Model Paper Set 9 with Answers in English

Question 14.
Consider the following distribution of daily wages of 50 workers of a factory. (2)
RBSE 10th Maths Model Paper Set 9 with Answers in English 9
Find the mean daily wages of the workers of the factory by using an appropriate method.
Answer:
Here, h = 20
RBSE 10th Maths Model Paper Set 9 with Answers in English 10

From the table, we have
Σfi = 50, Σfiui = – 12, a = 150, h = 20
∴ x̄ = a + \(\left(\frac{\Sigma f_{i} u_{i}}{\Sigma f_{i}}\right)\) × h
⇒ x̄ = 150 + \(\left(-\frac{12}{50}\right)\) × 20
⇒ x̄ = 150 – 4.80 = 145.20.
Hence, the mean daily wages of the workers = ₹ 145.20.

Question 15.
Find the mode of the following distribution. (2)
RBSE 10th Maths Model Paper Set 9 with Answers in English 11
Answer:
The class interval 35 – 40 has maximum frequency. So, it is the modal class.
∴ l = 35, f1 = 50, f0 = 34, f2 = 42, h = 5
Mode = l + \(\left(\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}}\right)\) × h
= 35 + \(\left(\frac{50-34}{2 \times 50-34-42}\right)\) × 5
= 35 + \(\frac{16 \times 5}{24}\)
= 35 + 3.33
= 38.33

Question 16.
A game consists of tossing a one rupee coin 3 times and noting its outcome each time. Hanif wins if all the tosses give the same result i.e.,three heads or three tails, and loses otherwise. Calculate the probability that Hanif will lose the game. (2)
Answer:
When a coin tossed three times, the the possible outcomes are
(H, H, H), (H, H, T), (H, T, T), (H, T, H), (T, H, H), (T, H, T), (T, T, H) and (T, T, T)
Total number of possible outcomes = 8
Number of cases when all the tosses give the same results are two i.e., (H, H, H) and (T, T, T)
It is given that Hanif wins when all the tosses give the same results, Number of cases in which Hanif will lose the game = 8 – 2 = 6
Number of favorable outcomes = 6
∴ P (Hanif will lose the game) = \(\frac{6}{8}\) = \(\frac{3}{4}\)

RBSE 10th Maths Model Paper Set 9 with Answers in English

Section – C

Question 17.
A manufactures of TV set produce 720 sets in fourth year and 1080 set in the sixth year, assuming that the production increase uniformly by a fixed number every year, then find total production in first 9 year. (3)
Or
Find the sum of n terms of the series: (3)
(4 – \(\frac{1}{n}\)) + (4 – \(\frac{2}{n}\)) + (4 – \(\frac{3}{n}\)) + ………
Answer:
Production of TV sets in 4th year = 720,
Production of TV sets in 6th year = 1080
Let a and d be first term and common difference respectively then,
a4 = a + (4 – 1)d
⇒ 720 = a + 3d
⇒ a + 3d = 720 …(1)
And a6 = a + (6 – 1)d
⇒ 1080 = a + 5 d
⇒ a + 5d = 1080 …(2)
Substracting equ. (2) from equ. (1), We get
RBSE 10th Maths Model Paper Set 9 with Answers in English 12
Substituting the value of d in equ. (1), We get
a + 3 × 180 = 720
⇒ a + 540 = 720
⇒ a = 720 – 540
⇒ a = 180
Production of TV sets increases uniformly by a fixed numbers. So,
Total productions of TV sets in first 9 year = a9
⇒ a9 = a + (9 – 1) 180
= 180 + 8 × 180
= 180 + 1440 = 1620
Hence, production in first 9th year = 1620.

Question 18.
In what ratio does the point P(\(\frac{24}{11}\), y) divide the line segment joining the points P(2, – 2) and Q(3, 7) ? (3)
Or
Find the value of P, if the points A(2, 3) B(4, P) and C(6, – 3) are collinear. (3)
Answer:
Let the required ratio be k : 1
Here, x1 = 2, y1 = -2, x2 = 3, y2 = 7, x = \(\frac{24}{11}\) y = y, m1 = k, m2 = 1
By section formula, we have
RBSE 10th Maths Model Paper Set 9 with Answers in English 13

RBSE 10th Maths Model Paper Set 9 with Answers in English

Question 19.
Prove that: \(\frac{1}{1+\sin ^{2} \theta}+\frac{1}{1+\cos ^{2} \theta}+\frac{1}{1+\sec ^{2} \theta}+\frac{1}{1+{cosec}^{2} \theta}\) = 2
Or
If sin θ + cos θ = √3, then prove that tan θ + cot θ = 1
Answer:
We have, LHS
RBSE 10th Maths Model Paper Set 9 with Answers in English 14

Question 20.
The table below shows the salaries of 280 persons:
RBSE 10th Maths Model Paper Set 9 with Answers in English 15
Calculate the median salary of the data.
Or
100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows :
RBSE 10th Maths Model Paper Set 9 with Answers in English 16
Find the mean number of letters in the surnames. (3)
Answer:
We prepare the cumulative frequency distribution table as given below:
RBSE 10th Maths Model Paper Set 9 with Answers in English 17
n = 280
⇒ \(\frac{n}{2}=\frac{280}{2}\) = 140
But 140 comes under the cumulative frequency 182 and class interval against the cumulative frequency 182 is 10 – 15. So, it is the median class.
∴ l = 10, cf = 49, f = 133 and h = 5.
Median = l + \(\left(\frac{\frac{n}{2}-c f}{f}\right)\) × h
= 10 + \(\left(\frac{140-49}{133}\right)\) × 5
= 10 + \(\frac{91 \times 5}{133}\)
= 10 + \(\frac{65}{19}\)
= 10 + 3.42
= 13.42 (thousands)
= ₹ 13.42

RBSE 10th Maths Model Paper Set 9 with Answers in English

Section – D

Question 21.
Akhila went to a fair in her village. She wanted to enjoy rides on the Giant Wheel and play Hoopla (a game in which you throw a ring on the items kept in a stall, and if the ring covers any object completely you get it.) The number of times she played Hoopla is half the number of rides she had on the Giant wheel. If each ride costs ₹ 3, and a game of Hoopla costs ₹ 4, how would you find out the number of rides she had and how many times she played Hoopla, provided she spent ₹ 20 ?
Represent this situation algebraically and graphically (Geometrically). (4)
Or
Cost of 3 balls and 2 bats is ₹ 100 and the cost of 3 balls and 4 bats is ? 150. Formulate the problem algebrically and solve it graphically.(4)
Answer:
Let the number of rides be x and number of times Hoopla played be y. According to question,
y = \(\frac{1}{2}\) x
⇒ x – 2y = 0
and 3x + 4y = 20
The algebraic representation of the given situation is:
x – 2y = 0 …(i)
3x + 4y = 20 ….(ii)
For representation of these equations graphically we draw the graph of these equations as follows:
x – 2y = 0
⇒ y = \(\frac{x}{2}\)
Put the different values of x in this equation then we get different values of y, and we prepare the table of x, y for the equation x – 2y = 0
RBSE 10th Maths Model Paper Set 9 with Answers in English 18
and 3x + 4y = 20
⇒ 4y = 20 – 3x ⇒ y = \(\frac{20-3 x}{4}\)
Now, we put the different values of x then different values of y and we prepare the table of x, y for the equation 3x + 4y = 20
RBSE 10th Maths Model Paper Set 9 with Answers in English 19
Now, we plot the values of x and y from table 1 & 2 on a graph paper and draw the graphs through these values. Observe that we get two straight lines which intersect each other at the point (4, 2).
Hence, number of rides she had taken on giant wheel is 4 and the number of times she played Hoopla is 2.
RBSE 10th Maths Model Paper Set 9 with Answers in English 20

RBSE 10th Maths Model Paper Set 9 with Answers in English

Question 22.
Construct an Isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are 1\(\frac{1}{2}\) times the corresponding sides of the isosceles triangle. (4)
Or
Draw a circle of radius 3 cm. From a point 5 cm away from its centre, construct the pair of tangents to the circle and measure their lengths. Also verify the measurement by actual calculation. (4)
Answer:
Steps of Construction :

  • Let ABC be the isosceles triangle with base AB = 8 cm and altitude CD = 4 cm.
  • Draw a line segment AB = 8 cm.
  • Draw the perpendicular bisector of AB intersecting AB at D.
  • With D as a centre and radius 4 cm draw an arc intersecting the perpendicular bisector at C.
  • Join AC and BC to get the isosceles triangle ABC.
    RBSE 10th Maths Model Paper Set 9 with Answers in English 21
  • produce AB to B’ such that
    AB’ = 1\(\frac{1}{2}\) of 8 = \(\frac{3}{2}\) × 8 = 12 cm.
  • Through B’ draw B’C’ || BC intersecting the produced line segment AC to C’. .
    Then AB’C’ is the required triangle each of whose side is 1\(\frac{1}{2}\) times of the corresponding sides of ∆ABC.

Justification: Since,
BC || B’C’
Therefore, ∆C’AB’ ~ ∆CAB
⇒ \(\frac{A C^{\prime}}{A C}=\frac{A B^{\prime}}{A B}=\frac{B^{\prime} C^{\prime}}{B C}=\frac{3}{2}\)

Question 23.
Change the following distribution to a more than type distribution. Hence, draw the more than type ogive for this distribution. (4)
RBSE 10th Maths Model Paper Set 9 with Answers in English 22
Or
A survey regarding the heights (in cm) of 51 girls of class X of a school was conducted and the following data was obtained: (4)
RBSE 10th Maths Model Paper Set 9 with Answers in English 23
Find the median height.
Answer:
We prepare the cumulative frequency table by more than method as given below:
RBSE 10th Maths Model Paper Set 9 with Answers in English 24
On the graph paper, we plot the points (20, 100), (30, 90), (40, 82), (50, 70), (60, 46), (70, 40), (80, 15). Joining these points with free hand to get more than curve as shown below in graph.
RBSE 10th Maths Model Paper Set 9 with Answers in English 25

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