Samacheer Kalvi 10th Maths Model Question Paper 5 English Medium

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Tamilnadu Samacheer Kalvi 10th Maths Model Question Paper 5 English Medium

Instructions

• The question paper comprises of four parts.
• You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
• All questions of Part I, II, III and IV are to be attempted separately.
• Question numbers 1 to 14 in Part I are Multiple Choice Quèstions of one-mark each. These are to be answered by choosing the most suitable answer from the given four alternatives and.writing the option code and the corresponding answer.
• Question numbers 15 to 28 in Part II àre two-marks questions. These are to be answered in about one or two sentences.
• Question numbers 29 to 42 in Part III are five-marks questions. These are to be answered in about three to five short sentences.
• Question numbers 43 to 44 in Part IV are eight-marks questions. These are to be answered in detail. Draw diagrams wherever necessary.

Time: 3 Hours
Max Marks: 100

PART-I

I. Choose the correct answer. Answer all the questions. [Answers are in bold] [14 × 1 = 14]

Question 1.
If n(A × B) = 6 and A = {1, 3} then n (B) is ………….. .
(1) 1
(2) 2
(3) 3
(4) 6
Answer:
(3) 3

Question 2.
Given F₁ = 1, F₂ = 3 and F_n = F_n-1+ F_n-2 then F₅ is ………….. .
(1) 3
(2) 5
(3) 8
(4) 11
Answer:
(4) 11

Question 3.
In an A.P., the first term is 1 and the common difference is 4. How many terms of the A.P must be taken for their sum to be equal to 120?
(1) 6
(2) 7
(3) 8
(4) 9
Answer:
(3) 8

Question 4.
f = {(2, a), (3, b), (4, b), (5, c)} is a ………….. .
(1) identity function
(2) one-one function
(3) many-one function
(4) constant function
Answer:
(3) many-one function

Question 5.
The number of points of intersection of quadratic polynomial x² + 4x + 4 with the x axis is ………….. .
(1) 0
(2) 1
(3) 0 (or) 1
(4) 2
Answer:
(2) 1

Question 6.
The non-diagonal elements in any unit matrix are ………….. .
(1) 0
(2) 1
(3) m
(4) n
Answer:
(1) 0

Question 7.
If A is a 2′ 3 matrix and B is a 3′ 4 matrix, how many columns does AB have?
(1) 3
(2) 4
(3) 2
(4) 5
Answer:
(2) 4

Question 8.
In figure CP and CQ are tangents to a circle with centre at O. ARB is another tangent touching the circle at R. If CP = 11 cm and BC = 7 cm then the length of BR is ………….. .

(1) 6 cm
(2) 5 cm
(3) 8 cm
(4) 4 cm
Answer:
(4) 4 cm

Question 9
The slope of the line joining (12, 3), (4, a) is \(\frac { 1 }{ 8 }\). The value of ‘a’ is ………….. .
(1) 1
(2) 4
(3) -5
(4) 2
Answer:
(4) 2

Question 10.
If x = a tan θ and y = b sec θ then ………….. .
(1) \(\frac{y^{2}}{b^{2}}-\frac{x^{2}}{a^{2}}=1\)
(2) \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\)
(3) \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\)
(4) \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=0\)
Answer:
(1) \(\frac{y^{2}}{b^{2}}-\frac{x^{2}}{a^{2}}=1\)

Question 11.
A letter is chosen at random from the letter of the word “PROBABILITY”. Find the probability that it is not a vowel.
(1) \(\frac{1}{5}\)
(2) \(\frac{2}{3}\)
(3) \(\frac{1}{3}\)
(4) \(\frac{3}{5}\)
Answer:
(2) \(\frac{2}{3}\)

Question 12.
The height of a right circular cone whose radius is 5 cm and slant height is 13 cm will be ………….. .
(1) 12 cm
(2) 10 cm
(3) 13 cm
(4) 5 cm
Answer:
(1) 12 cm

Question 13.
If the mean and co-efficient of variation of a data are 4 and 87.5% then the standard deviation is ………….. .
(1) 3.5
(2) 3
(3) 4.5
(4) 2.5
Answer:
(1) 3.5

Question 14.
Variance of first 20 natural numbers is ………….. .
(1) 32.25
(2) 44.25
(3) 33.25
(4) 30
Answer:
(3) 33.25

PART-II

II. Answer any ten questions. Question No. 28 is compulsory. [10 × 2 = 20]

Question 15.
Define a function.
Answer:
A relation “f” between two non-empty sets X and Y is called a function from X to Y if for each x ∈ X there exists only one y ∈ Y such that (x, y) ∈ f

Question 16.
Compute x such that 10⁴ ≡ x (mod 19)
Answer:
10² = 100 ≡ 5 (mod 19)
10⁴ = (10²)² ≡ 5² (mod 19)
10⁴ = 25 (mod 19)
10⁴ = 6 (mod 19)
(since 25 ≡ 6 (mod 19))
Therefore, x = 6

Question 7.
Simlify \(\frac{4 x^{2} y}{2 z^{2}} \times \frac{6 x z^{3}}{20 y^{4}}\)
Answer:

Question 18.
Pari needs 4 hours to complete the work. His friend Yuvan needs 6 hours to complete the work. How long will it take to complete if they work together?
Answer:
Let the work done by Pari and Yuvan together be x
Work done by Pari = \(\frac { 1 }{ 4 }\)
Work done by Yuvan = \(\frac { 1 }{ 6 }\)
By the given condition
\(\frac{1}{4}+\frac{1}{6}=\frac{1}{x} \Rightarrow \frac{3+2}{12}=\frac{1}{x}\)
\(\frac{5}{12}=\frac{1}{x}\)
5x = 12 ⇒ x = \(\frac{12}{5}\)
x = \(2 \frac{2}{5}\) hours (or) 2 hours 24 minutes

Question 19.
Find the values of x, y and z from the following equation \(\left( \begin{matrix} 12 & 3 \\ x & \frac { 3 }{ 2 } \end{matrix} \right) =\left( \begin{matrix} y & z \\ 3 & 5 \end{matrix} \right) \)
Answer:
Since the given matrices are equal then all the corresponding elements are equal. y = 12, z = 3, x = 3
The value of x = 3, y = 12 and z = 3

Question 20.
What length of ladder is needed to reach a height of 7 ft along the wall when the base of the ladder is 4 ft from the wall? A
Answer:

Let x be the length of the ladder. BC = 4 ft, AC = 7 ft.
By Pythagoras theorem we have, AB² = AC² + BC²
x² = 7² + 4² gives x² = 49 + 16
x² = 65. Hence, x = √65
The number √65 is between 8 and 8.1.
8² = 64 < 65 < 65.61 = 8.1²
Therefore, the length of the ladder is approximately 8.1 ft.

Question 21.
Prove that \(\sqrt{\frac{1+\cos \theta}{1-\cos \theta}}\) = cosec θ + cot θ
Answer:

Question 22.
The radius of a sphere increases by 25%. Find the percentage increase in its surface area.
Answer:
Let the radius of the be “r”
Surface area of the sphere = 4πr²sq.units …… (1)
If the radius is increased by 25%
New radius = \(\frac { 25 }{ 100 }\) × r + r
= \(\frac { r }{ 4 }\) + 4
= \(\frac{r+4 r}{4}=\frac{5 r}{4}\)
Surface area of the sphere
= 4π\(\left(\frac{5 r}{4}\right)^{2}\) sq.units
= 4 × π × \(\frac{25 r^{2}}{16}\)
= \(\frac{25 \pi r^{2}}{4}\) sq.units
Difference in surface area
= \(\frac{25 \pi r^{2}}{4}-4 \pi r^{2}\)
= \(\pi r^{2}\left(\frac{25}{4}-4\right)\)
= \(\pi r^{2}\left(\frac{25-16}{4}\right)\)
= \(\pi r^{2}\left(\frac{9}{4}\right)=\frac{9 \pi r^{2}}{4}\)
Percentage of increase in surface area

Percentage of increase in surface area = 56.25%

Question 23.
The standard deviation and mean of a data are 6.5 and 12.5 respectively. Find the coefficient of variation.
Answer:
Standard deviation of a data (σ) = 6.5
Mean of the data (x̄) = 12.5
Coefficient of variation = \(\frac{\sigma}{x} \times 100 \%\) ⇒ \(\frac{6.5}{12.5} \times 100 \%\) = 52%
Coefficient of variation = 52%

Question 24.
If f{x) = 3 + x, g(x) = x – 4, then check whether fog = gof
Answer:
f(x) = 3 + x ; g(x) = x – 4
fog = f[g(x)]
= f(x – 4)
= 3 + x – 4
= x – 1
gof = g [f(x)}
= g(3 + x)
= 3 + x – 4
= x – 1
∴ fog = gof

Question 25.
An organization plans to plant saplings in 25 streets in a town in such a way that one sapling for the first street, three for the second, nine for the third and so on. How many saplings are needed to complete the work?
Answer:
Here n = 25, a = 1, r = 3
S_n = \(a \frac{\left(r^{n}-1\right)}{r-1}\)
S₂₅ = \(\frac{1\left(3^{25}-1\right)}{3-1}\)
= \(\frac{3^{25}-1}{2}\)

Question 26.
Find the 19th term of an A.P. – 11, -15, -19,……..
Answer:
First term (a) = -11
Common difference (d) = -15 – (-11)
= -15 + 11 = -4
n = 19 .
t_n = a + (n – 1) d
t_n = -11 + 18(-4)
= -11 – 72
t₁₉ = -83
∴ 19th term of an A.P. is – 83

Question 27.
Find the value of ZB AC in the given triangle.
Answer:

In right triangle ABC

θ = tan^-1 \(\left(\frac{4}{5}\right)\) = tan^-1 (0.8)
θ = 38.7° (since tan 38.7° = 0.8011)
∠BAC = 38.7°

Question 28.
The vertices of a triangle are A(-1,3), B(-1, 1) and C(5, 1). Find the length of the median through the vertex C.
Answer:

Mid point of AB = \(\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)\)
= \(\left(\frac{-1+1}{2}, \frac{3-1}{2}\right)\)
= (0,1)
Length of the median CD = \(\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\)
= \(\sqrt{(5-0)^{2}+(1-1)^{2}}\)
= \(\sqrt{25}\) = 5

PART – III

III. Answer any ten questions. Question No. 42 is compulsory. [10 × 5 = 50]

Question 29.
Let f be a function f : N → N be defined by f(x) = 3x + 2, x ∈ N
(i) Find the images of 1, 2,3
(ii) Find the pre – images of 29, 53
(iii) Identify the type of function.
Answer:
The function f: N → N is defined by f(x) = 3x + 2
(i) If x = 1, f(1) = 3(1) + 2 = 5
If x = 2, f(2) = 3(2) + 2 = 8
If x = 3, f(3) = 3(3) + 2=11
The images of 1, 2, 3 are 5, 8, 11 respectively.

(ii) If x is the pre-image of 29, then f(x) = 29 . Hence 3x + 2 = 29
3x = 21 ⇒ x = 9.
Similarly, if x is the pre-image of 53, then f(x) = 53.
Hence 3x + 2 = 53 3x = 51 ⇒ x = 17.
Thus the pre-images of 29 and 53 are 9 and 17 respectively.

(iii) Since different elements of N have different images in the co-domain, the function f is one – one function.
The co-domain of f is N.
But the range of f = {5, 8, 11, 14, 17,…} is a proper subset of N.
Therefore f is not an onto function. That is, f is an into function.
Thus f is one – one and into function.

Question 30.
Let f: A → B be a function defined by f(x) = \(\frac { x }{ 2 }\) – 1, where A = {2,4,6,10,12} and B = {0,1,2, 4,5,9}. Represent f by
(i) set of ordered pairs (ii) a table (iii) an arrow diagram (iv) a graph
Answer:
A= {2, 4, 6, 10, 12} and B = {0, 1, 2, 4, 5, 9}
f(x) = \(\frac { x }{ 2 }\) – 1
f(2) = \(\frac { 2 }{ 2 }\) – 1 = 1 – 1 = 0
f(4) = \(\frac { 4 }{ 2 }\) – 1 = 2 – 1 = 1
f(6) = \(\frac { 6 }{ 2 }\) – 1 = 3 – 1 = 2
f(10) = \(\frac { 10 }{ 2 }\) – 1 = 5 – 1 = 4
f(12) = \(\frac { 12 }{ 2 }\) – 1 = 6 – 1 = 5

(i) Set of ordered pairs
f = {(2, 0) (4, 1) (6, 2) (10, 4) (12, 5}

(ii) Table

(iii) Arrow diagram

(iv) Graph

Question 31.
The ratio of 6^th and 8^th term of an A.P is 7:9. Find the ratio of 9^th term to 13^th terms.
Answer:
Given t₆ : t₈ = 7 : 9 (using t_n = a + (n – 1)d
a + 5d : a + 7d =7 : 9
9(a + 5d) = 7(a + 7d)
9a + 45d = 7a + 49d
9a – 7a = 49d – 45d
2a = 4d
a = 2d
To find t₉ : t₁₃
t₉ : t₁₃ = a + 8d : a + 12d
= 2d + 8d : 2d + 12d
= 10d : 14d
= 5 : 7
∴ t₉ : t₁₃ = 5 : 7

Question 32.
The sum of first n, In and 3n terms of an A.P. are S₁ S₂ and S₃ respectively. Prove that S₃ = 3 (S₂ – S₁).
Answer:
If S₁ S₇ and S₃ are sum of first n, 2n and 3n terms of an A.P. respectively then

Question 33.
Find the values of m and n if the exprtession \(\frac{1}{x^{4}}-\frac{6}{x^{3}}+\frac{13}{x^{2}}+\frac{m}{x}+n\) is a perfect a square.
Answer:

Since it is a perfect square
\(\frac { 1 }{ 2 }\)(m + 12) = 0
m + 12 = 0
m = -12
n – 4 = 0
n = 4
∴ The value of m = -12 and n = 4

Question 34.
If α, β are the roots of the equation 2x² – x – 1 = 0 then form the equation whose roots are α²β, β²α.
Answer:
2x² – x – 1 = 0 ⇒ Here a = 2, b = -1,c = -1
α + β = \(\frac{-b}{a}=\frac{-(-1)}{2}=\frac{1}{2}\)
α + β = \(\frac{c}{a}=\frac{-1}{2}\)
Given roots are α²β, β²α
Summ of the roots α²β + β²α = αβ (α + β) = \(-\frac{1}{2}\left(\frac{1}{2}\right)=-\frac{1}{4}\)
Product of the roots (α²β) × (β²α) = α³β³ = (αβ)³ =
\(\left(-\frac{1}{2}\right)^{3}=-\frac{1}{8}\)
The required equation is x² – (Sum of the roots) x + (Product of the roots) = 0
\(x^{2}-\left(-\frac{1}{4}\right) x-\frac{1}{8}=0\) gives 8x² + 2x – 1 = 0

Question 35.
P and Q are the mid-points of the sides CA and CB respectively of a ∆ABC, right angled at C. Prove that 4(AQ² + BP²) = 5AB².
Answer:

Since, ∆AQC is a right triangle at C, AQ² = AC² + QC² ……. (1)
Also, ∆BPC is a right triangle at C, BP² = BC² + CP² ……… (2)
From (1) and (2), AQ² + BP² = AC² + QC² + BC² + CP²
4(AQ² + BP²) = 4AC² + 4QC² + 4BC² + 4CP²
= 4AC² + (2QC)² + 4BC² + (2CP)²
= 4AC² + BC² + 4BC² + AC² (Since P and Q are mid points)
= 5(AC² + BC²)
4(AQ² + BP²) = 5AB² (By Pythagoras Theorem)

Question 36.
Find the equation of a straight line passing through (1, -4) and has intercepts which are in the ratio 2:5.
Answer:
Let the x-intercept be 2a and the y intercept 5a
The equation of a line is \(\frac{x}{a}+\frac{y}{a}=1\) ⇒ \(\frac{x}{2 a}+\frac{y}{5 a}=1\)
The line passes through the point (1, -4)
\(\frac{1}{2 a}+\frac{(-4)}{5 a}=1\) ⇒ \(\frac{1}{2 a}-\frac{4}{5 a}=1\)
Multiply by 10a
(L.C.M of 2a and 5a is 10a)
5 – 8 = 10a ⇒ -3 = 10a ⇒ a = \(\frac{-3}{10}\)
The equation of the line is \(\frac{x}{2(-3 / 10)}+\frac{y}{5(-3 / 10)}=1\)
\(\frac{x}{-3 / 5}+\frac{y}{-3 / 2}=1\) ⇒ \(\frac{5 x}{-3}+\frac{2 y}{-3}=1\)
\(\frac{-5 x}{3}-\frac{2 y}{3}=1\)
Multipy by 3
– 5x – 2y = 3 ⇒ – 5x – 2y – 3 = 0
5x + 2y + 3 = 0
The equation of a line is 5x + 2y + 3 = 0

Question 37.
From the top of the tower 60m high the angles of depression of the top and bottom of a vertical lamp post are observed to be 38° and 60° respectively. Find the height of the lamp post (tan 38° = 0.7813, √3 = 1.732)
Answer:

Let the height of the lamp post be “h”
The height of the tower (BC) = 60 m
∴ EC = 60 – h
Let AB be x
In the right ∆ ABC
tan 60° = \(\frac{B C}{A B}\)
√3 = \(\frac{60}{x}\)
x = \(\frac{60}{\sqrt{3}}\) …… (1)
In the right ∆ DEC, tan 38° = \(\frac{E C}{D E}\)
0.7813 = \(\frac{60-h}{x}\)
x = \(\frac{60-h}{0.7813}\) ….. (2)
Froom (1) and (2) we get
\(\frac{60}{\sqrt{3}}=\frac{60-h}{0.7813}\)
60 × 0.7813 = 60√3 – √3h
46.88 = 60√3 – √3h
√3h = 60√3 – 46.88
= 60 × 1.732 – 46.88
= 103.92 – 46.88
1.732 h = 57.04 ⇒ h = \(\frac{57.04}{1.732}\)
h = \(\frac{5704}{1732}\) = 32.93m
∴ Height of the lamp post = 32.93 m

Question 38.
Calculate the weight of a hollow brass sphere if the diameter is 14 cm and thickness is 1 mm, and whose density is 17.3 g/cm³.
Answer:
Let r and R be the inner and outer radii of the hollow sphere.
Given that, inner diameter d = 14 cm; inner radius r = 7 cm; thickness = 1 mm = \(\frac{1}{10}\) cm
Outer radius R = 7 + \(\frac{1}{10}=\frac{71}{10}\) = 7.1 cm
Volume of hollow sphere = \(\frac{4}{3}\) π (R³ – r³) cu. cm
= \(\frac{4}{3} \times \frac{22}{7}\)(357.91 -343) = 62.48 cm³
But, weight of brass in 1 cm³ = 17.3 gm
Total weight = 17.3 × 62.48 = 1080.90 gm
Therefore, total weight is 1080.90 grams.

Question 39.
Find the Co-efficient of variation of 24, 26,33,37,29,31
Answer:
Arrange in ascending order we get 24, 26,29, 31,33,37
Assumed mean = 29


\(\frac{4.32}{30} \times 100\) = \(\frac{432}{30}=14.4\)
Coefficient of variation = 14.4%

Question 40.
Two dice, one blue and one grey, are thrown at the same time. Write down all the possible outcomes. W hat is the probability that the sum of the two numbers appearing on the top of the dice is ………………. .
Answer:
When we roll two dice
Sample space = {(1. 1) (1,2) (1, 3) (1, 4) (1, 5) (1, 6)
(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)
(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)
(4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)
(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)
(6, 1) (6, 2) (6, 3) (6. 4) (6, 5) (6, 6)}
n(S) = 36
(i) Let A be the event of getting the sum of two number is 8
A = {(2, 6) (3, 5) (4, 4) (5, 3) (6, 2)} 5
n(A) = 5
P(A) = \(\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{5}{36}\)

(ii) Let B be the event of getting the sum of the two numbers is 13.
B = { }
n(B) = 0
P(B) = \(\frac{n(\mathrm{B})}{n(\mathrm{S})}=\frac{0}{36}=0\)

(iii) Let C be the event of getting the sum of the two numbers is less than 12 or equal to 12.
n(C) = 36
P(C) = \(\frac{n(\mathrm{C})}{n(\mathrm{S})}=\frac{36}{36}=1\)

Question 41.
Find two consecutive positive integers, sum of whose squares is 365.
Answer:
Let the two consecutive positive integers be “x” and x + 1
Sum of squares = 365
x² + (x+1)² = 365
2x² + 2x – 364 = 0
x² + x – 182 = 0

(x + 14) (x – 13) = 0
x + 14 = 0 or x – 13 = 0
x = -14 or x = 13 (rejecting -14. Given positive integer)
The consecutive terms are 13 and 14.

Question 42.
A cylindrical bucket of 32 cm high and with radius of base 18 cm, is filled with sand completely. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the heap.
Answer:
Cylindrical bucket
Height of the bucket (h) = 32 cm
Radius of the bucket (r) = 18 cm

Conical heap
Height of the cone (H) = 24 cm
Let the radius of the conical heap be “P ”
By the given condition
Volume of the conical heap = Volume of the cylindrical bucket
\(\frac { 1 }{ 3 }\)πR²H = πr²
\(\frac { 1 }{ 3 }\) × R² × 24 = r²H
\(\frac { 1 }{ 3 }\) × R² × 24 = 18 × 18 × 32
R² = \(\frac{18 \times 18 \times 32 \times 3}{24}\)
= \(\frac{18 \times 18 \times 4 \times 3}{3}\)
= 18 × 18 × 4
= 36 cm
Slant height of the cone (l) = \(\sqrt{H^{2}+R^{2}}\)
= \(\sqrt{24^{2}+36^{2}}\)
= \(\sqrt{(12 \times 2)^{2}+(12 \times 3)^{2}}\)
= 12\(\sqrt{2^{2}+3^{2}}\)
= 12√13
Slant height of the cone = 12√13 cm
Radius of the cone = 36 cm

PART – IV

IV. Answer all the questions. [2 × 8 = 16]

Question 43.
(a) PQ is a chord of length 8 cm to a circle of radius 5 cm. The tangents at P and Q intersect at a point T. Find the length of the tangent TP.
Answer:

Let TR = y. Since, OT is perpendicular bisector of PQ
PR = QR = 4 cm
In ∆ORP, OP² = OR² + PR²
QR² = OP² – PR²
OR² = 5² – 4² = 25 – 16 = 9
OR = 3 cm
OT = OR + RT = 3 + y …. (1)
In ∆PRT , TP² = TR² + PR² …. (2)
and ∆OPT we have, OT² = TP² + OP²
OT² = (TR² + PR²) + OP² (substitute for TP² from (2)
(3 + y)² = y² + 4² + 5²(substitute for OT from (1))
9 + 6y + y² = y² + 16 + 5 There fore y = TR = \(\frac{16}{3}\)
6y = 41 – 9 we get y = \(\frac{16}{3}\)
From (2) TP² = TR² + PR²
TP² = \(\left(\frac{16}{3}\right)^{2}+4^{2}=\frac{256}{9}+16=\frac{400}{9} \mathrm{so}, \mathrm{TP}=\frac{20}{3} \mathrm{cm}\)

[OR]

(b) Draw a triangle ABC of base BC = 8 cm, ∠A = 60° and the bisector of ∠A meets BC at D such that BD = 6 cm.
Answer:


Step 1 : Draw a line segment BC = 8 cm.
Step 2 : At B, draw’ BE such that ∠CBE = 60° .
Step 3 : At B, draw BF such that ∠EBF = 90° .
Step 4 : Draw the perpendicular bisector to BC, w’hich intersects BF at O and BC at G.
Step 5 : With O as centre and OB as radius draw a circle.
Step 6 : From B, mark an arc of 6 cm on BC at D.
Step 7 : The perpendicular bisector intersects the circle at I. Joint ID.
Step 8 : ID produced meets the circle at A. Now join AB and AC.
Then ∆ABC is the required triangle.

Question 44.
(a) Draw the graph of y = x² + 3x – 4 and hence use it to solve x² + 3x – 4 = 0.
Answer:
Let y = x² + 3x – 4
(i) Draw the graph of y = x² + 3x – 4

(ii) Plot the points (-5, 6), (-4, 0), (-3, -4), (-2, -6), (-1, -6), (0, -4), (1, 0), (2, 6), (3, 14) on the graph using suitable scale.
(iii) Join the points by a free hand smooth curve.
The smooth curve is the graph of y = x² + 3x – 4
(iv) To solve x² + 3x – 4 = 0, subtract x² + 3x – 4 = 0 from y = x² + 3x – 4.
y = o
∴ The point of intersection with the X – axis is the solution set.
The solution set is -4 and 1.

[OR]

(b) A motor boat whose speed is 18 km/hr in still water takes 1 hour more to go to 24 km upstream than to return downstream to the same spot. Find the speed of the stream.
Answer:
Let the speed of the stream be “x” km/hr
Speed of the motor boat to go for upstream = (18 – x) km/hr
Speed of the motor boat to go for down stream = (18 + x) km/hr
Time taken to go for upstream = \(\frac{24}{18-x}\) hour
Time taken to go for down stream = \(\frac{24}{18-x}\)
By the given condition
\(\frac{24}{18-x}-\frac{24}{18+x}=1\)
\(\frac{24(18+x)-24(18-x)}{(18-x)(18+x)}=1\)
432 + 24x + 432 + 24x = (18 – x)(18 – x)
48x = 324 – x²
x² + 48x – 324 = 0
x² + 54x – 6x – 324 = 0
x(x + 54) – 6 (x + 54) = 0
(x + 54) (x – 6) = 0
x + 54 = 0 or x – 6 = 0
x = -54 or x = 6 (speed cannot be negative)
∴ Speed of the boat = 6 km/hr