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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 F_{1} = 1, F_{2} = 3 and F_{n} = F_{n-1}+ F_{n-2} then F_{5} 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 x^{2} + 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 10^{4} ≡ x (mod 19)

Answer:

10^{2} = 100 ≡ 5 (mod 19)

10^{4} = (10^{2})^{2} ≡ 5^{2} (mod 19)

10^{4} = 25 (mod 19)

10^{4} = 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, AB^{2} = AC^{2} + BC^{2}

x^{2} = 7^{2} + 4^{2} gives x^{2} = 49 + 16

x^{2} = 65. Hence, x = √65

The number √65 is between 8 and 8.1.

8^{2} = 64 < 65 < 65.61 = 8.1^{2}

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 = 4πr^{2}sq.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} = \(a \frac{\left(r^{n}-1\right)}{r-1}\)

S_{25} = \(\frac{1\left(3^{25}-1\right)}{3-1}\)

= \(\frac{3^{25}-1}{2}\)

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

t_{19} = -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 t_{6} : t_{8} = 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 t_{9} : t_{13}

t_{9 }: t_{13} = a + 8d : a + 12d

= 2d + 8d : 2d + 12d

= 10d : 14d

= 5 : 7

∴ t_{9 }: t_{13} = 5 : 7

Question 32.

The sum of first n, In and 3n terms of an A.P. are S_{1} S_{2} and S_{3} respectively. Prove that S_{3} = 3 (S_{2} – S_{1}).

Answer:

If S_{1} S_{7} and S_{3} 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 2x^{2} – x – 1 = 0 then form the equation whose roots are α^{2}β, β^{2}α.

Answer:

2x^{2} – 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 α^{2}β, β^{2}α

Summ of the roots α^{2}β + β^{2}α = αβ (α + β) = \(-\frac{1}{2}\left(\frac{1}{2}\right)=-\frac{1}{4}\)

Product of the roots (α^{2}β) × (β^{2}α) = α^{3}β^{3} = (αβ)^{3} =

\(\left(-\frac{1}{2}\right)^{3}=-\frac{1}{8}\)

The required equation is x^{2} – (Sum of the roots) x + (Product of the roots) = 0

\(x^{2}-\left(-\frac{1}{4}\right) x-\frac{1}{8}=0\) gives 8x^{2} + 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(AQ^{2} + BP^{2}) = 5AB^{2}.

Answer:

Since, ∆AQC is a right triangle at C, AQ^{2} = AC^{2} + QC^{2} ……. (1)

Also, ∆BPC is a right triangle at C, BP^{2} = BC^{2} + CP^{2} ……… (2)

From (1) and (2), AQ^{2} + BP^{2} = AC^{2} + QC^{2} + BC^{2} + CP^{2}

4(AQ^{2} + BP^{2}) = 4AC^{2} + 4QC^{2} + 4BC^{2} + 4CP^{2}

= 4AC^{2} + (2QC)^{2} + 4BC^{2} + (2CP)^{2}

= 4AC^{2} + BC^{2} + 4BC^{2} + AC^{2} (Since P and Q are mid points)

= 5(AC^{2} + BC^{2})

4(AQ^{2} + BP^{2}) = 5AB^{2} (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/cm^{3}.

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 (R^{3} – r^{3}) cu. cm

= \(\frac{4}{3} \times \frac{22}{7}\)(357.91 -343) = 62.48 cm^{3}

But, weight of brass in 1 cm^{3} = 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

x^{2} + (x+1)^{2} = 365

2x^{2} + 2x – 364 = 0

x^{2} + 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

\(\frac { 1 }{ 3 }\)πR^{2}H = πr^{2}

\(\frac { 1 }{ 3 }\) × R^{2} × 24 = r^{2}H

\(\frac { 1 }{ 3 }\) × R^{2} × 24 = 18 × 18 × 32

R^{2} = \(\frac{18 \times 18 \times 32 \times 3}{24}\)

= \(\frac{18 \times 18 \times 4 \times 3}{3}\)

= 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, OP^{2} = OR^{2} + PR^{2}

QR^{2} = OP^{2} – PR^{2}

OR^{2} = 5^{2} – 4^{2} = 25 – 16 = 9

OR = 3 cm

OT = OR + RT = 3 + y …. (1)

In ∆PRT , TP^{2} = TR^{2} + PR^{2} …. (2)

and ∆OPT we have, OT^{2} = TP^{2} + OP^{2}

OT^{2} = (TR^{2} + PR^{2}) + OP^{2} (substitute for TP^{2} from (2)

(3 + y)^{2} = y^{2} + 4^{2} + 5^{2}(substitute for OT from (1))

9 + 6y + y^{2} = y^{2} + 16 + 5 There fore y = TR = \(\frac{16}{3}\)

6y = 41 – 9 we get y = \(\frac{16}{3}\)

From (2) TP^{2} = TR^{2} + PR^{2}

TP^{2} = \(\left(\frac{16}{3}\right)^{2}+4^{2}=\frac{256}{9}+16=\frac{400}{9} \mathrm{so}, \mathrm{TP}=\frac{20}{3} \mathrm{cm}\)

[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 = x^{2} + 3x – 4 and hence use it to solve x^{2} + 3x – 4 = 0.

Answer:

Let y = x^{2} + 3x – 4

(i) Draw the graph of y = x^{2} + 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 = x^{2} + 3x – 4

(iv) To solve x^{2} + 3x – 4 = 0, subtract x^{2} + 3x – 4 = 0 from y = x^{2} + 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 – x^{2}

x^{2} + 48x – 324 = 0

x^{2} + 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