10th Standard Latest Syllabus Solutions, Official Question Papers, Answer Key, Tamil Nadu, Original Question Paper | Kanchipuram District | Mr. P. Kanagavel (English Medium)

Standard X Mathematics - First Mid Term Test 2024 | Solved Paper

Standard X Mathematics - Mid Term Test 2024 (Solved)

FIRST MID TERM TEST - 2024 Marks: 50 Time: 1.30 hrs

Part - I

I. Choose the correct answer:

  1. If \(A = \{a, b, p\}\), \(B = \{2, 3\}\), \(C = \{p, q, r, s\}\), then \(n[(A \cup C) \times B]\) is
    1. 8
    2. 20
    3. 12
    4. 16
  2. If \(f: A \to B\) is a bijective function and if \(n(B) = 7\), then \(n(A)\) is equal to
    1. 7
    2. 49
    3. 1
    4. 14
  3. \(f(x) = (x + 1)^3 - (x - 1)^3\) represents a function which is
    1. linear
    2. cubic
    3. reciprocal
    4. quadratic
  4. The least number that is divisible by all the numbers from 1 to 10 (both inclusive) is
    1. 2025
    2. 5220
    3. 5025
    4. 2520
  5. Given \(F_1 = 1, F_2 = 3, F_n = F_{n-1} + F_{n-2}\), then \(F_5\) is
    1. 3
    2. 5
    3. 8
    4. 11
  6. The solution of the system \(x + y - 3z = -6, -7y + 7z = 7, 3z = 9\) is
    1. x = 1, y = 2, z = 3
    2. x = -1, y = 2, z = 3
    3. x = -1, y = -2, z = 3
    4. x = -1, y = -2, z = -3
  7. In \(\Delta LMN\), \(\angle L = 60^\circ, \angle M = 50^\circ\). If \(\Delta LMN \sim \Delta PQR\), then the value of \(\angle R\) is
    1. \(40^\circ\)
    2. \(70^\circ\)
    3. \(30^\circ\)
    4. \(110^\circ\)

Part - II (5 x 2 = 10)

II. Answer any 5 questions. (Q.No. 14 is compulsory)

  1. If \(B \times A = \{(-2,3), (-2,4), (0,3), (0,4), (3,3), (3,4)\}\), find A and B.
  2. A relation 'f' is defined by \(f(x) = x^2 - 2\) where \(x \in \{-2, -1, 0, 3\}\).
    1. List the elements of f.
    2. Is f a function?
  3. 'a' and 'b' are two positive integers such that \(a^b \times b^a = 800\). Find 'a' and 'b'.
  4. Find the sum to infinity of \(9 + 3 + 1 + \dots\)
  5. Find \(a_6\) and \(a_{13}\) of the sequence whose \(n^{th}\) term is given by \(a_n = \frac{5n}{n+2}\).
  6. If \(\Delta ABC \sim \Delta DEF\) such that BC = 3 cm, EF = 4 cm and area of \(\Delta ABC = 54 \text{ cm}^2\), find the area of \(\Delta DEF\).
  7. Find the sum of \(1 + 8 + 27 + \dots + 1000\).

Part - III (5 x 5 = 25)

III. Answer any 5 questions. (Q.No. 21 is compulsory)

  1. Given \(A = \{1,2,3\}, B = \{2,3,5\}, C = \{3,4\}, D = \{1,3,5\}\), check if \((A \cap C) \times (B \cap D) = (A \times B) \cap (C \times D)\) is true.
  2. If \(f(x) = 2x + 3, g(x) = 1 - 2x\) and \(h(x) = 3x\), prove that \(fo(goh) = (fog)oh\).
  3. Use Euclid's division algorithm to find the HCF of 84, 90 and 120.
  4. Find the sum to n terms of the series: \(5 + 55 + 555 + \dots\)
  5. Solve the following system of linear equations in three variables: \(3x - 2y + z = 2, 2x + 3y - z = 5, x + y + z = 6\).
  6. The sum of three consecutive terms that are in A.P is 27 and their product is 288. Find the three terms.
  7. A function \(f: [-5, 9] \to \mathbb{R}\) is defined as follows: \[ f(x) = \begin{cases} 6x+1 & \text{if } -5 \le x < 2 \\ 5x^2-1 & \text{if } 2 \le x < 6 \\ 3x-4 & \text{if } 6 \le x \le 9 \end{cases} \] Find i) \(f(7) - f(1)\) and ii) \(\frac{2f(-2) - f(6)}{f(4) + f(-2)}\).

Part - IV (1 x 8 = 8)

IV. Answer the following question.

  1. a) Construct a triangle similar to a given triangle PQR with its sides equal to \(\frac{3}{5}\) of the corresponding sides of the triangle PQR (Scale factor \(\frac{3}{5} < 1\)).

    (OR)

    b) Construct a triangle similar to a given triangle PQR with its sides equal to \(\frac{7}{3}\) of the corresponding sides of the triangle PQR (Scale factor \(\frac{7}{3} > 1\)).

Solutions

Part - I Solutions

1. Solution

Given sets are \(A = \{a, b, p\}\), \(B = \{2, 3\}\), and \(C = \{p, q, r, s\}\).

Step 1: Find the union \(A \cup C\).
\(A \cup C = \{a, b, p\} \cup \{p, q, r, s\} = \{a, b, p, q, r, s\}\).
Step 2: Find the number of elements \(n(A \cup C)\).
The number of elements in \(A \cup C\) is \(n(A \cup C) = 6\).
Step 3: Find the number of elements \(n(B)\).
The number of elements in \(B\) is \(n(B) = 2\).
Step 4: Calculate \(n[(A \cup C) \times B]\).
The formula is \(n(X \times Y) = n(X) \times n(Y)\).
\(n[(A \cup C) \times B] = n(A \cup C) \times n(B) = 6 \times 2 = 12\).
Correct answer: c) 12

2. Solution

A bijective function (or one-to-one correspondence) is a function that is both injective (one-to-one) and surjective (onto). A key property of a bijective function between two finite sets is that the sets must have the same number of elements.

Given \(f: A \to B\) is a bijective function. This implies \(n(A) = n(B)\).

We are given \(n(B) = 7\). Therefore, \(n(A) = 7\).

Correct answer: a) 7

3. Solution

We need to simplify the expression for \(f(x)\).

Given: \(f(x) = (x + 1)^3 - (x - 1)^3\).

Using the algebraic identities \((a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3\) and \((a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3\):

\((x+1)^3 = x^3 + 3(x^2)(1) + 3(x)(1^2) + 1^3 = x^3 + 3x^2 + 3x + 1\)
\((x-1)^3 = x^3 - 3(x^2)(1) + 3(x)(1^2) - 1^3 = x^3 - 3x^2 + 3x - 1\)

Now, subtract the second expression from the first:

\(f(x) = (x^3 + 3x^2 + 3x + 1) - (x^3 - 3x^2 + 3x - 1)\)
\(f(x) = x^3 + 3x^2 + 3x + 1 - x^3 + 3x^2 - 3x + 1\)
\(f(x) = (x^3 - x^3) + (3x^2 + 3x^2) + (3x - 3x) + (1 + 1)\)
\(f(x) = 6x^2 + 2\)

This is a polynomial of degree 2, which represents a quadratic function.

Correct answer: d) quadratic

4. Solution

We need to find the Least Common Multiple (LCM) of the numbers from 1 to 10: LCM(1, 2, 3, 4, 5, 6, 7, 8, 9, 10).

Step 1: Write the prime factorization of each number.
1 = 1
2 = 2
3 = 3
4 = \(2^2\)
5 = 5
6 = 2 \(\times\) 3
7 = 7
8 = \(2^3\)
9 = \(3^2\)
10 = 2 \(\times\) 5
Step 2: Take the highest power of each prime factor.
Highest power of 2 is \(2^3\).
Highest power of 3 is \(3^2\).
Highest power of 5 is \(5^1\).
Highest power of 7 is \(7^1\).
Step 3: Multiply these highest powers together.
LCM = \(2^3 \times 3^2 \times 5 \times 7 = 8 \times 9 \times 5 \times 7 = 72 \times 35 = 2520\).
Correct answer: d) 2520

5. Solution

The sequence is defined by the recurrence relation \(F_n = F_{n-1} + F_{n-2}\) with initial values \(F_1 = 1\) and \(F_2 = 3\).

\(F_3 = F_{2} + F_{1} = 3 + 1 = 4\)

\(F_4 = F_{3} + F_{2} = 4 + 3 = 7\)

\(F_5 = F_{4} + F_{3} = 7 + 4 = 11\)

Correct answer: d) 11

6. Solution

We are given the system of equations:
(1) \(x + y - 3z = -6\)
(2) \(-7y + 7z = 7\)
(3) \(3z = 9\)

Step 1: Solve equation (3) for z.
\(3z = 9 \implies z = \frac{9}{3} = 3\).
Step 2: Substitute z = 3 into equation (2) to find y.
\(-7y + 7(3) = 7 \implies -7y + 21 = 7 \implies -7y = 7 - 21 \implies -7y = -14 \implies y = \frac{-14}{-7} = 2\).
Step 3: Substitute y = 2 and z = 3 into equation (1) to find x.
\(x + (2) - 3(3) = -6 \implies x + 2 - 9 = -6 \implies x - 7 = -6 \implies x = -6 + 7 \implies x = 1\).
The solution is x = 1, y = 2, z = 3. Correct answer: a)

7. Solution

In \(\Delta LMN\), the sum of angles is \(180^\circ\). We are given \(\angle L = 60^\circ\) and \(\angle M = 50^\circ\).

Step 1: Find the third angle, \(\angle N\).
\(\angle N = 180^\circ - (\angle L + \angle M) = 180^\circ - (60^\circ + 50^\circ) = 180^\circ - 110^\circ = 70^\circ\).
Step 2: Use the property of similar triangles.
Given \(\Delta LMN \sim \Delta PQR\). This means corresponding angles are equal.
\(\angle L = \angle P\)
\(\angle M = \angle Q\)
\(\angle N = \angle R\)
Step 3: Find \(\angle R\).
Since \(\angle N = 70^\circ\), then \(\angle R = 70^\circ\).
Correct answer: b) \(70^\circ\)

Part - II Solutions

8. Solution

Given the Cartesian product \(B \times A = \{(-2,3), (-2,4), (0,3), (0,4), (3,3), (3,4)\}\).

In a Cartesian product \(B \times A\), the first element of each ordered pair belongs to set B, and the second element belongs to set A.

Set B: The set of all first elements is \(\{-2, 0, 3\}\).
Set A: The set of all second elements is \(\{3, 4\}\).
A = {3, 4} and B = {-2, 0, 3}.

9. Solution

The relation is defined by \(f(x) = x^2 - 2\) for the domain \(x \in \{-2, -1, 0, 3\}\).

i) List the elements of f:
We calculate the output for each input value:
For \(x = -2\), \(f(-2) = (-2)^2 - 2 = 4 - 2 = 2\). The ordered pair is (-2, 2).
For \(x = -1\), \(f(-1) = (-1)^2 - 2 = 1 - 2 = -1\). The ordered pair is (-1, -1).
For \(x = 0\), \(f(0) = (0)^2 - 2 = 0 - 2 = -2\). The ordered pair is (0, -2).
For \(x = 3\), \(f(3) = (3)^2 - 2 = 9 - 2 = 7\). The ordered pair is (3, 7).
The set of elements of f is \(\{(-2, 2), (-1, -1), (0, -2), (3, 7)\}.
ii) Is f a function?
Yes, f is a function. For every element \(x\) in the domain \(\{-2, -1, 0, 3\}\), there is exactly one unique image (output value). No input value maps to more than one output value.
i) Elements: \(\{(-2, 2), (-1, -1), (0, -2), (3, 7)\}\)
ii) Yes, it is a function.

10. Solution

We are given \(a^b \times b^a = 800\), where 'a' and 'b' are positive integers.

Step 1: Prime factorize 800.
\(800 = 8 \times 100 = 2^3 \times 10^2 = 2^3 \times (2 \times 5)^2 = 2^3 \times 2^2 \times 5^2 = 2^5 \times 5^2\).
Step 2: Match the form \(a^b \times b^a\) with \(2^5 \times 5^2\).
Let's try setting \(a=2\) and \(b=5\).
Then \(a^b \times b^a = 2^5 \times 5^2\).
\(2^5 = 32\) and \(5^2 = 25\).
\(32 \times 25 = 800\). This matches the given equation.
The positive integers are a = 2 and b = 5 (or a = 5 and b = 2).

11. Solution

The series is \(9 + 3 + 1 + \dots\). This is a Geometric Progression (GP).

Step 1: Identify the first term (a) and common ratio (r).
First term, \(a = 9\).
Common ratio, \(r = \frac{3}{9} = \frac{1}{3}\).
Step 2: Check the condition for the sum to infinity.
The sum to infinity of a GP exists if \(|r| < 1\). Here, \(|1/3| < 1\), so the sum exists.
Step 3: Apply the formula for the sum to infinity.
The formula is \(S_\infty = \frac{a}{1 - r}\).
\(S_\infty = \frac{9}{1 - 1/3} = \frac{9}{2/3} = 9 \times \frac{3}{2} = \frac{27}{2}\).
The sum to infinity is \(\frac{27}{2}\) or 13.5.

12. Solution

The \(n^{th}\) term of the sequence is given by the formula \(a_n = \frac{5n}{n+2}\).

Find \(a_6\):
Substitute \(n=6\) into the formula.
\(a_6 = \frac{5(6)}{6+2} = \frac{30}{8} = \frac{15}{4}\).
Find \(a_{13}\):
Substitute \(n=13\) into the formula.
\(a_{13} = \frac{5(13)}{13+2} = \frac{65}{15} = \frac{13}{3}\).
\(a_6 = \frac{15}{4}\) and \(a_{13} = \frac{13}{3}\).

13. Solution

The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

Given: \(\Delta ABC \sim \Delta DEF\), BC = 3 cm, EF = 4 cm, Area(\(\Delta ABC\)) = 54 cm\(^2\).

Set up the ratio of areas to the ratio of sides.
\(\frac{\text{Area}(\Delta ABC)}{\text{Area}(\Delta DEF)} = \left(\frac{BC}{EF}\right)^2\)
Substitute the given values.
\(\frac{54}{\text{Area}(\Delta DEF)} = \left(\frac{3}{4}\right)^2 = \frac{9}{16}\)
Solve for Area(\(\Delta DEF\)).
\(\text{Area}(\Delta DEF) = 54 \times \frac{16}{9}\)
\(\text{Area}(\Delta DEF) = 6 \times 16 = 96\) cm\(^2\).
The area of \(\Delta DEF\) is 96 cm\(^2\).

14. Solution (Compulsory)

The series is \(1 + 8 + 27 + \dots + 1000\).

Step 1: Recognize the pattern.
The terms are cubes of natural numbers: \(1^3 + 2^3 + 3^3 + \dots + 10^3\). The last term is \(1000 = 10^3\), so we are summing the cubes of the first 10 natural numbers (\(n=10\)).
Step 2: Apply the formula for the sum of the first n cubes.
The formula is \(\sum_{k=1}^{n} k^3 = \left(\frac{n(n+1)}{2}\right)^2\).
Step 3: Substitute n = 10 and calculate.
Sum = \(\left(\frac{10(10+1)}{2}\right)^2 = \left(\frac{10 \times 11}{2}\right)^2 = \left(\frac{110}{2}\right)^2 = (55)^2\).
\(55^2 = 3025\).
The sum of the series is 3025.

Part - III Solutions

15. Solution

Given \(A = \{1,2,3\}, B = \{2,3,5\}, C = \{3,4\}, D = \{1,3,5\}\).

We need to check if \((A \cap C) \times (B \cap D) = (A \times B) \cap (C \times D)\).

Step 1: Calculate the Left Hand Side (LHS).
\(A \cap C = \{1,2,3\} \cap \{3,4\} = \{3\}\).
\(B \cap D = \{2,3,5\} \cap \{1,3,5\} = \{3,5\}\).
LHS = \((A \cap C) \times (B \cap D) = \{3\} \times \{3,5\} = \{(3,3), (3,5)\}\).
Step 2: Calculate the Right Hand Side (RHS).
\(A \times B = \{(1,2), (1,3), (1,5), (2,2), (2,3), (2,5), (3,2), (3,3), (3,5)\}\).
\(C \times D = \{(3,1), (3,3), (3,5), (4,1), (4,3), (4,5)\}\).
RHS = \((A \times B) \cap (C \times D)\). We find the common ordered pairs:
RHS = \(\{(3,3), (3,5)\}\).
Step 3: Compare LHS and RHS.
LHS = \(\{(3,3), (3,5)\}\) and RHS = \(\{(3,3), (3,5)\}\). Since LHS = RHS, the statement is true.
The statement \((A \cap C) \times (B \cap D) = (A \times B) \cap (C \times D)\) is true.

16. Solution

Given \(f(x) = 2x+3, g(x) = 1-2x, h(x) = 3x\). We need to prove \(fo(goh) = (fog)oh\).

Step 1: Calculate the LHS: \(fo(goh)\)
First, find \(goh(x)\):
\(goh(x) = g(h(x)) = g(3x) = 1 - 2(3x) = 1 - 6x\).
Now, find \(fo(goh)(x)\):
\(fo(goh)(x) = f(goh(x)) = f(1-6x) = 2(1-6x) + 3 = 2 - 12x + 3 = 5 - 12x\).
Step 2: Calculate the RHS: \((fog)oh\)
First, find \(fog(x)\):
\(fog(x) = f(g(x)) = f(1-2x) = 2(1-2x) + 3 = 2 - 4x + 3 = 5 - 4x\).
Now, find \((fog)oh(x)\):
\((fog)oh(x) = fog(h(x)) = fog(3x) = 5 - 4(3x) = 5 - 12x\).
Step 3: Compare LHS and RHS.
LHS = \(5 - 12x\) and RHS = \(5 - 12x\). Since LHS = RHS, the property is proven.
Hence, \(fo(goh) = (fog)oh\) is proved.

17. Solution

We use Euclid's division algorithm to find the HCF of 84, 90, and 120.

Step 1: Find HCF of 84 and 90.
\(90 = 84 \times 1 + 6\)
\(84 = 6 \times 14 + 0\)
The last non-zero remainder is 6. So, HCF(84, 90) = 6.
Step 2: Find HCF of the result (6) and the remaining number (120).
\(120 = 6 \times 20 + 0\)
The last non-zero remainder is 6. So, HCF(6, 120) = 6.
The HCF of 84, 90, and 120 is 6.

18. Solution

Let \(S_n = 5 + 55 + 555 + \dots\) to n terms.

Step 1: Factor out 5.
\(S_n = 5(1 + 11 + 111 + \dots \text{ to n terms})\)
Step 2: Multiply and divide by 9.
\(S_n = \frac{5}{9}(9 + 99 + 999 + \dots \text{ to n terms})\)
Step 3: Express terms as powers of 10.
\(S_n = \frac{5}{9}[(10-1) + (100-1) + (1000-1) + \dots \text{ to n terms}]\)
Step 4: Separate the terms into two series.
\(S_n = \frac{5}{9}[(10 + 10^2 + 10^3 + \dots \text{ to n terms}) - (1 + 1 + 1 + \dots \text{ n times})]\)
Step 5: Sum the GP and the constant series.
The first part is a GP with \(a=10, r=10\). Its sum is \(\frac{a(r^n-1)}{r-1} = \frac{10(10^n-1)}{10-1} = \frac{10}{9}(10^n-1)\).
The second part is simply \(n\).
Step 6: Combine and simplify.
\(S_n = \frac{5}{9}\left[\frac{10}{9}(10^n-1) - n\right]\)
\(S_n = \frac{50}{81}(10^n-1) - \frac{5n}{9}\)
The sum to n terms is \(S_n = \frac{50}{81}(10^n - 1) - \frac{5n}{9}\).

19. Solution

(1) \(3x - 2y + z = 2\)
(2) \(2x + 3y - z = 5\)
(3) \(x + y + z = 6\)

Step 1: Eliminate 'z' using equations (1) and (2).
Add (1) and (2): \((3x-2y+z) + (2x+3y-z) = 2+5\)
\(\implies 5x + y = 7\) --- (4)
Step 2: Eliminate 'z' using equations (2) and (3).
Add (2) and (3): \((2x+3y-z) + (x+y+z) = 5+6\)
\(\implies 3x + 4y = 11\) --- (5)
Step 3: Solve the new system of two equations (4) and (5).
From (4), \(y = 7 - 5x\). Substitute this into (5):
\(3x + 4(7 - 5x) = 11\)
\(3x + 28 - 20x = 11\)
\(-17x = 11 - 28 \implies -17x = -17 \implies x = 1\).
Step 4: Substitute x=1 back to find y.
Using \(y = 7 - 5x\): \(y = 7 - 5(1) = 2\).
Step 5: Substitute x=1 and y=2 back into original equation (3) to find z.
\(1 + 2 + z = 6 \implies 3 + z = 6 \implies z = 3\).
The solution is x = 1, y = 2, z = 3.

20. Solution

Let the three consecutive terms in an Arithmetic Progression (A.P) be \(a-d, a, a+d\).

Step 1: Use the sum of the terms.
Sum = \((a-d) + a + (a+d) = 27\)
\(3a = 27 \implies a = 9\).
Step 2: Use the product of the terms.
Product = \((a-d)(a)(a+d) = 288\).
Substitute \(a=9\): \((9-d)(9)(9+d) = 288\).
\((9-d)(9+d) = \frac{288}{9} = 32\).
Step 3: Solve for the common difference 'd'.
Using the difference of squares formula: \(9^2 - d^2 = 32\).
\(81 - d^2 = 32 \implies d^2 = 81 - 32 = 49 \implies d = \pm 7\).
Step 4: Find the three terms.
If \(d=7\), the terms are \(9-7, 9, 9+7\), which are 2, 9, 16.
If \(d=-7\), the terms are \(9-(-7), 9, 9-7\), which are 16, 9, 2.
In both cases, the set of terms is the same.
The three terms are 2, 9, and 16.

21. Solution (Compulsory)

The function is defined as: \[ f(x) = \begin{cases} 6x+1 & \text{if } -5 \le x < 2 \\ 5x^2-1 & \text{if } 2 \le x < 6 \\ 3x-4 & \text{if } 6 \le x \le 9 \end{cases} \]

i) Find \(f(7) - f(1)\)
To find \(f(7)\), we note that \(6 \le 7 \le 9\), so we use \(f(x) = 3x-4\). \(f(7) = 3(7) - 4 = 21 - 4 = 17\).
To find \(f(1)\), we note that \(-5 \le 1 < 2\), so we use \(f(x) = 6x+1\). \(f(1) = 6(1) + 1 = 7\).
\(f(7) - f(1) = 17 - 7 = 10\).
ii) Find \(\frac{2f(-2) - f(6)}{f(4) + f(-2)}\)
First, find the value of each function term:
\(f(-2)\): Since \(-5 \le -2 < 2\), use \(f(x) = 6x+1\). \(f(-2) = 6(-2)+1 = -12+1 = -11\).
\(f(6)\): Since \(6 \le 6 \le 9\), use \(f(x) = 3x-4\). \(f(6) = 3(6)-4 = 18-4 = 14\).
\(f(4)\): Since \(2 \le 4 < 6\), use \(f(x) = 5x^2-1\). \(f(4) = 5(4^2)-1 = 5(16)-1 = 80-1 = 79\).
Now substitute these values into the expression:
Numerator: \(2f(-2) - f(6) = 2(-11) - 14 = -22 - 14 = -36\).
Denominator: \(f(4) + f(-2) = 79 + (-11) = 68\).
Fraction: \(\frac{-36}{68} = \frac{-9 \times 4}{17 \times 4} = -\frac{9}{17}\).
i) \(f(7) - f(1) = 10\)
ii) \(\frac{2f(-2) - f(6)}{f(4) + f(-2)} = -\frac{9}{17}\)

Part - IV Solutions

22. Solution (Construction)

These questions require geometric construction. The steps are described below.

a) Construction of a similar triangle with scale factor \(\frac{3}{5} < 1\).
  1. Draw the given triangle \(\Delta PQR\).
  2. Draw a ray QX starting from Q, making an acute angle with the side QR.
  3. Mark 5 (the larger of 3 and 5) equally spaced points on the ray QX, namely \(Q_1, Q_2, Q_3, Q_4, Q_5\).
  4. Join the last point, \(Q_5\), to R.
  5. From the 3rd point (for the numerator), \(Q_3\), draw a line parallel to \(Q_5R\). This line will intersect QR at a new point, R'.
  6. From R', draw a line parallel to PR. This line will intersect PQ at a new point, P'.
  7. The triangle \(\Delta P'QR'\) is the required triangle, which is similar to \(\Delta PQR\) and its sides are \(\frac{3}{5}\) of the corresponding sides of \(\Delta PQR\).
b) Construction of a similar triangle with scale factor \(\frac{7}{3} > 1\).
  1. Draw the given triangle \(\Delta PQR\).
  2. Extend the sides QP and QR.
  3. Draw a ray QX starting from Q, making an acute angle with the side QR.
  4. Mark 7 (the larger of 7 and 3) equally spaced points on the ray QX, namely \(Q_1, Q_2, \dots, Q_7\).
  5. Join the 3rd point (for the denominator), \(Q_3\), to R.
  6. From the 7th point (for the numerator), \(Q_7\), draw a line parallel to \(Q_3R\). This line will intersect the extended side QR at a new point, R'.
  7. From R', draw a line parallel to PR. This line will intersect the extended side QP at a new point, P'.
  8. The triangle \(\Delta P'QR'\) is the required triangle, which is similar to \(\Delta PQR\) and its sides are \(\frac{7}{3}\) of the corresponding sides of \(\Delta PQR\).