OMTEX CLASSES: 10th Standard Latest Syllabus Solutions, Official Question Papers, Answer Key, Tamil Nadu, Paper With Answer Key | Thiruvallur District

Std. X Maths - First Mid Term Test 2024 - Solved Paper

Time: 1.30 hrs Marks: 50

Part - I

Choose the correct answer. (7 x 1 = 7)

1. If A = {a,b,p}, B = {2,3}, C = {p,q,r,s}, then n[(A∪C) × B] is

a) 8 b) 20 c) 12 d) 16

Step 1: Find the union of sets A and C.

$A = \{a, b, p\}$

$C = \{p, q, r, s\}$

$A \cup C = \{a, b, p\} \cup \{p, q, r, s\} = \{a, b, p, q, r, s\}$

Step 2: Find the number of elements in (A∪C).

$n(A \cup C) = 6$

Step 3: Find the number of elements in B.

$B = \{2, 3\}$

$n(B) = 2$

Step 4: Calculate n[(A∪C) × B].

The number of elements in the Cartesian product of two sets is the product of the number of elements in each set.

$n[(A \cup C) \times B] = n(A \cup C) \times n(B) = 6 \times 2 = 12$

Correct answer: c) 12

2. If f : A→B is a bijective function and if n(B) = 7, then n(A) is equal to

a) 7 b) 49 c) 1 d) 14

A bijective function is a function that is both one-to-one (injective) and onto (surjective).

One-to-one (Injective): Every element in the domain A maps to a unique element in the codomain B.
Onto (Surjective): Every element in the codomain B is mapped to by at least one element from the domain A.

For a function between two finite sets A and B to be bijective, the number of elements in both sets must be equal.

Given $n(B) = 7$ and f is bijective, it must be that $n(A) = n(B)$.

Therefore, $n(A) = 7$.

Correct answer: a) 7

3. f(x) = (x + 1)³ – (x – 1)³ represents a function which is

a) linear b) cubic c) reciprocal d) quadratic

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

We use the algebraic identities:

$(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$

$(a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$

Let a = x and b = 1.

$f(x) = (x^3 + 3x^2(1) + 3x(1)^2 + 1^3) - (x^3 - 3x^2(1) + 3x(1)^2 - 1^3)$

$f(x) = (x^3 + 3x^2 + 3x + 1) - (x^3 - 3x^2 + 3x - 1)$

$f(x) = x^3 + 3x^2 + 3x + 1 - x^3 + 3x^2 - 3x + 1$

Combine like terms:

$f(x) = (x^3 - x^3) + (3x^2 + 3x^2) + (3x - 3x) + (1 + 1)$

$f(x) = 6x^2 + 2$

The simplified function is a polynomial of degree 2. This is the definition of a quadratic function.

Correct answer: d) quadratic

4. The least number that is divisible by all the numbers from 1 to 10 (both inclusive) is

a) 2025 b) 5220 c) 5025 d) 2520

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

Step 1: Find 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: Find the highest power of each prime factor.

The prime factors involved are 2, 3, 5, and 7.

Highest power of 2 is $2^3$ (from 8).
Highest power of 3 is $3^2$ (from 9).
Highest power of 5 is $5^1$ (from 5 and 10).
Highest power of 7 is $7^1$ (from 7).

Step 3: Multiply these highest powers together.

$LCM = 2^3 \times 3^2 \times 5^1 \times 7^1 = 8 \times 9 \times 5 \times 7$

$LCM = 72 \times 35 = 2520$

Correct answer: d) 2520

5. Given F₁ = 1, F₂ = 3, Fₙ = Fₙ₋₁ + Fₙ₋₂ then F₅ is

a) 3 b) 5 c) 8 d) 11

We are given a recurrence relation $F_n = F_{n-1} + F_{n-2}$ with initial values $F_1 = 1$ and $F_2 = 3$. We need to find $F_5$.

$F_1 = 1$

$F_2 = 3$

For n=3: $F_3 = F_{3-1} + F_{3-2} = F_2 + F_1 = 3 + 1 = 4$

For n=4: $F_4 = F_{4-1} + F_{4-2} = F_3 + F_2 = 4 + 3 = 7$

For n=5: $F_5 = F_{5-1} + F_{5-2} = F_4 + F_3 = 7 + 4 = 11$

Correct answer: d) 11

6. The solution of the system x + y - 3z = -6, -7y + 7z = 7, 3z = 9 is

a) x=1, y=2, z=3 b) x=-1, y=2, z=3 c) x=-1, y=-2, z=3 d) x=-1, y=-2, z=-3

We have a system of three linear equations:

$x + y - 3z = -6$
$-7y + 7z = 7$
$3z = 9$

We can solve this system by back-substitution, starting from the last equation.

Step 1: Solve for z from Equation (3).

$3z = 9 \implies z = \frac{9}{3} = 3$

Step 2: Substitute z = 3 into Equation (2) to find y.

$-7y + 7(3) = 7$

$-7y + 21 = 7$

$-7y = 7 - 21$

$-7y = -14$

$y = \frac{-14}{-7} = 2$

Step 3: Substitute y = 2 and z = 3 into Equation (1) to find x.

$x + (2) - 3(3) = -6$

$x + 2 - 9 = -6$

$x - 7 = -6$

$x = -6 + 7 = 1$

The solution is x=1, y=2, z=3.

Correct answer: a) x = 1, y = 2, z = 3

7. In ΔLMN, ∠L = 60°, ∠M = 50°. If ΔLMN ~ ΔPQR, then the value of ∠R is

a) 40° b) 70° c) 30° d) 110°

Step 1: Find the third angle in ΔLMN.

The sum of angles in a triangle is 180°.

$\angle L + \angle M + \angle N = 180°$

$60° + 50° + \angle N = 180°$

$110° + \angle N = 180°$

$\angle N = 180° - 110° = 70°$

Step 2: Use the property of similar triangles.

Given that $\Delta LMN \sim \Delta PQR$, the corresponding angles are equal.

$\angle L = \angle P$

$\angle M = \angle Q$

$\angle N = \angle R$

Since we found $\angle N = 70°$, then $\angle R$ must also be 70°.

Correct answer: b) 70°

Part - II

Answer any 5 questions. (Q.No.14 is compulsory) (5 x 2 = 10)

8. If B × A = {(-2,3), (-2,4), (0,3), (0,4), (3,3), (3,4)}, find A and B.

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

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

To find set B: Collect all the first elements of the ordered pairs.

The first elements are -2, 0, and 3.

So, $B = \{-2, 0, 3\}$.

To find set A: Collect all the second elements of the ordered pairs.

The second elements are 3 and 4.

So, $A = \{3, 4\}$.

A = {3, 4} and B = {-2, 0, 3}

9. A relation 'f' is defined by f(x) = x² – 2 where x ∈ {−2, -1, 0, 3}.

i) List the elements of f

ii) Is f a function?

i) List the elements of f

We calculate the output f(x) for each value of x in the domain {−2, -1, 0, 3}.

For x = -2: $f(-2) = (-2)^2 - 2 = 4 - 2 = 2$
For x = -1: $f(-1) = (-1)^2 - 2 = 1 - 2 = -1$
For x = 0: $f(0) = (0)^2 - 2 = 0 - 2 = -2$
For x = 3: $f(3) = (3)^2 - 2 = 9 - 2 = 7$

The elements of f are the set of ordered pairs (x, f(x)):

f = {(-2, 2), (-1, -1), (0, -2), (3, 7)}

ii) Is f a function?

A relation is a function if every element in the domain has exactly one image in the codomain.

In this case, each element of the domain {−2, -1, 0, 3} is mapped to a single, unique output value.

-2 maps to 2
-1 maps to -1
0 maps to -2
3 maps to 7

Since each input has only one output, the relation f is a function.

Yes, f is a function.

10. 'a' and 'b' are two positive integers such that $a^b \times b^a = 800$. Find 'a' and 'b'.

We are given the equation $a^b \times b^a = 800$.

Step 1: Find the prime factorization of 800.

$800 = 8 \times 100 = 8 \times 10^2 = 2^3 \times (2 \times 5)^2 = 2^3 \times 2^2 \times 5^2 = 2^{3+2} \times 5^2 = 2^5 \times 5^2$

Step 2: Compare the factored form with the given equation.

$a^b \times b^a = 2^5 \times 5^2$

By matching the bases and exponents, we can hypothesize a solution.

Let's try setting $a = 2$ and $b = 5$.

Substitute these values back into the expression $a^b \times b^a$:

$2^5 \times 5^2 = 32 \times 25 = 800$

This matches the given value.

Alternatively, if we try $a = 5$ and $b = 2$, we get:

$5^2 \times 2^5 = 25 \times 32 = 800$

This also works. Since the question does not specify an order for a and b, both solutions are valid.

a = 2, b = 5 (or a = 5, b = 2)

11. Find the sum to infinity of 9 + 3 + 1 + ...

The given series is a Geometric Progression (GP).

Step 1: Identify the first term (a) and the common ratio (r).

First term, $a = 9$.

Common ratio, $r = \frac{\text{second term}}{\text{first term}} = \frac{3}{9} = \frac{1}{3}$. We can check this with the next pair: $\frac{1}{3}$.

Step 2: Check if the sum to infinity exists.

The sum to infinity of a GP exists only if the absolute value of the common ratio is less than 1 (i.e., $|r| < 1$).

Here, $|r| = |\frac{1}{3}| = \frac{1}{3}$, which is less than 1. So the sum exists.

Step 3: Apply the formula for the sum to infinity.

The formula is $S_\infty = \frac{a}{1 - r}$.

$S_\infty = \frac{9}{1 - \frac{1}{3}} = \frac{9}{\frac{3}{3} - \frac{1}{3}} = \frac{9}{\frac{2}{3}}$

$S_\infty = 9 \times \frac{3}{2} = \frac{27}{2}$

The sum to infinity is $\frac{27}{2}$ or 13.5

12. Find $a_6$ and $a_{13}$ of the sequence whose nth term is given by $a_n = \frac{5n}{n+2}$

We are given the formula for the nth term of a sequence: $a_n = \frac{5n}{n+2}$.

To find $a_6$:

Substitute n = 6 into the formula.

$a_6 = \frac{5(6)}{6+2} = \frac{30}{8}$

Simplify the fraction by dividing the numerator and denominator by their greatest common divisor, which is 2.

$a_6 = \frac{15}{4}$

To find $a_{13}$:

Substitute n = 13 into the formula.

$a_{13} = \frac{5(13)}{13+2} = \frac{65}{15}$

Simplify the fraction by dividing the numerator and denominator by their greatest common divisor, which is 5.

$a_{13} = \frac{13}{3}$

$a_6 = \frac{15}{4}$ and $a_{13} = \frac{13}{3}$

13. If ΔABC ~ ΔDEF such that BC = 3 cm, EF = 4 cm and area of ΔABC = 54 cm², Find the area of ΔDEF.

We use the theorem that states that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

Given $\Delta ABC \sim \Delta DEF$, the corresponding sides are BC and EF.

The formula is: $\frac{\text{Area}(\Delta ABC)}{\text{Area}(\Delta DEF)} = \left(\frac{BC}{EF}\right)^2$

Substitute the given values:

$\frac{54}{\text{Area}(\Delta DEF)} = \left(\frac{3}{4}\right)^2$

$\frac{54}{\text{Area}(\Delta DEF)} = \frac{9}{16}$

Now, solve for Area(ΔDEF) by cross-multiplication.

$\text{Area}(\Delta DEF) = 54 \times \frac{16}{9}$

$\text{Area}(\Delta DEF) = \frac{54}{9} \times 16 = 6 \times 16 = 96$

The area of ΔDEF is 96 cm².

14. (Compulsory) Find the sum of 1 + 8 + 27 + ... + 1000.

First, we recognize the pattern in the series.

$1 = 1^3$

$8 = 2^3$

$27 = 3^3$

... and the last term is $1000 = 10^3$.

So, the series is the sum of the cubes of the first 10 natural numbers:

$S = 1^3 + 2^3 + 3^3 + \dots + 10^3$

We use the formula for the sum of the first n cubes: $\sum_{k=1}^{n} k^3 = \left[\frac{n(n+1)}{2}\right]^2$.

In this case, n = 10.

Substitute n = 10 into the formula:

$S = \left[\frac{10(10+1)}{2}\right]^2 = \left[\frac{10 \times 11}{2}\right]^2$

$S = \left[5 \times 11\right]^2 = (55)^2$

$S = 3025$

The sum of the series is 3025.

Part - III

Answer any 5 questions. (Q.No.21 is compulsory) (5 x 5 = 25)

15. Given A={1,2,3}, B={2,3,5}, C={3,4}, D={1,3,5}, check if (A∩C) × (B∩D) = (A × B) ∩ (C × D) is true.

We need to evaluate both the Left-Hand Side (LHS) and the Right-Hand Side (RHS) to check if they are equal.

LHS: (A∩C) × (B∩D)

Step 1: Find A ∩ C

$A \cap C = \{1,2,3\} \cap \{3,4\} = \{3\}$

Step 2: Find B ∩ D

$B \cap D = \{2,3,5\} \cap \{1,3,5\} = \{3,5\}$

Step 3: Find the Cartesian product

$(A \cap C) \times (B \cap D) = \{3\} \times \{3,5\} = \{(3,3), (3,5)\}$

RHS: (A × B) ∩ (C × D)

Step 1: Find A × B

$A \times B = \{(1,2), (1,3), (1,5), (2,2), (2,3), (2,5), (3,2), (3,3), (3,5)\}$

Step 2: Find C × D

$C \times D = \{(3,1), (3,3), (3,5), (4,1), (4,3), (4,5)\}$

Step 3: Find the intersection of the two sets

$(A \times B) \cap (C \times D) = \{(3,3), (3,5)\}$

Conclusion:

Since LHS = $\{(3,3), (3,5)\}$ and RHS = $\{(3,3), (3,5)\}$, the statement is true.

LHS = RHS, so the statement is true.

16. f(x) = 2x + 3, g(x) = 1 - 2x and h(x) = 3x, prove that fo(goh) = (fog)oh

This question asks us to prove the associative property of function composition.

LHS: fo(goh)

Step 1: Find the composite function goh(x).

$goh(x) = g(h(x)) = g(3x)$

Since $g(x) = 1 - 2x$, we replace x with 3x:

$g(3x) = 1 - 2(3x) = 1 - 6x$

So, $goh(x) = 1 - 6x$.

Step 2: Find fo(goh)(x).

$fo(goh)(x) = f(goh(x)) = f(1 - 6x)$

Since $f(x) = 2x + 3$, we replace x with (1 - 6x):

$f(1 - 6x) = 2(1 - 6x) + 3 = 2 - 12x + 3 = 5 - 12x$

So, $LHS = 5 - 12x$.

RHS: (fog)oh

Step 1: Find the composite function fog(x).

$fog(x) = f(g(x)) = f(1 - 2x)$

Since $f(x) = 2x + 3$, we replace x with (1 - 2x):

$f(1 - 2x) = 2(1 - 2x) + 3 = 2 - 4x + 3 = 5 - 4x$

So, $fog(x) = 5 - 4x$.

Step 2: Find (fog)oh(x).

$(fog)oh(x) = fog(h(x)) = fog(3x)$

Since $fog(x) = 5 - 4x$, we replace x with 3x:

$fog(3x) = 5 - 4(3x) = 5 - 12x$

So, $RHS = 5 - 12x$.

Conclusion:

Since LHS = 5 - 12x and RHS = 5 - 12x, we have proved that fo(goh) = (fog)oh.

Hence Proved.

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

Euclid's division algorithm is based on the principle that the greatest common divisor of two numbers does not change if the larger number is replaced by its difference with the smaller number. We use the lemma $a = bq + r$, where HCF(a, b) = HCF(b, r).

Step 1: Find the HCF of the first two numbers, 84 and 90.

Let a = 90 and b = 84.

$90 = 84 \times 1 + 6$. The remainder is 6.

Now, let a = 84 and b = 6.

$84 = 6 \times 14 + 0$. The remainder is 0.

The last non-zero remainder is the HCF. So, HCF(84, 90) = 6.

Step 2: Find the HCF of the result from Step 1 and the third number, 120.

We need to find HCF(6, 120).

Let a = 120 and b = 6.

$120 = 6 \times 20 + 0$. The remainder is 0.

The last non-zero remainder (in this case, the divisor b) is the HCF. So, HCF(6, 120) = 6.

The HCF of 84, 90, and 120 is 6.

18. Find the sum to n terms of the series: 5 + 55 + 555 + ...

Let $S_n$ be the sum of the first n terms of the series.

$S_n = 5 + 55 + 555 + \dots + n \text{ terms}$

Step 1: Factor out the common term.

$S_n = 5(1 + 11 + 111 + \dots + n \text{ terms})$

Step 2: Multiply and divide by 9 to create a pattern of 9s.

$S_n = \frac{5}{9} (9 \times (1 + 11 + 111 + \dots)) = \frac{5}{9} (9 + 99 + 999 + \dots)$

Step 3: Express each term as a power of 10 minus 1.

$S_n = \frac{5}{9} ((10 - 1) + (100 - 1) + (1000 - 1) + \dots + n \text{ terms})$

$S_n = \frac{5}{9} ((10^1 - 1) + (10^2 - 1) + (10^3 - 1) + \dots + (10^n - 1))$

Step 4: Separate the terms into two groups.

$S_n = \frac{5}{9} [ (10 + 10^2 + 10^3 + \dots + 10^n) - (1 + 1 + 1 + \dots + n \text{ times}) ]$

Step 5: Calculate the sum of each group.

The first group is a Geometric Progression (GP) with first term $a=10$, common ratio $r=10$, and $n$ terms. The sum is $\frac{a(r^n - 1)}{r-1} = \frac{10(10^n - 1)}{10-1} = \frac{10}{9}(10^n - 1)$.

The second group is the sum of 1 repeated n times, which is simply $n$.

Step 6: Combine the results.

$S_n = \frac{5}{9} \left[ \frac{10}{9}(10^n - 1) - n \right]$

Distribute the outer term to get the final form:

$S_n = \frac{50}{81}(10^n - 1) - \frac{5n}{9}$

19. Solve the following system of linear equations in three variables :

3x - 2y + z = 2 ... (1)

2x + 3y - z = 5 ... (2)

x + y + z = 6 ... (3)

We use the method of elimination to solve the system.

Step 1: Eliminate 'z' using equations (1) and (2).

Add equation (1) and equation (2):

$(3x - 2y + z) + (2x + 3y - z) = 2 + 5$

$5x + y = 7$ ... (4)

Step 2: Eliminate 'z' using equations (2) and (3).

Add equation (2) and equation (3):

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

$3x + 4y = 11$ ... (5)

Step 3: Solve the new system of two equations (4) and (5).

We have:

5x + y = 7 ... (4)

3x + 4y = 11 ... (5)

From equation (4), we can express y: $y = 7 - 5x$.

Substitute this into equation (5):

$3x + 4(7 - 5x) = 11$

$3x + 28 - 20x = 11$

$-17x = 11 - 28$

$-17x = -17 \implies x = 1$

Step 4: Find 'y' by substituting x=1 back into equation (4).

$5(1) + y = 7 \implies 5 + y = 7 \implies y = 2$

Step 5: Find 'z' by substituting x=1 and y=2 back into equation (3).

$(1) + (2) + z = 6 \implies 3 + z = 6 \implies z = 3$

The solution is x = 1, y = 2, z = 3.

20. The sum of three consecutive terms that are in A.P is 27 and their product is 288. Find the three terms.

Let the three consecutive terms in an Arithmetic Progression (A.P.) be $a - d$, $a$, and $a + d$, where 'a' is the middle term and 'd' is the common difference.

Step 1: Use the sum to find 'a'.

Sum = $(a - d) + a + (a + d) = 27$

$3a = 27 \implies a = \frac{27}{3} = 9$

Step 2: Use the product to find 'd'.

Product = $(a - d) \times a \times (a + d) = 288$

Substitute $a = 9$:

$(9 - d) \times 9 \times (9 + d) = 288$

Divide both sides by 9:

$(9 - d)(9 + d) = \frac{288}{9} = 32$

Using the difference of squares formula $(x-y)(x+y) = x^2 - y^2$:

$9^2 - d^2 = 32$

$81 - d^2 = 32$

$d^2 = 81 - 32 = 49$

$d = \pm\sqrt{49} \implies d = \pm 7$

Step 3: Find the terms for both values of d.

Case 1: If d = 7

The terms are $a - d, a, a + d$ = $9 - 7, 9, 9 + 7$ = 2, 9, 16.

Case 2: If d = -7

The terms are $a - d, a, a + d$ = $9 - (-7), 9, 9 + (-7)$ = 16, 9, 2.

In both cases, the set of terms is the same.

The three terms are 2, 9, and 16.

21. (Compulsory) A function f: [-5, 9] → 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) ii) $\frac{2f(-2)-f(6)}{f(4)+f(-2)}$

We first need to evaluate the function at the required points by choosing the correct definition based on the value of x.

i) Find f(7) - f(1)

To find f(7): Since $6 \le 7 \le 9$, we use the third definition: $f(x) = 3x - 4$.

$f(7) = 3(7) - 4 = 21 - 4 = 17$.

To find f(1): Since $-5 \le 1 < 2$, we use the first definition: $f(x) = 6x + 1$.

$f(1) = 6(1) + 1 = 7$.

$f(7) - f(1) = 17 - 7 = 10$.

i) f(7) - f(1) = 10

ii) Find $\frac{2f(-2)-f(6)}{f(4)+f(-2)}$

We need to find f(-2), f(6), and f(4).

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: $f(4) + f(-2) = 79 + (-11) = 68$.

The expression becomes $\frac{-36}{68}$. Simplify by dividing the numerator and denominator by their greatest common divisor, 4.

$\frac{-36 \div 4}{68 \div 4} = \frac{-9}{17}$

ii) $\frac{2f(-2)-f(6)}{f(4)+f(-2)} = -\frac{9}{17}$

Part - IV

Answer the following question. (1 x 8 = 8)

22. 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 $\frac{7}{3} > 1$).

Here are the steps for both constructions.

a) Construction for Scale Factor $\frac{3}{5} < 1$

Draw the Base Triangle: Construct any triangle ΔPQR.
Draw an Acute Ray: From vertex Q, draw a ray QX making an acute angle with the side QR (on the side opposite to P).
Mark Points: The scale factor is $\frac{3}{5}$. The larger number is 5. Using a compass, mark 5 points (Q₁, Q₂, Q₃, Q₄, Q₅) on the ray QX such that all segments are equal: $QQ_1 = Q_1Q_2 = \dots = Q_4Q_5$.
Join the Denominator Point: The denominator of the factor is 5. Join the 5th point (Q₅) to the vertex R.
Draw a Parallel Line: The numerator is 3. From the 3rd point (Q₃), draw a line parallel to Q₅R. This line will intersect the side QR at a new point, R'. (To draw a parallel line, use a compass to copy the angle at Q₅ to Q₃).
Complete the Triangle: From R', draw a line parallel to the side PR. This line will intersect the side PQ at a new point, P'.
Result: The triangle ΔP'QR' is the required similar triangle, where each side is $\frac{3}{5}$ of the corresponding side of ΔPQR.

b) Construction for Scale Factor $\frac{7}{3} > 1$

Draw the Base Triangle: Construct any triangle ΔPQR.
Extend the Base Sides: Extend the sides QR and QP from the vertex Q.
Draw an Acute Ray: From vertex Q, draw a ray QX making an acute angle with the extended line of QR.
Mark Points: The scale factor is $\frac{7}{3}$. The larger number is 7. Using a compass, mark 7 points (Q₁, Q₂, ..., Q₇) on the ray QX such that all segments are equal.
Join the Denominator Point: The denominator of the factor is 3. Join the 3rd point (Q₃) to the vertex R.
Draw a Parallel Line: The numerator is 7. From the 7th point (Q₇), draw a line parallel to Q₃R. This line will intersect the extended side QR at a new point, R'.
Complete the Triangle: From R', draw a line parallel to the side PR. This line will intersect the extended side QP at a new point, P'.
Result: The triangle ΔP'QR' is the required similar triangle, where each side is $\frac{7}{3}$ of the corresponding side of ΔPQR.

10th Standard Latest Syllabus Solutions, Official Question Papers, Answer Key, Tamil Nadu, Paper With Answer Key | Thiruvallur District | (English Medium)

FIRST MID TERM TEST - 2024

Part - I

Part - II

Part - III

Part - IV

a) Construction for Scale Factor \(\frac{3}{5} < 1\)

b) Construction for Scale Factor \(\frac{7}{3} > 1\)