9th Standard Maths | கணிதம் - First Mid Term Exam 2025 - Question Papers, Answer Keys Official Original QP (EM) | Thoothukudi | Mr. D. Jenis

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

THOOTHUKUDI DISTRICT - FIRST MID TERM TEST 2024

Standard IX | MATHEMATICS | Fully Solved

Question Paper

Part - I

Choose the correct answer. (4 x 1 = 4)

1. Which of the following is correct?

  • a) $\phi \subseteq \{a, b\}$
  • b) $\phi \in \{a,b\}$
  • c) $\{a\} \in \{a,b\}$
  • d) $a \subseteq \{a,b\}$

2. Let $A = \{\phi\}$ and $B = P(A)$ then $A \cap B$ is

  • a) $\{\phi, \{\phi\}\}$
  • b) $\{\phi\}$
  • c) $\phi$
  • d) $\{0\}$

3. When $(2\sqrt{5} - \sqrt{2})^2$ is simplified we get

  • a) $4\sqrt{5} + 2\sqrt{2}$
  • b) $22 - 4\sqrt{10}$
  • c) $8 - 4\sqrt{10}$
  • d) $2\sqrt{10} - 9$

4. If $\frac{1}{7} = 0.\overline{142857}$, then the value of $\frac{5}{7}$ is

  • a) $0.142857$
  • b) $0.\overline{714285}$
  • c) $0.571428$
  • d) $0.714285$

Part - II

Answer any five questions. (5 x 2 = 10)

5. Write the set of letters of the following words in Roster form.

a) ASSESSMENT      b) PRINCIPAL

6. If $A = \{a, \{a,b\}\}$, write all the subsets of A.

7. If $n(A) = 36$, $n(B) = 10$, $n(A \cup B) = 40$ and $n(A') = 27$, find $n(U)$ and $n(A \cap B)$.

8. Convert the following decimal numbers in the form of $\frac{p}{q}$, where p and q are integers and $q \neq 0$.

i) 35      ii) 2.176

9. Find the value of $(81)^{5/4}$.

10. Simplify: $3\sqrt{75} + 5\sqrt{48} - \sqrt{243}$.

11. Represent the following numbers in scientific notation: 2000.57

Part - III

Answer any 4 questions. (4 x 5 = 20)

12. Draw Venn diagram and shade the region representing the following sets:

i) $A'$      ii) $(A - B)'$      iii) $(A \cup B)'$

13. Verify the associative property of intersection of sets for $A = \{-11, \sqrt{2}, \sqrt{5}, 7\}$, $B = \{\sqrt{3}, \sqrt{5}, 6, 13\}$ and $C = \{\sqrt{2}, \sqrt{3}, \sqrt{5}, 9\}$.

14. In a class, all students take part in either music or drama or both. 25 students take part in music, 30 students take part in drama and 8 students take part in both music and drama. Find:

i) The number of students who take part in only music

ii) The number of students who take part in only drama

iii) The total number of students in the class

15. If $U = \{4, 7, 8, 10, 11, 12, 15, 16\}$, $A = \{7, 8, 11, 12\}$ and $B = \{4, 8, 12, 15\}$, then verify De Morgan's laws for complementation.

16. Without actual division, classify the decimal expansion of the following numbers as terminating or non-terminating & recurring.

i) $\frac{13}{64}$      ii) $\frac{-71}{125}$      iii) $\frac{43}{375}$      iv) $\frac{31}{400}$

17. If $\sqrt{2}=1.414$, $\sqrt{3}=1.732$, $\sqrt{5}=2.236$, $\sqrt{10}=3.162$, then find the values of the following correct to 3 places of decimals.

i) $\sqrt{40} - \sqrt{20}$      ii) $\sqrt{300} + \sqrt{90} - \sqrt{8}$

18. Find the value of a and b if $\frac{\sqrt{7}-2}{\sqrt{7}+2} = a\sqrt{7} + b$.

Part - IV

Answer all the questions. (2 x 8 = 16)

19. a) Represent $\sqrt{9.3}$ on a number line.

(OR)

b) Represent the number $6.\bar{4}$ on the number line upto 3 decimal places.

20. a) Construct the centroid of $\triangle PQR$ whose sides are $PQ = 8$ cm, $QR = 6$ cm, $RP = 7$ cm.

(OR)

b) Draw an equilateral triangle of sides 6.5 cm and locate its orthocentre.

Solutions

Part - I Solutions

1. Which of the following is correct?

Solution

Let's analyze the options:
  • a) $\phi \subseteq \{a, b\}$: The empty set ($\phi$) is a subset of every set. This statement is correct.
  • b) $\phi \in \{a,b\}$: The elements of the set are 'a' and 'b', not the empty set. This is incorrect.
  • c) $\{a\} \in \{a,b\}$: The elements are 'a' and 'b', not the set {a}. This is incorrect.
  • d) $a \subseteq \{a,b\}$: 'a' is an element, not a set. The subset notation ($\subseteq$) is used for sets. The correct notation would be $a \in \{a,b\}$ or $\{a\} \subseteq \{a,b\}$. This is incorrect.
Correct answer: a) $\phi \subseteq \{a, b\}$

2. Let $A = \{\phi\}$ and $B = P(A)$ then $A \cap B$ is

Solution

Given: $A = \{\phi\}$. The set A contains one element, which is the empty set $\phi$.
The power set of A, $P(A)$, is the set of all subsets of A. The subsets of A are $\phi$ and $\{\phi\}$.
So, $B = P(A) = \{\phi, \{\phi\}\}$.
We need to find the intersection of A and B: $A \cap B$.
$A \cap B = \{\phi\} \cap \{\phi, \{\phi\}\}$.
The common element in both sets is $\phi$.
Therefore, $A \cap B = \{\phi\}$.
Correct answer: b) $\{\phi\}$

3. When $(2\sqrt{5} - \sqrt{2})^2$ is simplified we get

Solution

We use the algebraic identity $(a - b)^2 = a^2 - 2ab + b^2$.
Here, $a = 2\sqrt{5}$ and $b = \sqrt{2}$.
$(2\sqrt{5} - \sqrt{2})^2 = (2\sqrt{5})^2 - 2(2\sqrt{5})(\sqrt{2}) + (\sqrt{2})^2$
$= (4 \times 5) - 4(\sqrt{5 \times 2}) + 2$
$= 20 - 4\sqrt{10} + 2$
$= 22 - 4\sqrt{10}$
Correct answer: b) $22 - 4\sqrt{10}$

4. If $\frac{1}{7} = 0.\overline{142857}$, then the value of $\frac{5}{7}$ is

Solution

We are given $\frac{1}{7} = 0.\overline{142857}$.
To find $\frac{5}{7}$, we can multiply $\frac{1}{7}$ by 5.
$\frac{5}{7} = 5 \times \frac{1}{7} = 5 \times 0.142857142857...$
Let's perform the multiplication: $$ \begin{array}{@{}c@{\,}c@{}c@{}c@{}c@{}c@{}c} & & 1 & 4 & 2 & 8 & 5 & 7 \\ \times & & & & & & & 5 \\ \hline & 7 & 1 & 4 & 2 & 8 & 5 \\ \end{array} $$
$5 \times 0.142857 = 0.714285$. Since the original decimal is recurring, the result will also be recurring.
So, $\frac{5}{7} = 0.\overline{714285}$.
Correct answer: b) $0.\overline{714285}$

Part - II Solutions

5. Write the set of letters of the following words in Roster form.

Solution

In roster form, we list all unique elements of the set.

a) ASSESSMENT

The letters are A, S, S, E, S, S, M, E, N, T.
The unique letters are A, S, E, M, N, T.
The set in roster form is {A, S, E, M, N, T}.

b) PRINCIPAL

The letters are P, R, I, N, C, I, P, A, L.
The unique letters are P, R, I, N, C, A, L.
The set in roster form is {P, R, I, N, C, A, L}.

6. If $A = \{a, \{a,b\}\}$, write all the subsets of A.

Solution

The set A has two elements: the first element is 'a' and the second element is the set '{a,b}'.
Let $x = a$ and $y = \{a,b\}$. So, $A = \{x, y\}$.
The number of subsets is $2^n$, where $n$ is the number of elements. Here $n=2$, so there are $2^2 = 4$ subsets.
The subsets are:
  1. The empty set: $\phi$
  2. Subsets with one element: $\{x\}$, $\{y\}$. Substituting back: $\{a\}$, $\{\{a,b\}\}$
  3. Subsets with two elements: $\{x, y\}$. Substituting back: $\{a, \{a,b\}\}$
The subsets of A are: $\phi$, $\{a\}$, $\{\{a,b\}\}$, $\{a, \{a,b\}\}$.

7. If $n(A) = 36$, $n(B) = 10$, $n(A \cup B) = 40$ and $n(A') = 27$, find $n(U)$ and $n(A \cap B)$.

Solution

To find $n(U)$ (the number of elements in the universal set):

We use the formula: $n(U) = n(A) + n(A')$.
Given: $n(A) = 36$ and $n(A') = 27$.
$n(U) = 36 + 27 = 63$.
$n(U) = 63$.

To find $n(A \cap B)$ (the number of elements in the intersection of A and B):

We use the formula: $n(A \cup B) = n(A) + n(B) - n(A \cap B)$.
Given: $n(A) = 36$, $n(B) = 10$, $n(A \cup B) = 40$.
$40 = 36 + 10 - n(A \cap B)$
$40 = 46 - n(A \cap B)$
$n(A \cap B) = 46 - 40 = 6$.
$n(A \cap B) = 6$.

8. Convert the following decimal numbers in the form of $\frac{p}{q}$.

Solution

i) 35

Any integer 'p' can be written in the form $\frac{p}{q}$ by setting q=1.
$35 = \frac{35}{1}$

ii) 2.176

This is a terminating decimal. To convert it, we write the number without the decimal point as the numerator and a power of 10 as the denominator. The power of 10 corresponds to the number of decimal places.
There are 3 decimal places, so the denominator is $10^3 = 1000$.
$2.176 = \frac{2176}{1000}$
Now, we simplify the fraction by dividing both numerator and denominator by their greatest common divisor (which is 8).
$\frac{2176 \div 8}{1000 \div 8} = \frac{272}{125}$
$2.176 = \frac{272}{125}$

9. Find the value of $(81)^{5/4}$.

Solution

We can write $(81)^{5/4}$ as $(\sqrt[4]{81})^5$.
First, find the prime factors of 81.
$81 = 9 \times 9 = 3 \times 3 \times 3 \times 3 = 3^4$.
So, $(81)^{5/4} = (3^4)^{5/4}$.
Using the power rule $(a^m)^n = a^{m \times n}$:
$(3^4)^{5/4} = 3^{4 \times \frac{5}{4}} = 3^5$.
$3^5 = 3 \times 3 \times 3 \times 3 \times 3 = 243$.
The value is 243.

10. Simplify: $3\sqrt{75} + 5\sqrt{48} - \sqrt{243}$.

Solution

First, we simplify each surd by factoring out perfect squares.
$\sqrt{75} = \sqrt{25 \times 3} = \sqrt{25}\sqrt{3} = 5\sqrt{3}$.
$\sqrt{48} = \sqrt{16 \times 3} = \sqrt{16}\sqrt{3} = 4\sqrt{3}$.
$\sqrt{243} = \sqrt{81 \times 3} = \sqrt{81}\sqrt{3} = 9\sqrt{3}$.
Now substitute these back into the expression:
$3(5\sqrt{3}) + 5(4\sqrt{3}) - 9\sqrt{3}$
$= 15\sqrt{3} + 20\sqrt{3} - 9\sqrt{3}$
Since all terms are like surds, we can combine the coefficients:
$(15 + 20 - 9)\sqrt{3} = (35 - 9)\sqrt{3} = 26\sqrt{3}$.
The simplified form is $26\sqrt{3}$.

11. Represent 2000.57 in scientific notation.

Solution

Scientific notation is a way of writing numbers in the form $a \times 10^n$, where $1 \le |a| < 10$ and n is an integer.
The given number is 2000.57.
To get a number between 1 and 10, we need to move the decimal point 3 places to the left.
$2000.57 \rightarrow 2.00057$
Since we moved the decimal 3 places to the left, the power of 10 will be positive 3.
$2000.57 = 2.00057 \times 10^3$.

Part - III Solutions

12. Draw Venn diagram and shade the region representing the following sets:

Solution

In a Venn diagram, the universal set U is represented by a rectangle, and subsets A and B are represented by circles inside the rectangle.

i) A' (Complement of A)

A' represents all elements in the universal set U that are NOT in set A.
Shading: Shade the entire area inside the rectangle (U) but outside the circle A.

ii) (A - B)' (Complement of A-B)

A - B represents the elements that are in A but not in B. This is the part of circle A that does not overlap with circle B.
(A - B)' represents everything in U EXCEPT the A-B region.
Shading: Shade the entire rectangle including all of circle B and the area outside both circles. The only unshaded part is the portion of circle A that does not intersect B.

iii) (A U B)' (Complement of A union B)

A U B represents all elements that are in A, or in B, or in both. This is the entire area covered by both circles.
(A U B)' represents all elements in U that are NOT in A U B. This is also known as A' $\cap$ B' by De Morgan's Law.
Shading: Shade the area inside the rectangle (U) but outside both circles A and B.

13. Verify the associative property of intersection of sets for $A = \{-11, \sqrt{2}, \sqrt{5}, 7\}$, $B = \{\sqrt{3}, \sqrt{5}, 6, 13\}$ and $C = \{\sqrt{2}, \sqrt{3}, \sqrt{5}, 9\}$.

Solution

The associative property of intersection is $A \cap (B \cap C) = (A \cap B) \cap C$.
Given sets:
  • $A = \{-11, \sqrt{2}, \sqrt{5}, 7\}$
  • $B = \{\sqrt{3}, \sqrt{5}, 6, 13\}$
  • $C = \{\sqrt{2}, \sqrt{3}, \sqrt{5}, 9\}$

Step 1: Calculate the Left Hand Side (LHS): $A \cap (B \cap C)$

First, find $B \cap C$:
$B \cap C = \{\sqrt{3}, \sqrt{5}, 6, 13\} \cap \{\sqrt{2}, \sqrt{3}, \sqrt{5}, 9\} = \{\sqrt{3}, \sqrt{5}\}$
Now, find $A \cap (B \cap C)$:
$A \cap (B \cap C) = \{-11, \sqrt{2}, \sqrt{5}, 7\} \cap \{\sqrt{3}, \sqrt{5}\} = \{\sqrt{5}\}$
LHS = $\{\sqrt{5}\}$

Step 2: Calculate the Right Hand Side (RHS): $(A \cap B) \cap C$

First, find $A \cap B$:
$A \cap B = \{-11, \sqrt{2}, \sqrt{5}, 7\} \cap \{\sqrt{3}, \sqrt{5}, 6, 13\} = \{\sqrt{5}\}$
Now, find $(A \cap B) \cap C$:
$(A \cap B) \cap C = \{\sqrt{5}\} \cap \{\sqrt{2}, \sqrt{3}, \sqrt{5}, 9\} = \{\sqrt{5}\}$
RHS = $\{\sqrt{5}\}$

Step 3: Compare LHS and RHS

Since LHS = RHS = $\{\sqrt{5}\}$, the associative property of intersection is verified.

14. In a class, all students take part in either music or drama or both. 25 students take part in music, 30 students take part in drama and 8 students take part in both music and drama. Find...

Solution

Let M be the set of students who take part in music, and D be the set of students who take part in drama.
Given:
  • Number of students in music, $n(M) = 25$
  • Number of students in drama, $n(D) = 30$
  • Number of students in both, $n(M \cap D) = 8$

i) The number of students who take part in only music

This is given by $n(M) - n(M \cap D)$.
Number of students in only music = $25 - 8 = 17$.
17 students take part in only music.

ii) The number of students who take part in only drama

This is given by $n(D) - n(M \cap D)$.
Number of students in only drama = $30 - 8 = 22$.
22 students take part in only drama.

iii) The total number of students in the class

Since all students take part in either music or drama or both, the total number of students is $n(M \cup D)$.
Using the formula: $n(M \cup D) = n(M) + n(D) - n(M \cap D)$.
$n(M \cup D) = 25 + 30 - 8 = 55 - 8 = 47$.
The total number of students in the class is 47.

15. If $U = \{4, 7, 8, 10, 11, 12, 15, 16\}$, $A = \{7, 8, 11, 12\}$ and $B = \{4, 8, 12, 15\}$, then verify De Morgan's laws for complementation.

Solution

De Morgan's laws for complementation are:
  1. $(A \cup B)' = A' \cap B'$
  2. $(A \cap B)' = A' \cup B'$
First, let's find the complements A' and B'.
  • $A' = U - A = \{4, 10, 15, 16\}$
  • $B' = U - B = \{7, 10, 11, 16\}$

Verification of Law 1: $(A \cup B)' = A' \cap B'$

LHS: $(A \cup B)'$
$A \cup B = \{7, 8, 11, 12\} \cup \{4, 8, 12, 15\} = \{4, 7, 8, 11, 12, 15\}$
$(A \cup B)' = U - (A \cup B) = \{4, 7, 8, 10, 11, 12, 15, 16\} - \{4, 7, 8, 11, 12, 15\} = \{10, 16\}$
RHS: $A' \cap B'$
$A' \cap B' = \{4, 10, 15, 16\} \cap \{7, 10, 11, 16\} = \{10, 16\}$
Since LHS = RHS, the first law is verified.

Verification of Law 2: $(A \cap B)' = A' \cup B'$

LHS: $(A \cap B)'$
$A \cap B = \{7, 8, 11, 12\} \cap \{4, 8, 12, 15\} = \{8, 12\}$
$(A \cap B)' = U - (A \cap B) = \{4, 7, 10, 11, 15, 16\}$
RHS: $A' \cup B'$
$A' \cup B' = \{4, 10, 15, 16\} \cup \{7, 10, 11, 16\} = \{4, 7, 10, 11, 15, 16\}$
Since LHS = RHS, the second law is verified.
Both De Morgan's laws are verified.

16. Without actual division, classify the decimal expansion of the following numbers as terminating or non-terminating & recurring.

Solution

A rational number $\frac{p}{q}$ has a terminating decimal expansion if the prime factorization of the denominator $q$ is of the form $2^n 5^m$, where n and m are non-negative integers. Otherwise, it is non-terminating and recurring.

i) $\frac{13}{64}$

Denominator $q = 64$. Prime factorization: $64 = 2^6$. This is in the form $2^n 5^m$ (with $n=6, m=0$).
Terminating.

ii) $\frac{-71}{125}$

Denominator $q = 125$. Prime factorization: $125 = 5^3$. This is in the form $2^n 5^m$ (with $n=0, m=3$).
Terminating.

iii) $\frac{43}{375}$

Denominator $q = 375$. Prime factorization: $375 = 3 \times 125 = 3 \times 5^3$.
Since the denominator has a prime factor of 3 (other than 2 and 5), the decimal expansion is non-terminating and recurring.
Non-terminating & recurring.

iv) $\frac{31}{400}$

Denominator $q = 400$. Prime factorization: $400 = 4 \times 100 = 2^2 \times 10^2 = 2^2 \times (2 \times 5)^2 = 2^2 \times 2^2 \times 5^2 = 2^4 \times 5^2$. This is in the form $2^n 5^m$.
Terminating.

17. If $\sqrt{2}=1.414$, $\sqrt{3}=1.732$, $\sqrt{5}=2.236$, $\sqrt{10}=3.162$, find the values correct to 3 decimal places.

Solution

i) $\sqrt{40} - \sqrt{20}$

Simplify the surds:
$\sqrt{40} = \sqrt{4 \times 10} = 2\sqrt{10}$
$\sqrt{20} = \sqrt{4 \times 5} = 2\sqrt{5}$
Substitute the given values:
$2\sqrt{10} - 2\sqrt{5} = 2(3.162) - 2(2.236)$
$= 6.324 - 4.472$
$= 1.852$
1.852

ii) $\sqrt{300} + \sqrt{90} - \sqrt{8}$

Simplify the surds:
$\sqrt{300} = \sqrt{100 \times 3} = 10\sqrt{3}$
$\sqrt{90} = \sqrt{9 \times 10} = 3\sqrt{10}$
$\sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2}$
Substitute the given values:
$10\sqrt{3} + 3\sqrt{10} - 2\sqrt{2} = 10(1.732) + 3(3.162) - 2(1.414)$
$= 17.320 + 9.486 - 2.828$
$= 26.806 - 2.828$
$= 23.978$
23.978

18. Find the value of a and b if $\frac{\sqrt{7}-2}{\sqrt{7}+2} = a\sqrt{7} + b$.

Solution

To simplify the left side, we rationalize the denominator by multiplying the numerator and denominator by the conjugate of the denominator, which is $\sqrt{7}-2$.
$\frac{\sqrt{7}-2}{\sqrt{7}+2} \times \frac{\sqrt{7}-2}{\sqrt{7}-2} = \frac{(\sqrt{7}-2)^2}{(\sqrt{7})^2 - (2)^2}$
Numerator: $(\sqrt{7}-2)^2 = (\sqrt{7})^2 - 2(\sqrt{7})(2) + (2)^2 = 7 - 4\sqrt{7} + 4 = 11 - 4\sqrt{7}$.
Denominator: $(\sqrt{7})^2 - (2)^2 = 7 - 4 = 3$.
So, the expression becomes $\frac{11 - 4\sqrt{7}}{3}$.
We can write this as $\frac{11}{3} - \frac{4}{3}\sqrt{7}$.
Now, we compare this with $a\sqrt{7} + b$. To match the format, we write it as $-\frac{4}{3}\sqrt{7} + \frac{11}{3}$.
$-\frac{4}{3}\sqrt{7} + \frac{11}{3} = a\sqrt{7} + b$
By comparing the coefficients of $\sqrt{7}$ and the constant terms:
$a = -\frac{4}{3}$ and $b = \frac{11}{3}$.

Part - IV Solutions

19. a) Represent $\sqrt{9.3}$ on a number line. (OR) b) Represent $6.\bar{4}$ on the number line upto 3 decimal places.

Solution for 19(a): Represent $\sqrt{9.3}$ on a number line

Steps of Construction:
  1. Draw a line and mark a point A on it.
  2. Mark a point B on the line such that AB = 9.3 units.
  3. From B, mark a point C on the same line such that BC = 1 unit.
  4. Find the midpoint of AC. Let it be O. (To do this, draw the perpendicular bisector of AC).
  5. With O as the center and OA (or OC) as the radius, draw a semicircle.
  6. Draw a line perpendicular to AC passing through B. Let it intersect the semicircle at D.
  7. The length of the line segment BD is $\sqrt{9.3}$ units.
  8. To represent this on the number line, consider the line AC as the number line with B as the origin (0). With B as the center and BD as the radius, draw an arc that intersects the number line at a point E.
  9. The point E represents the number $\sqrt{9.3}$ on the number line.

Solution for 19(b): Represent $6.\bar{4}$ on the number line

We need to represent $6.\bar{4} = 6.444...$ on the number line up to 3 decimal places (i.e., we locate 6.444) using the method of successive magnification.
  1. Step 1: The number 6.444 lies between 6 and 7. Draw a number line from 6 to 7 and divide it into 10 equal parts (6.1, 6.2, ..., 6.9). The number lies between 6.4 and 6.5.
  2. Step 2: Magnify the interval [6.4, 6.5]. Draw a new number line representing this interval and divide it into 10 equal parts (6.41, 6.42, ..., 6.49). The number 6.444 lies between 6.44 and 6.45.
  3. Step 3: Magnify the interval [6.44, 6.45]. Draw a third number line for this interval and divide it into 10 equal parts (6.441, 6.442, ..., 6.449).
  4. Step 4: Locate the point 6.444 on this third number line. This is the representation of $6.\bar{4}$ up to 3 decimal places.

20. a) Construct the centroid of $\triangle PQR$. (OR) b) Draw an equilateral triangle and locate its orthocentre.

Solution for 20(a): Construct the centroid of $\triangle PQR$

Given sides: $PQ = 8$ cm, $QR = 6$ cm, $RP = 7$ cm.
The centroid is the point of intersection of the medians of a triangle. A median connects a vertex to the midpoint of the opposite side.
Steps of Construction:
  1. Draw the triangle PQR with the given measurements. (Draw QR = 6 cm. With Q as center, draw an arc of 8 cm. With R as center, draw an arc of 7 cm. The intersection point is P).
  2. Find the midpoint of any two sides, say QR and RP.
    • To find the midpoint of QR, draw the perpendicular bisector of QR. Let the midpoint be D.
    • To find the midpoint of RP, draw the perpendicular bisector of RP. Let the midpoint be E.
  3. Draw the medians. A median is a line segment joining a vertex to the midpoint of its opposite side.
    • Draw the median PD (from vertex P to midpoint D of QR).
    • Draw the median QE (from vertex Q to midpoint E of RP).
  4. The point where the two medians PD and QE intersect is the centroid of the triangle. Label this point G.
The point G is the required centroid.

Solution for 20(b): Draw an equilateral triangle and locate its orthocentre

Given: Equilateral triangle with sides 6.5 cm.
The orthocentre is the point of intersection of the altitudes of a triangle. An altitude is a perpendicular line from a vertex to the opposite side.
Steps of Construction:
  1. Draw an equilateral triangle ABC with side length 6.5 cm. (Draw base BC = 6.5 cm. With B and C as centers, draw arcs of 6.5 cm to intersect at A).
  2. Construct the altitude from any two vertices, say A and B.
    • Altitude from A to BC: With A as the center, draw an arc that intersects BC at two points, say X and Y. Now, with X and Y as centers, draw two arcs that intersect below BC. Draw a line from A through this intersection point to BC. Let this line meet BC at D. AD is the altitude from A.
    • Altitude from B to AC: Repeat the process. With B as the center, draw an arc intersecting AC at two points. From these points, draw arcs that intersect. Draw a line from B through this new intersection to AC. Let this line meet AC at E. BE is the altitude from B.
  3. The point where the two altitudes AD and BE intersect is the orthocentre of the triangle. Label this point H.
The point H is the required orthocentre. (Note: For an equilateral triangle, the centroid, circumcenter, incenter, and orthocentre are all the same point).