Squares and Square Roots — Questions and Answers
Welcome to HSLC Guru. This page gives the complete, step-by-step solutions to ASSEB Class 8 General Mathematics Chapter 6 “Squares and Square Roots” — every textbook question of Exercises 6.1 to 6.4, the in-text “Try These / Try yourself” activities, worked examples and additional practice questions with fully worked answers.
Summary
When a number is multiplied by itself, the product is called the square of that number, e.g. $2 \times 2 = 2^2 = 4$ and $3 \times 3 = 3^2 = 9$. A number that can be written as the square of another number is a square number, and the squares of integers $1, 4, 9, 16, 25, 36, 49, \dots$ are called perfect squares.
Square numbers have several useful properties — a square number can end only in $0, 1, 4, 5, 6$ or $9$ at the unit place, never in $2, 3, 7$ or $8$. The square of an odd number is odd and the square of an even number is even. The sum of the first $n$ consecutive odd numbers is always $n^2$, and the sum of two consecutive triangular numbers is a square number.
The inverse operation of squaring is the square root, denoted by $\sqrt{\ }$; for example $\sqrt{121} = 11$. Square roots are found by the prime-factorisation method and the long-division method, and the division method also works for decimal numbers. The chapter also covers Pythagorean triplets $a^2 + b^2 = c^2$ and a quick trick to square numbers ending in $5$.
Summary: This is ASSEB Class 8 General Mathematics Chapter 6, Squares and Square Roots. It explains square numbers, perfect squares, their unit-digit and odd/even properties, the sum of consecutive odd numbers as $n^2$, triangular numbers, Pythagorean triplets, and finding square roots of whole and decimal numbers by prime factorisation and long division, with fully worked answers to Exercises 6.1, 6.2, 6.3 and 6.4.
Textbook Questions and Answers
Exercise 6.1
1. How many perfect square numbers are there in between 1 and 200?
Answer: Since $14^2 = 196$ and $15^2 = 225 > 200$, the perfect squares from 1 to 200 are $1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196$ — that is, $14$ perfect square numbers.
2. What will be the digit in the unit place of each of the squares of the following numbers? (i) 51 (ii) 99 (iii) 205 (iv) 3400 (v) 1987
Answer: The unit digit of a square depends only on the square of its own unit digit.
- (i) $51$ ends in $1$; $1^2 = 1$ → unit digit $1$.
- (ii) $99$ ends in $9$; $9^2 = 81$ → unit digit $1$.
- (iii) $205$ ends in $5$; $5^2 = 25$ → unit digit $5$.
- (iv) $3400$ ends in $0$; $0^2 = 0$ → unit digit $0$.
- (v) $1987$ ends in $7$; $7^2 = 49$ → unit digit $9$.
3. Write the reasons why the following numbers are not square numbers. (i) 4347 (ii) 24832 (iii) 35493 (iv) 403388 (v) 182000
Answer: A number ending in $2, 3, 7$ or $8$ is definitely not a perfect square.
- (i) $4347$ ends in $7$ — not a square number.
- (ii) $24832$ ends in $2$ — not a square number.
- (iii) $35493$ ends in $3$ — not a square number.
- (iv) $403388$ ends in $8$ — not a square number.
- (v) $182000$ ends in $3$ zeros (an odd number of zeros). A perfect square always ends in an even number of zeros, so $182000$ is not a square number.
4. (i) Write five numbers whose squares are even. (ii) Write five numbers whose squares are odd.
Answer: (i) The square of an even number is even, so $2, 4, 6, 8, 10$ (squares $4, 16, 36, 64, 100$ — all even).
(ii) The square of an odd number is odd, so $1, 3, 5, 7, 9$ (squares $1, 9, 25, 49, 81$ — all odd).
5. Without adding directly, find the sum (using relevant properties).
Answer: The sum of the first $n$ consecutive odd numbers $= n^2$.
$$1 + 3 + 5 + \dots + (2n-1) = n^2$$
- (i) $1 + 3 + 5 + 7$ — first $4$ odd numbers, sum $= 4^2 = 16$.
- (ii) $1 + 3 + 5 + 7 + 9 + 11 + 13 + 15$ — first $8$ odd numbers, sum $= 8^2 = 64$.
- (iii) $1 + 3 + \dots + 25$ — first $13$ odd numbers, sum $= 13^2 = 169$.
6. Can 36 be expressed as the sum of 6 consecutive odd numbers?
Answer: Yes. $36 = 6^2$, and the sum of the first $6$ odd numbers is $6^2 = 36$.
$$1 + 3 + 5 + 7 + 9 + 11 = 36$$
7. Write 5 consecutive triangular numbers greater than 15. From these numbers consider two consecutive numbers and examine whether their sums are square numbers or not.
Answer: The triangular numbers are $1, 3, 6, 10, 15, 21, 28, 36, 45, 55, \dots$. Five consecutive triangular numbers greater than $15$ are $21, 28, 36, 45, 55$. Taking sums of two consecutive ones —
- $21 + 28 = 49 = 7^2$
- $28 + 36 = 64 = 8^2$
- $36 + 45 = 81 = 9^2$
- $45 + 55 = 100 = 10^2$
Each sum is a square number. So the sum of two consecutive triangular numbers is always a square number.
Exercise 6.2
1. Find the squares of the following numbers. (i) 35 (ii) 55 (iii) 95
Answer: For a number ending in $5$, its square $= (\text{tens digit} \times \text{next number})$ hundred $+ 25$.
- (i) $35^2 = (3 \times 4)$ hundred $+ 25 = 1200 + 25 = 1225$.
- (ii) $55^2 = (5 \times 6)$ hundred $+ 25 = 3000 + 25 = 3025$.
- (iii) $95^2 = (9 \times 10)$ hundred $+ 25 = 9000 + 25 = 9025$.
2. Write three Pythagorean triplets.
Answer: Three triplets satisfying $a^2 + b^2 = c^2$ are —
- $(3, 4, 5)$: $3^2 + 4^2 = 9 + 16 = 25 = 5^2$.
- $(6, 8, 10)$: $6^2 + 8^2 = 36 + 64 = 100 = 10^2$.
- $(5, 12, 13)$: $5^2 + 12^2 = 25 + 144 = 169 = 13^2$.
3. Find a Pythagorean triplet whose smallest term is 10.
Answer: For $p > 1$, $2p,\ p^2 – 1,\ p^2 + 1$ form a Pythagorean triplet. Taking the smallest term $2p = 10$ gives $p = 5$; then $p^2 – 1 = 24$ and $p^2 + 1 = 26$. So the triplet is $(10, 24, 26)$.
$$10^2 + 24^2 = 100 + 576 = 676 = 26^2$$
Exercise 6.3
1. Answer whether the following are true or false.
- (i) The square root of an even perfect square number is even. — True ($\sqrt{16}=4$, $\sqrt{36}=6$, $\sqrt{100}=10$).
- (ii) The sum $1 + 3 + 5 + 7 + 9 + 11 + 13$ is a square number. — True (sum of the first $7$ odd numbers $= 7^2 = 49$).
- (iii) A number has 8 at its unit place, therefore the number may be a square number. — False (a number ending in $2, 3, 7$ or $8$ is never a square number).
- (iv) A square number has 1 at the unit place, therefore its square root may have 1 or 9 at its unit place. — True ($1^2=1$, $9^2=81$; both squares end in $1$).
2. What will be the possible digits in the unit places of the square roots of the following numbers? (i) 8281 (ii) 5476 (iii) 172225 (iv) 12100
- (i) $8281$ ends in $1$ → square root ends in $1$ or $9$ ($\sqrt{8281}=91$).
- (ii) $5476$ ends in $6$ → square root ends in $4$ or $6$ ($\sqrt{5476}=74$).
- (iii) $172225$ ends in $5$ → square root ends in $5$ ($\sqrt{172225}=415$).
- (iv) $12100$ ends in $0$ → square root ends in $0$ ($\sqrt{12100}=110$).
3. Find the square roots of the following numbers by prime factorisation method. (i) 256 (ii) 729 (iii) 1764 (iv) 5184 (v) 7744 (vi) 5929 (vii) 8836 (viii) 4225
Answer: Break the number into prime factors, pair the equal factors, and multiply one factor from each pair.
- (i) $256 = 2^8 = (2^4)^2$; $\sqrt{256} = 2^4 = 16$.
- (ii) $729 = 3^6 = (3^3)^2$; $\sqrt{729} = 3^3 = 27$.
- (iii) $1764 = 2^2 \times 3^2 \times 7^2$; $\sqrt{1764} = 2 \times 3 \times 7 = 42$.
- (iv) $5184 = 2^6 \times 3^4 = (2^3 \times 3^2)^2$; $\sqrt{5184} = 8 \times 9 = 72$.
- (v) $7744 = 2^6 \times 11^2$; $\sqrt{7744} = 2^3 \times 11 = 88$.
- (vi) $5929 = 7^2 \times 11^2$; $\sqrt{5929} = 7 \times 11 = 77$.
- (vii) $8836 = 2^2 \times 47^2$; $\sqrt{8836} = 2 \times 47 = 94$.
- (viii) $4225 = 5^2 \times 13^2$; $\sqrt{4225} = 5 \times 13 = 65$.
4. What is the smallest number by which each of the following numbers must be multiplied to make it a perfect square? (i) 15 (ii) 45 (iii) 150 (iv) 175
Answer: After prime factorisation, multiply by the unpaired factors.
- (i) $15 = 3 \times 5$; both unpaired → multiply by $3 \times 5 = 15$ ($15 \times 15 = 225 = 15^2$). Smallest number $15$.
- (ii) $45 = 3^2 \times 5$; $5$ unpaired → multiply by $5$ ($45 \times 5 = 225 = 15^2$).
- (iii) $150 = 2 \times 3 \times 5^2$; $2$ and $3$ unpaired → multiply by $2 \times 3 = 6$ ($150 \times 6 = 900 = 30^2$). Smallest number $6$.
- (iv) $175 = 5^2 \times 7$; $7$ unpaired → multiply by $7$ ($175 \times 7 = 1225 = 35^2$).
5. (i) Find the smallest perfect square number divisible by 8, 15 and 20. (ii) Find the least perfect square number divisible by 12, 20 and 25.
Answer: (i) $8 = 2^3$, $15 = 3 \times 5$, $20 = 2^2 \times 5$; LCM $= 2^3 \times 3 \times 5 = 120$. Here $2, 3, 5$ have odd exponents, so multiply by $2 \times 3 \times 5 = 30$.
$$120 \times 30 = 3600 = 60^2$$
So the smallest perfect square is $3600$.
(ii) $12 = 2^2 \times 3$, $20 = 2^2 \times 5$, $25 = 5^2$; LCM $= 2^2 \times 3 \times 5^2 = 300$. Here only $3$ has an odd exponent, so multiply by $3$.
$$300 \times 3 = 900 = 30^2$$
So the least perfect square is $900$.
6. (i) By what least number is 4032 to be divided to get a perfect square number? Find the square root of the quotient. (ii) By what smallest number is 14112 to be multiplied to get a perfect square number? Find the square root of the product.
Answer: (i) $4032 = 2^6 \times 3^2 \times 7$. Here $7$ is unpaired, so divide by $7$.
$$4032 \div 7 = 576 = 2^6 \times 3^2 = (2^3 \times 3)^2 \Rightarrow \sqrt{576} = 24$$
(ii) $14112 = 2^5 \times 3^2 \times 7^2$. Here one $2$ is unpaired, so multiply by $2$.
$$14112 \times 2 = 28224 = 2^6 \times 3^2 \times 7^2 = (2^3 \times 3 \times 7)^2 \Rightarrow \sqrt{28224} = 168$$
7. In a school there are all total 1024 students. In the morning prayer they are asked to make rows such that the number of rows and the number of students in each row are the same. Find the number of rows and the number of students in each row.
Answer: Number of rows must equal number in each row, so we find $\sqrt{1024}$. Since $1024 = 2^{10} = (2^5)^2$, $\sqrt{1024} = 32$. So there are $32$ rows with $32$ students in each row.
8. In a tea garden, the number of rows of tea plants and the number of tea plants in each row are the same. 835 plants are supplied, but it is seen that more plants are required to satisfy the condition. How many additional tea plants are required?
Answer: $28^2 = 784$ and $29^2 = 841$, so $835$ is not a perfect square; the next perfect square after $835$ is $841 = 29^2$. Additional plants required $= 841 – 835 = $ $6$.
Exercise 6.4
1. Without finding the square root, find how many digits are there in the square roots of the following numbers. (i) 100 (ii) 21904 (iii) 17850625
Answer: For an $n$-digit perfect square, the square root has $\frac{n}{2}$ digits if $n$ is even, and $\frac{n+1}{2}$ digits if $n$ is odd.
- (i) $100$ — $3$ digits (odd) → $\frac{3+1}{2} = $ $2$ digits ($\sqrt{100}=10$).
- (ii) $21904$ — $5$ digits (odd) → $\frac{5+1}{2} = $ $3$ digits ($\sqrt{21904}=148$).
- (iii) $17850625$ — $8$ digits (even) → $\frac{8}{2} = $ $4$ digits ($\sqrt{17850625}=4225$).
2. Using the division method, find the square root of each of the following numbers.
Answer: Placing a bar over each pair of digits from the unit place and applying the long-division method —
- (i) $\sqrt{676} = 26$
- (ii) $\sqrt{841} = 29$
- (iii) $\sqrt{1156} = 34$
- (iv) $\sqrt{2025} = 45$
- (v) $\sqrt{2704} = 52$
- (vi) $\sqrt{4489} = 67$
- (vii) $\sqrt{8100} = 90$
- (viii) $\sqrt{14641} = 121$
- (ix) $\sqrt{15129} = 123$
- (x) $\sqrt{21904} = 148$
3. Find the square roots of the following decimal numbers. (i) 51.84 (ii) 79.21 (iii) 98.01 (iv) 1.44 (v) 6.25 (vi) 973.44
Answer: Placing bars from the unit place in the integral part and from the decimal point in the decimal part, then dividing —
- (i) $\sqrt{51.84} = 7.2$
- (ii) $\sqrt{79.21} = 8.9$
- (iii) $\sqrt{98.01} = 9.9$
- (iv) $\sqrt{1.44} = 1.2$
- (v) $\sqrt{6.25} = 2.5$
- (vi) $\sqrt{973.44} = 31.2$
4. For a rectangular field the length is 35 metres and the breadth is 12 metres. What will be the length of one diagonal?
Answer: By the Pythagoras theorem, if the diagonal is $d$ then $d^2 = 35^2 + 12^2 = 1225 + 144 = 1369$.
$$d = \sqrt{1369} = 37 \text{ metres}$$
So the length of the diagonal is $37$ metres.
5. In a school there are 1089 students. At the flag hoisting on the first day of the annual sports, they are asked to stand so that the number of rows and the number of students in each row are the same. How many rows can be formed?
Answer: The number of rows and columns must be equal, so we find $\sqrt{1089}$. Since $1089 = 3^2 \times 11^2 = (3 \times 11)^2$, $\sqrt{1089} = 33$. So $33$ rows can be formed (with $33$ students in each).
6. Find the smallest number to be added to each of the following numbers to get a perfect square. (i) 1220 (ii) 1750 (iii) 5451 (iv) 1015
Answer: Subtract the number from the next perfect square above it.
- (i) $34^2 = 1156$, $35^2 = 1225$; add $1225 – 1220 = $ $5$.
- (ii) $41^2 = 1681$, $42^2 = 1764$; add $1764 – 1750 = $ $14$.
- (iii) $73^2 = 5329$, $74^2 = 5476$; add $5476 – 5451 = $ $25$.
- (iv) $31^2 = 961$, $32^2 = 1024$; add $1024 – 1015 = $ $9$.
7. Find the smallest number to be subtracted from each of the following numbers to get a perfect square, and find the square root of the number so obtained. (i) 825 (ii) 1450 (iii) 3250 (iv) 6262
Answer: Subtract the largest perfect square that is less than or equal to the number.
- (i) $28^2 = 784$; subtract $825 – 784 = $ $41$; square root $= 28$.
- (ii) $38^2 = 1444$; subtract $1450 – 1444 = $ $6$; square root $= 38$.
- (iii) $57^2 = 3249$; subtract $3250 – 3249 = $ $1$; square root $= 57$.
- (iv) $79^2 = 6241$; subtract $6262 – 6241 = $ $21$; square root $= 79$.
8. What is the nearest perfect square number of 4612?
Answer: $67^2 = 4489$ and $68^2 = 4624$. Since $4612 – 4489 = 123$ but $4624 – 4612 = 12$, the number $4624$ is nearer. So the nearest perfect square is $4624$ ($= 68^2$).
9. Give 5 examples of each of the following — (i) Square number whose unit-place digit is 4. (ii) Square number whose unit-place digit is 9. (iii) Square number whose unit-place digit is 0.
- (i) Unit digit $4$: $4\,(2^2),\ 64\,(8^2),\ 144\,(12^2),\ 324\,(18^2),\ 484\,(22^2)$.
- (ii) Unit digit $9$: $9\,(3^2),\ 49\,(7^2),\ 169\,(13^2),\ 289\,(17^2),\ 529\,(23^2)$.
- (iii) Unit digit $0$: $100\,(10^2),\ 400\,(20^2),\ 900\,(30^2),\ 1600\,(40^2),\ 2500\,(50^2)$.
Try These and Worked Examples
Activity: Find the square numbers between 10 and 100.
Answer: $4^2 = 16$, $5^2 = 25$, $6^2 = 36$, $7^2 = 49$, $8^2 = 64$, $9^2 = 81$. So the square numbers between $10$ and $100$ are $16, 25, 36, 49, 64, 81$.
Observe: Is $\frac{1}{4}$ a square number?
Answer: $\frac{1}{4} = \left(\frac{1}{2}\right)^2$, so $\frac{1}{4}$ is the square of a rational number and hence a square number in that sense; but it is not a perfect square of an integer.
Try yourself: Which of the following are Pythagorean triplets? (i) 15, 20, 25 (ii) 7, 24, 25 (iii) 5, 10, 13
- (i) $15^2 + 20^2 = 225 + 400 = 625 = 25^2$ — Pythagorean triplet.
- (ii) $7^2 + 24^2 = 49 + 576 = 625 = 25^2$ — Pythagorean triplet.
- (iii) $5^2 + 10^2 = 25 + 100 = 125$ but $13^2 = 169$; $125 \ne 169$ — not a Pythagorean triplet.
Try yourself: Find the square roots of the following numbers. (i) 256 (ii) 2304 (iii) 74529
- (i) $256 = 2^8 = (2^4)^2$; $\sqrt{256} = 16$.
- (ii) $2304 = 2^8 \times 3^2 = (2^4 \times 3)^2$; $\sqrt{2304} = 48$.
- (iii) $74529 = 3^2 \times 7^2 \times 13^2 = (3 \times 7 \times 13)^2$; $\sqrt{74529} = 273$.
Example (Pythagoras theorem): If the sides containing a right angle are $4$ units and $3$ units, what is the length of the hypotenuse?
$$3^2 + 4^2 = 9 + 16 = 25 = 5^2 \Rightarrow \text{hypotenuse} = 5 \text{ units}$$
Example: If the square of a number is 5184, what is the number?
Answer: We must find $\sqrt{5184}$. Since $5184 = 2^6 \times 3^4 = (2^3 \times 3^2)^2$, $\sqrt{5184} = 8 \times 9 = 72$. The number is $72$.
Example: By what smallest number must 180 be multiplied so that the product becomes a perfect square? What is the square root of the number thus obtained?
Answer: $180 = 2^2 \times 3^2 \times 5$; here $5$ is alone (unpaired), so multiply by $5$. Then $180 \times 5 = 900 = 30^2$ and $\sqrt{900} = 3 \times 2 \times 5 = 30$.
Example: By what smallest number must 2645 be divided so that the quotient becomes a perfect square? What is the square root of the number so obtained?
Answer: $2645 = 5 \times 23 \times 23$; here $5$ is alone, so divide by $5$. Then $2645 \div 5 = 529 = 23^2$ and $\sqrt{529} = 23$.
Example: The length and breadth of a small rectangular park are 15 metres and 8 metres. Find the diagonal of that rectangular park.
Answer: $\triangle ABC$ is right angled; taking the diagonal $AC = x$, $x^2 = AB^2 + BC^2 = 15^2 + 8^2 = 225 + 64 = 289$, so $x = \sqrt{289} = 17$. The diagonal is $17$ metres.
Example: Find the smallest square number that is divisible by each of 8, 12 and 18.
Answer: LCM of $8, 12, 18 = 2^2 \times 3^2 \times 2 = 72$. In $72 = 2^3 \times 3^2$, the power of $2$ is odd, so multiply by $2$: $72 \times 2 = 144 = 12^2$. So the smallest such square number is $144$.
Example: Find the square root. (i) $\sqrt{2^8}$ (ii) $\sqrt{9 \times 36}$
Answer: (i) $2^8 = (2^4)^2$, so $\sqrt{2^8} = 2^4 = 16$.
(ii) $9 \times 36 = 3^2 \times 6^2 = (3 \times 6)^2$, so $\sqrt{9 \times 36} = 3 \times 6 = 18$.
Additional Questions and Answers
Multiple Choice Questions (MCQ)
1. Which of the following is a perfect square number? (a) 122 (b) 144 (c) 150 (d) 200
Answer: (b) $144$ ($= 12^2$).
2. Which of the following digits cannot be in the unit place of a square number? (a) 4 (b) 5 (c) 6 (d) 7
Answer: (d) $7$.
3. $\sqrt{225} = ?$ (a) 15 (b) 25 (c) 45 (d) 12
Answer: (a) $15$.
4. The sum of the first $n$ odd numbers is (a) $n$ (b) $2n$ (c) $n^2$ (d) $n^2 + 1$
Answer: (c) $n^2$.
5. Which of the following is a Pythagorean triplet? (a) (2,3,4) (b) (6,8,10) (c) (1,2,3) (d) (4,5,6)
Answer: (b) $(6,8,10)$ ($6^2 + 8^2 = 100 = 10^2$).
6. $90^2 = ?$ (a) 810 (b) 8100 (c) 8000 (d) 9000
Answer: (b) $8100$.
7. $\sqrt{1.44} = ?$ (a) 1.2 (b) 12 (c) 0.12 (d) 1.44
Answer: (a) $1.2$.
8. The square root of a 6-digit perfect square number has how many digits? (a) 2 (b) 3 (c) 4 (d) 6
Answer: (b) $3$.
9. $45^2 = ?$ (a) 2025 (b) 2005 (c) 1625 (d) 2500
Answer: (a) $2025$.
10. $36 = 6^2$ is the sum of the first how many consecutive odd numbers? (a) 5 (b) 6 (c) 7 (d) 8
Answer: (b) $6$ ($1+3+5+7+9+11 = 36$).
Fill in the Blanks
- The square of an odd number is an _______ number. (odd)
- The symbol $\sqrt{\ }$ denotes the _______ square root of a number. (positive)
- $1 + 3 + 5 + 7 + 9 = $ _______. ($25$)
- If a perfect square ends in $0$, it must end in an _______ number of zeros. (even)
- $(3, 4, 5)$ is a _______ triplet. (Pythagorean)
True or False
- The square of every even number is even. — True
- A number ending in $2, 3, 7$ or $8$ can be a perfect square. — False
- A perfect square number has two square roots. — True
- $\sqrt{100} = 10$. — True
- Every number is a perfect square number. — False
Short Answer Questions
1. What is a perfect square number? Give an example.
Answer: A number that can be expressed as the square of an integer is called a perfect square number. For example, $49 = 7^2$ is a perfect square number.
2. What is a Pythagorean triplet? Give an example.
Answer: Three natural numbers in which the sum of the squares of two equals the square of the third, i.e. $a^2 + b^2 = c^2$, form a Pythagorean triplet. For example, $(3, 4, 5)$.
3. Find $\sqrt{1296}$ by the prime factorisation method.
Answer: $1296 = 2^4 \times 3^4 = (2^2 \times 3^2)^2 = 36^2$, so $\sqrt{1296} = 36$.
4. Using the formula $2p,\ p^2 – 1,\ p^2 + 1$, write a Pythagorean triplet for $p = 3$.
Answer: For $p = 3$: $2p = 6$, $p^2 – 1 = 8$, $p^2 + 1 = 10$; so the triplet is $(6, 8, 10)$. Check: $6^2 + 8^2 = 36 + 64 = 100 = 10^2$.
Key Terms
| Term | Meaning |
|---|---|
| Square | The product obtained by multiplying a number by itself |
| Square number | A number expressible as the square of a number |
| Perfect square | A number that is the square of an integer |
| Square root | The inverse operation of squaring |
| Prime factor | A factor that is a prime number |
| Division method | The long-division technique to find a square root |
| Hypotenuse | The longest side of a right-angled triangle |
| Pythagorean triplet | Three numbers satisfying $a^2 + b^2 = c^2$ |
| Triangular number | Numbers of the form $1, 3, 6, 10, \dots$ |
| Unit place | The rightmost digit position of a number |
| Decimal number | A number containing a decimal point |
| Consecutive odd numbers | Odd numbers that follow one another |