This chapter teaches two concurrency properties of a triangle (incentre and circumcentre) and how to construct a triangle with ruler and compass when enough sides and angles are given. Always draw a rough figure first.
Key points
The three angle bisectors of a triangle are concurrent. The point is the incentre (I); it is always inside the triangle and equally far from all three sides.
The three perpendicular bisectors of the sides are concurrent. The point is the circumcentre (C); it is equally far from all three vertices.
Circumcentre lies inside an acute triangle, on the hypotenuse (its midpoint) of a right triangle, and outside an obtuse triangle.
Two sides alone cannot fix a unique triangle; we need three suitable measures.
three sides (SSS),
two sides and the included angle (SAS),
two angles and the included side (ASA),
the hypotenuse and one side of a right-angled triangle.
A unique triangle can be constructed when we know:
If two angles and a non-included side are given, find the third angle (sum of angles = 180°) and use ASA.
Rough figure: show the longest side as longest, an obtuse angle as obtuse, and mark all given measures.
Important terms
Concurrent lines – three or more lines passing through one point.
Angle bisector – ray that divides an angle into two equal parts.
Perpendicular bisector – line through the midpoint of a segment and perpendicular to it.
Incentre – point of concurrence of angle bisectors.
Circumcentre – point of concurrence of perpendicular bisectors of sides.
Formulas
Sum of angles of a triangle = 180°
For a triangle to exist: sum of any two sides > third side
5 important questions
Q1. Construct ΔXYZ with XY = 6 cm, YZ = 4 cm, XZ = 5 cm. Write the steps.
Draw base XY = 6 cm. 2. With centre X and radius 5 cm, draw an arc. 3. With centre Y and radius 4 cm, draw an arc cutting the first arc at Z. 4. Join XZ and YZ.
Q2. Construct ΔPQR with PQ = 5.5 cm, ∠P = 50°, PR = 5 cm. Write the steps.
Draw PQ = 5.5 cm. 2. At P draw ray PG making 50° with PQ. 3. With centre P and radius 5 cm cut ray PG at R. 4. Join QR.
Q3. In ΔABC, ∠A = 60°, ∠B = 40°, AC = 6 cm. What else is needed to draw it?
∠C = 180° − (60° + 40°) = 80°. Now AC is the side between ∠A and ∠C, so draw AC = 6 cm, make 60° at A and 80° at C (ASA).
Q4. Three friends live at three different places and a toy shop is equidistant from their houses. Which construction shows the shop?
Join the three houses to form a triangle and draw the perpendicular bisectors of its sides. Their point of concurrence, the circumcentre, is equidistant from the three vertices – that is the shop.
Q5. Where does the circumcentre of a right-angled triangle lie?
On the hypotenuse, at its midpoint.
Exam tip
Always draw and label a rough figure before the actual construction, and do not erase construction arcs.
Integers include positive numbers, negative numbers and zero. This chapter teaches the sign rules for multiplying and dividing integers and how to write quotients of integers as fractions.
This chapter revises prime and co-prime numbers, introduces twin primes, and teaches how to factorise numbers into primes. Using prime factors (and the division method), we find the HCF and LCM of numbers and use them in word problems.
Key points
1 is neither prime nor composite. 2 is the only even prime number.
Co-prime numbers have only 1 as a common factor, e.g. (10, 21), (4, 9). Any two consecutive natural numbers are co-prime.
Twin primes are prime numbers with a difference of 2, e.g. (3, 5), (11, 13), (29, 31).
Euclid's rule: every composite number can be written as a product of primes. Example: 24 = 2 × 2 × 2 × 3; 250 = 2 × 5 × 5 × 5.
HCF (GCD): the greatest number that divides the given numbers exactly. Multiply the common prime factors.
LCM: the smallest number that is a multiple of all the given numbers. Multiply common factors once and all remaining factors.
HCF can also be found by the division method: divide the bigger number by the smaller, then the divisor by the remainder, until the remainder is 0; the last divisor is the HCF.
HCF of co-prime numbers is 1 and their LCM is their product.
HCF helps reduce fractions to lowest terms and in "greatest length/size" problems; LCM is used in "smallest number/same time again" problems.
Important terms
Prime number – has exactly two factors, 1 and itself.
Composite number – has more than two factors.
Prime factorisation – writing a number as a product of primes.
HCF – highest common factor. LCM – lowest common multiple.
Formulas
HCF × LCM = product of the two numbers
Other number = (HCF × LCM) ÷ given number
5 important questions
Q1. Find the HCF and LCM of 36 and 60 by prime factors.
This chapter is about the interior and exterior of an angle and different pairs of angles – adjacent, complementary, supplementary, linear pair and vertically opposite angles. It also proves that an exterior angle of a triangle equals the sum of its remote interior angles.
Key points
Points not on the arms of an angle are either in its interior or its exterior.
Adjacent angles: common vertex, common arm and separate interiors.
Complementary angles: two angles whose sum is 90° (e.g. 40° and 50°).
Supplementary angles: two angles whose sum is 180° (e.g. 110° and 70°).
Opposite rays: two rays with a common origin forming a straight line (180°).
Linear pair: adjacent angles whose non-common arms are opposite rays. Angles in a linear pair are supplementary.
When two lines intersect, vertically opposite angles are equal.
Exterior angle of a triangle: formed by extending one side; it forms a linear pair with the adjacent interior angle.
Exterior angle property: exterior angle = sum of the two remote (opposite) interior angles.
Sum of the three angles of a triangle = 180°.
Important terms
Vertex – common point of the two arms of an angle.
Complement of x° = (90 − x)°.
Supplement of x° = (180 − x)°.
Remote interior angles – the two interior angles not adjacent to the exterior angle.
Formulas
Complementary: ∠A + ∠B = 90°
Supplementary: ∠A + ∠B = 180°
Linear pair: ∠1 + ∠2 = 180°
Exterior angle ∠PRS = ∠P + ∠Q
5 important questions
Q1. Find the complement of 35° and the supplement of 112°.
A rational number can be written as m/n where m and n are integers and n ≠ 0. This chapter covers operations on rational numbers, multiplicative inverses, comparing rational numbers, their decimal form (terminating or recurring), and simplifying expressions using BODMAS.
Key points
Every natural number, whole number and integer is a rational number: 4 = 12/3, −3 = −3/1, 0 = 0/2.
Operations on rational numbers are done like operations on fractions.
Multiplicative inverse (reciprocal): two numbers whose product is 1, e.g. 5/6 and 6/5; (−4/7) and (−7/4).
−11/9 and 9/11 are NOT multiplicative inverses because their product is −1.
To divide by a number, multiply by its multiplicative inverse.
Zero has no multiplicative inverse; we cannot divide by zero.
Natural numbers are closed under + and ×; integers under +, − and ×; rational numbers under +, −, × and ÷ (except by zero).
Between any two rational numbers there are infinitely many rational numbers.
Decimal form: some rational numbers have a terminating decimal (7/4 = 1.75), others a recurring decimal (2/11 = 0.1818…, 5/6 = 0.833…).
A bar over digits shows repetition.
Simplify expressions using BODMAS: Brackets, Of, Division, Multiplication, Addition, Subtraction.
Important terms
Rational number – number of the form m/n, n ≠ 0.
Reciprocal – multiplicative inverse.
Terminating decimal – decimal that ends.
Recurring decimal – decimal in which a digit or group of digits repeats endlessly.
Formulas
a/b + c/d = (ad + bc)/bd
a/b × c/d = ac/bd
a/b ÷ c/d = a/b × d/c (c ≠ 0)
To compare a/b and c/d (b, d positive): compare ad and bc
5 important questions
Q1. Add 5/7 + 9/11.
LCM of 7 and 11 = 77. 5/7 = 55/77, 9/11 = 63/77. Sum = 118/77.
Q2. Divide (−3/5) by (4/7).
(−3/5) ÷ (4/7) = (−3/5) × (7/4) = −21/20.
Q3. Write the multiplicative inverse of −2/7 and 13.
Of −2/7 it is −7/2; of 13 it is 1/13.
Q4. Write 5/12 in decimal form. Is it terminating?
5 ÷ 12 = 0.41666… = 0.416 with 6 repeating. It is a recurring decimal.
Q5. Compare −7/9 and −4/5.
Cross-multiply: (−7) × 5 = −35 and (−4) × 9 = −36. Since −35 > −36, −7/9 > −4/5.
Exam tip
In BODMAS questions solve brackets first and show each step on a new line.
When a number is multiplied by itself many times, we write it in index form, like 2⁸. This chapter explains base and index, the laws of indices, and how to find square roots by prime factorisation.
Key points
In aᵐ, a is the base and m is the index (exponent): aᵐ = a × a × … (m times). Read as "a raised to the power m".
5⁴ = 5 × 5 × 5 × 5 = 625.
First power of any number is the number itself: a¹ = a.
Second power = square (3² = 9); third power = cube (5³ = 125).
A negative base with an even index gives a positive answer; with an odd index gives a negative answer: (−7)⁴ = 2401, (−6)³ = −216.
Any non-zero number raised to 0 is 1: a⁰ = 1.
a⁻ᵐ = 1/aᵐ (reciprocal).
Square root by prime factors: factorise, make pairs of equal factors, take one from each pair and multiply. √144 = 2 × 2 × 3 = 12.
Square roots of non-perfect squares can be approximated, e.g. √137 lies between 11.7 and 11.8.
Important terms
Base – the number being multiplied repeatedly.
Index / Exponent / Power – number of times the base is multiplied.
Perfect square – square of a whole number (1, 4, 9, 16, …).
Square root (√) – number whose square is the given number.
A joint bar graph shows two (or more) sets of related data side by side in one graph. It saves space and makes comparison easy. This chapter teaches how to read and draw joint bar graphs.
Key points
A simple bar graph shows one set of data; a joint bar graph shows two related sets together, e.g. wheat production of Ajay and Vijay in the same years.
Each pair of bars stands on the same mark on the X-axis (year, day, class) and touches each other; there is a gap between two pairs.
Use different colours or shading for each set and show them in an index (key).
Choose a suitable scale on the Y-axis, e.g. 1 cm = 10 quintals or 1 cm = 5°C, and write it on the graph.
Write a title and label both axes.
From a joint bar graph we can compare values, find the highest/lowest, and find differences.
Example: maximum and minimum temperature of Pune over five days – difference on a day = height of maximum bar − height of minimum bar.
Important terms
X-axis – horizontal line; usually shows categories (years, days).
Y-axis – vertical line; shows quantity.
Scale – the value one unit of length represents.
Index / Key – box showing which colour stands for which data.
Formulas
Length of bar = value ÷ value of 1 cm on the scale
Example: scale 1 cm = 10 quintals; 45 quintals → bar of 4.5 cm
5 important questions
Q1. What is a joint bar graph? Why is it useful?
A bar graph that shows two or more related sets of data side by side. It shows more information in less space and makes comparison easy.
Q2. What steps do you follow to draw a joint bar graph?
Draw X and Y axes; choose a suitable scale for the Y-axis; draw pairs of bars of equal width for each category with no gap within a pair; shade them differently; give an index, title and axis labels.
Q3. On a scale of 1 cm = 5°C, how long will bars for 35°C and 22°C be?
35 ÷ 5 = 7 cm; 22 ÷ 5 = 4.4 cm.
Q4. Maximum temperature on Thursday was 34°C and minimum was 21°C. Find the difference.
Difference = 34 − 21 = 13°C.
Q5. Ajay produced 50 quintals and Vijay 40 quintals in 2014. Who produced more and by how much?
Ajay produced 50 − 40 = 10 quintals more.
Exam tip
Never forget the scale and the index on your graph – marks are given for them.
Letters (variables) are used to stand for numbers that can change. This chapter explains algebraic expressions, terms, coefficients, like and unlike terms, types of expressions, and how to add, subtract and multiply them and find their value.
Key points
Number of sticks needed for n squares in a row = 3n + 1. Here n is a variable and 3n + 1 is an algebraic expression.
A term is a part of an expression in which only multiplication is used: in 4x² − 2y + xz, the terms are 4x², −2y, xz.
In 3x, 3 is the coefficient of x; in −15t the coefficient is −15.
Like terms have the same variables with the same powers (5x and −2x). Unlike terms do not (3x and 3y).
Types: monomial (one term: 7xy), binomial (two terms: a + b), trinomial (three terms: x² + x + 1), polynomial (one or more terms with whole-number powers).
Only like terms can be added or subtracted – add or subtract their coefficients.
To subtract, change the sign of every term of the expression being subtracted and add.
To multiply: multiply coefficients, and for the same variable add the powers. Multiply each term of one expression by each term of the other.
To find the value, substitute the given numbers for the variables.
Two quantities are in direct proportion if both increase or decrease together in the same ratio, and in inverse proportion if one increases when the other decreases so that their product stays constant. The chapter also teaches sharing profit in a business partnership.
Key points
Direct proportion: the ratio of the two quantities stays constant. E.g. number of children and notebooks (15 : 90 = 12 : 72 = 1 : 6), petrol filled and distance travelled.
Inverse proportion: the product of the two quantities stays constant. E.g. number of workers and days needed; speed and time for a fixed distance.
Sudha's 48 problems: 8 a day → 6 days, 12 a day → 4 days; 8 × 6 = 12 × 4 = 48.
In direct proportion, x/y = k; in inverse proportion, x × y = k.
Partnership: two or more people invest capital in a business. Profit or loss is shared in the ratio of their investments.
To share an amount in ratio a : b, take shares as ax and bx, so ax + bx = total.
Important terms
Ratio – comparison of two quantities by division.
Proportion – equality of two ratios.
Capital – money invested to start a business.
Partnership – business started jointly by two or more people.
Formulas
Direct: x₁/y₁ = x₂/y₂
Inverse: x₁ × y₁ = x₂ × y₂
Share of a partner = (his part of the ratio ÷ sum of ratio parts) × total profit
5 important questions
Q1. 15 workers take 8 hours to build a wall. How long will 12 workers take?
Inverse proportion: 12 × x = 15 × 8 = 120, so x = 10 hours.
Q2. If 5 kg of rice costs ₹225, what is the cost of 8 kg?
Direct proportion: 225/5 = x/8, so x = 45 × 8 = ₹360.
Q3. Jhelum and Atharva invested ₹2100 and ₹2800 and earned a profit of ₹3500. Share it.
Banks are government-recognised institutions that keep our money safe and give loans. This chapter explains types of bank accounts, passbook entries, ATM, debit and credit cards, and how to calculate simple interest and the amount.
Key points
Current account: for traders; money can be deposited or withdrawn any number of times; no interest is paid.
Savings account: interest is paid on the daily balance; some limits on withdrawals; passbook and cheque book given.
Recurring deposit (RD): a fixed amount is deposited every month; higher interest than savings; compulsory saving.
Fixed deposit (FD): a sum is deposited for a fixed period; higher interest; senior citizens get a slightly higher rate.
A joint account can be operated by more than one person (husband–wife, partners, societies).
ATM card withdraws cash without going to the bank; debit/credit cards allow cashless payments.
A passbook records date, particulars, cheque number, amount withdrawn, amount deposited and balance.
Principal (P): money borrowed or deposited. Rate (R): interest on ₹100 for one year, written "p.c.p.a." (per cent per annum). Time (N) in years.
Interest: extra money paid for using someone's money. Amount = principal + interest.
Important terms
Principal – the original sum of money.
Rate of interest – interest on ₹100 for 1 year.
Simple interest – interest calculated only on the principal.
Amount – total money returned (P + I).
Formulas
Simple interest I = (P × R × N) ÷ 100
Amount A = P + I
P = (I × 100) ÷ (R × N); R = (I × 100) ÷ (P × N); N = (I × 100) ÷ (P × R)
5 important questions
Q1. Neha took a loan of ₹50,000 at 12 p.c.p.a. What amount will she repay after one year?
Q4. Distinguish between a current account and a savings account.
A current account is for traders, allows unlimited transactions and earns no interest. A savings account earns interest on the balance but has some limits on withdrawals.
Q5. What is a fixed deposit?
Money deposited in a bank for a fixed period. It earns a higher rate of interest than a savings account, and senior citizens get a slightly higher rate.
Exam tip
Write P, R, N clearly before substituting in I = PRN/100, and convert months into years (6 months = ½ year).
This chapter explains the circumference of a circle and the constant π, and then arcs of a circle – minor arc, major arc, semicircular arc – and how to measure them using central angles.
Key points
Parts of a circle: centre, radius, chord, diameter (longest chord), circumference.
Diameter d = 2 × radius r.
For every circle, circumference ÷ diameter is a constant a little more than 3, called π (pi).
π is not a rational number; we use π = 22/7 or 3.14. If not given, take 22/7.
One full rotation of a wheel covers a distance equal to its circumference.
A chord divides a circle into two arcs: the smaller is the minor arc, the bigger is the major arc. A diameter divides it into two semicircular arcs.
Central angle: angle with its vertex at the centre of the circle.
Measure of a minor arc = its central angle.
Measure of a major arc = 360° − measure of the corresponding minor arc.
Measure of a semicircular arc = 180°; measure of a complete circle = 360°.
Important terms
Circumference – length of the boundary of a circle.
Arc – part of the circumference.
Corresponding arcs – a minor and major arc with the same end points that together make the full circle.
Central angle – angle subtended by an arc at the centre.
Formulas
Circumference c = πd = 2πr
d = c ÷ π ; r = c ÷ 2π
Number of rotations = distance ÷ circumference
m(major arc) = 360° − m(minor arc)
5 important questions
Q1. The circumference of a circle is 198 cm. Find its radius and diameter.
198 = 2 × 22/7 × r, so r = 198 × 7 ÷ 44 = 31.5 cm. Diameter = 63 cm.
Q2. The circumference of a circle is 62.8 cm. Taking π = 3.14, find its diameter.
d = 62.8 ÷ 3.14 = 20 cm.
Q3. A wheel has a diameter of 0.7 m. How many rotations will it make in 22 km?
Circumference = 22/7 × 0.7 = 2.2 m. 22 km = 22000 m. Rotations = 22000 ÷ 2.2 = 10000.
Q4. A minor arc measures 110°. What is the measure of its major arc?
360° − 110° = 250°.
Q5. Find the circumference of a circle of radius 7 cm.
c = 2 × 22/7 × 7 = 44 cm.
Exam tip
Convert all lengths to the same unit (km to m) before dividing distance by circumference.
Perimeter is the length of the boundary of a closed figure and area is the region it covers. This chapter revises perimeter and area of rectangles and squares, finds the area of a triangle, solves path and tile problems, and introduces the surface area of a cuboid and a cube.
Key points
Perimeter of a polygon = sum of all its sides.
Perimeter of a square = 4 × side; of a rectangle = 2(l + b).
Area of a rectangle = l × b; of a square = side².
Two identical right-angled triangles make a rectangle, so area of a right triangle = ½ × product of the sides forming the right angle.
Area of any triangle = ½ × base × height (height is the perpendicular from the opposite vertex to the base, even if it falls outside).
Path inside a garden: area of path = area of garden − area of inner rectangle (subtract 2 × width from both length and breadth).
Path outside a ground: area of path = area of outer rectangle (add 2 × width) − area of ground.
Number of tiles = area to be covered ÷ area of one tile (same units!).
Surface area of a 3-D object = sum of the areas of all its faces.
A cuboid has 6 rectangular faces; opposite faces are equal. A cube has 6 equal square faces.
Units: 1 m = 100 cm, so 1 sq m = 10,000 sq cm.
Important terms
Perimeter – total length of the boundary.
Area – amount of surface covered, in square units.
Height of a triangle – perpendicular distance from a vertex to the opposite side.
Surface area – total area of all faces of a solid.
Formulas
Area of triangle = ½ × b × h
Total surface area of cuboid = 2(lb + bh + lh)
Total surface area of cube = 6l²
Area of path (outside) = (l + 2w)(b + 2w) − lb
5 important questions
Q1. The perimeter of a rectangle is 64 cm and its length is 17 cm. Find its breadth.
2(17 + x) = 64, so 17 + x = 32, x = 15 cm.
Q2. Find the area of a triangle with base 5.6 cm and height 4.5 cm.
Area = ½ × 5.6 × 4.5 = 12.6 sq cm.
Q3. A garden is 40 m by 30 m. A 2 m wide path runs inside along the boundary. Find the area of the path.
Inner rectangle = 36 m × 26 m = 936 sq m. Garden = 1200 sq m. Path = 1200 − 936 = 264 sq m.
Q4. A playground is 65 m by 30 m with a 1.5 m wide path outside it. Find the area of the path.
Outer rectangle = 68 m × 33 m = 2244 sq m. Ground = 1950 sq m. Path = 2244 − 1950 = 294 sq m.
Q5. How much sheet metal is needed for a closed box 1.5 m × 1.2 m × 1.3 m?
In a right-angled triangle, the side opposite the right angle is the hypotenuse. Pythagoras' theorem connects the three sides: the square of the hypotenuse equals the sum of the squares of the other two sides. We use it to find a missing side and to check whether a triangle is right-angled.
Key points
The hypotenuse is the side opposite the right angle and is the longest side.
Pythagoras' theorem: in a right-angled triangle, (hypotenuse)² = (side 1)² + (side 2)².
Pythagoras was a Greek mathematician of the 6th century BCE. The same principle is given in the ancient Indian book Shulvasutra; Pythagoras was the first to prove it, so it is named after him.
If the sides of a triangle satisfy a² + b² = c² (c the longest), the triangle is right-angled (converse).
The product of algebraic expressions written out fully is called its expansion. This chapter gives the formulae for the square of a sum, the square of a difference and the product (a + b)(a − b), and uses them to calculate quickly and to factorise expressions.
Key points
(a + b)² can be shown as a square of side (a + b) divided into a², b² and two rectangles ab.
(a − b)² is shown by removing parts from a square of side a.
These formulae make mental calculation easy: 102² = (100 + 2)², 97² = (100 − 3)².
(a + b)(a − b) = a² − b²; so 102 × 98 = 100² − 2² = 9996.
Factorising means writing an expression as a product of its factors. Take out the common factor first: 9a²bc + 12abc² = 3abc(3a + 4c).
A difference of two squares factorises as a² − b² = (a + b)(a − b).
Important terms
Expansion – the product of algebraic expressions written as a sum of terms.
Binomial – expression with two terms.
Factor – an expression that divides another exactly.
Factorisation – writing an expression as a product of factors.
Statistics helps us collect and understand data. In this chapter we learn to find the average (mean) of scores and to arrange data in a frequency distribution table using tally marks.
Key points
The values collected in data are called readings or scores.
Average (mean) = sum of all scores ÷ number of scores. It gives a single value that represents the data.
The average may be a fraction even when the scores are whole numbers (e.g. average skips 60.42 per minute).
Convert all scores to the same unit before averaging (e.g. km to m).
Frequency: the number of times a score occurs in the data.
Frequency distribution table has three columns: score, tally marks, frequency.
Write scores in ascending order; for each reading put a tally mark; the fifth tally crosses the first four (bundle of five) for easy counting.
The sum of all frequencies is written as N; it equals the total number of scores.
Important terms
Data – collection of numerical information.
Score / Reading – each value in the data.
Average / Mean – sum of scores divided by their number.
Frequency – number of times a score appears.
Tally mark – a stroke used to count each occurrence.
Formulas
Average = (sum of all scores) ÷ (total number of scores)
Sum of scores = average × number of scores
N = sum of all frequencies
5 important questions
Q1. Asmita took 20, 20, 22, 18, 18, 20 minutes to reach school on six days. Find the average.
Sum = 118. Average = 118 ÷ 6 = 19⅔ minutes.
Q2. Daily rainfall in a week was 9, 11, 8, 20, 10, 16, 12 mm. Find the average rainfall.
Sum = 86 mm. Average = 86 ÷ 7 ≈ 12.29 mm.
Q3. The average of 5 numbers is 24. Find their sum.
Sum = 24 × 5 = 120.
Q4. Make a frequency table: 1, 3, 2, 4, 5, 4, 1, 3, 4, 5, 6, 4, 6, 4, 6.
Distance (km)
Frequency
1
2
2
1
3
2
4
5
5
2
6
3
N = 15.
Q5. Distances of students' homes are 950 m, 800 m, 700 m, 1.5 km, 1 km, 750 m. Find the average.
In metres: 950 + 800 + 700 + 1500 + 1000 + 750 = 5700. Average = 5700 ÷ 6 = 950 m.
Exam tip
Check that the total of the frequency column (N) equals the number of scores – it catches counting mistakes.