This chapter revises natural numbers, whole numbers, integers and rational numbers, and then shows how to mark rational numbers on a number line, compare them, write them in decimal form, and represent the irrational number √2 on a number line. All numbers shown on a number line are real numbers.
A rational number (परिमेय संख्या) is of the form m/n, where m and n are integers and n ≠ 0.
There are infinitely many rational numbers between any two rational numbers.
To show 7/3 on a number line, divide each unit into 3 equal parts and count 7 parts to the right of 0 (7/3 = 2 + 1/3). −2/3 is the same distance to the left of 0.
On a number line, a number to the left is always smaller.
Multiplying numerator and denominator by the same non-zero number does not change the value: a/b = ka/kb.
A negative number is always less than a positive number. If a < b, then −a > −b.
Decimal form: if division ends, it is a terminating decimal (7/4 = 1.75); otherwise it is non-terminating recurring (7/6 = 1.1666…).
Every rational number can be written as a non-terminating recurring decimal.
Irrational numbers (अपरिमेय संख्या) cannot be written as m/n; their decimal form is non-terminating and non-recurring, e.g. √2, π.
22/7 and 3.14 are only rational approximations of π.
Rational and irrational numbers together form the real numbers.
Important terms
Rational number – number of the form m/n, n ≠ 0.
Irrational number – number that cannot be written as m/n.
Terminating decimal – decimal form in which division ends.
Recurring decimal – decimal in which a digit or group of digits repeats endlessly.
Real numbers – all numbers that can be shown on a number line.
Formulas
For b, d positive: if a × d < b × c then a/b < c/d; if a × d = b × c then a/b = c/d; if a × d > b × c then a/b > c/d.
√2 on number line: right triangle with sides 1 and 1 gives hypotenuse √(1² + 1²) = √2.
5 important questions
Q1. Compare 5/4 and 2/3.
Cross-multiply: 5 × 3 = 15 and 4 × 2 = 8. Since 15 > 8, 5/4 > 2/3.
Q2. Compare −7/3 and −5/2.
First compare 7/3 and 5/2: 7 × 2 = 14, 3 × 5 = 15, so 7/3 < 5/2. For negatives the order reverses: −7/3 > −5/2.
Q3. Write 7/6 and 22/7 in decimal form.
7/6 = 1.1666… = 1.16 with bar on 6. 22/7 = 3.142857142857… = 3.142857 with bar on 142857. Both are non-terminating recurring.
Q4. Show √2 on a number line.
Mark A at 1. Draw AP ⊥ number line with AP = 1 unit. Join OP; by Pythagoras OP² = 1² + 1² = 2, so OP = √2. With centre O and radius OP draw an arc cutting the number line at Q. Q shows √2.
Q5. Is 3/5 = 6/10? Check.
3 × 10 = 30 and 5 × 6 = 30. Products are equal, so 3/5 = 6/10.
Exam tip
For comparing negatives, compare the positive parts first and then reverse the sign.
A line that cuts two or more lines in distinct points is called a transversal. This chapter names the angles formed by a transversal (corresponding, alternate and interior angles), states their properties when the lines are parallel, and gives the tests to decide whether two lines are parallel.
Key points
A transversal (छेदिका) intersects two lines in two distinct points and forms 8 angles.
Corresponding angles: arms on the transversal point in the same direction and the other arms are on the same side of the transversal. There are 4 pairs.
Interior angles: a pair on the same side of the transversal, inside the two lines. There are 2 pairs.
Alternate angles: on opposite sides of the transversal – 2 pairs of interior alternate and 2 pairs of exterior alternate angles.
When two parallel lines are cut by a transversal:
each pair of corresponding angles is congruent;
each pair of alternate angles is congruent;
each pair of interior angles is supplementary (sum 180°).
Tests for parallel lines (converse): if a transversal makes congruent corresponding angles, or congruent alternate angles, or supplementary interior angles, then the two lines are parallel.
Vertically opposite angles are always equal; angles in a linear pair add up to 180°.
Two lines in a plane perpendicular to the same line are parallel to each other.
Important terms
Transversal – line intersecting two or more lines at distinct points.
Corresponding angles – angles in matching positions at the two intersection points.
Alternate angles – angles on opposite sides of the transversal, both inside or both outside the lines.
Interior angles – angles on the same side of the transversal between the two lines.
Supplementary angles – two angles whose sum is 180°.
Formulas
For parallel lines with a transversal:
Corresponding angles equal
Alternate angles equal
Interior angles: ∠1 + ∠2 = 180°
Linear pair: sum = 180°; vertically opposite angles equal
5 important questions
Q1. Line AB ∥ line CD and a transversal makes ∠MNQ = 70° at N on line CD. Find ∠AON at O on line AB, where ∠AON and ∠ONP are interior angles.
∠ONP = ∠MNQ = 70° (vertically opposite). Interior angles are supplementary: ∠AON + 70° = 180°, so ∠AON = 110°.
Q2. Two parallel lines are cut by a transversal. One interior angle is (x + 15)° and the other on the same side is (2x + 15)°. Find x.
Interior angles are supplementary: x + 15 + 2x + 15 = 180 → 3x + 30 = 180 → 3x = 150 → x = 50. Angles are 65° and 115°.
Q3. State the properties of angles formed by two parallel lines and a transversal.
Corresponding angles are congruent, alternate angles are congruent and interior angles are supplementary.
Q4. If one angle formed by two parallel lines and a transversal is 75°, find all other angles.
Four angles are 75° (the angle, its vertically opposite, corresponding and alternate angles) and four are 180° − 75° = 105°.
Q5. How can you prove two lines are parallel using a transversal?
Show that a pair of corresponding angles or alternate angles is congruent, or that a pair of interior angles is supplementary.
Exam tip
Always write the reason in brackets (corresponding angles, alternate angles etc.) beside each step – marks are given for reasons.
This chapter revises the laws of indices and extends them to rational indices like 1/n and m/n. It then explains cubes and cube roots and how to find the cube root of a perfect cube by the factor method.
Key points
In 2⁵, 2 is the base, 5 is the index (घातांक) and 2⁵ is the index form.
Laws of indices hold for all integers m and n (see Formulas).
a^(1/n) means the nth root of a: 25^(1/2) = √25 = 5; 64^(1/3) = ∛64 = 4.
a^(m/n) can be read in two ways: the nth root of the mth power, (aᵐ)^(1/n), or the mth power of the nth root, (a^(1/n))ᵐ. Both give the same value.
Example: 8^(2/3) = (8²)^(1/3) = 64^(1/3) = 4, and also (8^(1/3))² = 2² = 4.
The cube of a number is the number multiplied by itself three times: a³ = a × a × a.
If a³ = b, then a is the cube root of b: ∛b = a.
Cube root by factor method: write the number as a product of prime factors, make groups of three equal factors, and take one factor from each group.
A number is a perfect cube only if all its prime factors can be grouped in threes.
Cube of a negative number is negative, so the cube root of a negative number is negative: ∛(−27) = −3.
Important terms
Base – the number that is multiplied repeatedly.
Index (exponent) – how many times the base is multiplied.
Cube – product of a number taken three times.
Cube root – the number whose cube gives the given number.
Perfect cube – a number whose cube root is an integer.
This chapter introduces two kinds of segments in a triangle – altitudes and medians. The three altitudes meet at the orthocentre, and the three medians meet at the centroid. It also shows where these points lie in acute, right and obtuse triangles and the special 2 : 1 property of the centroid.
Key points
Recall: angle bisectors of a triangle meet at the incentre; perpendicular bisectors of sides meet at the circumcentre.
An altitude (शिरोलंब) is the perpendicular segment drawn from a vertex to the opposite side (or the line containing it).
Every triangle has three altitudes, and they are concurrent. Their point of concurrence is the orthocentre, usually shown by O.
Orthocentre position: inside an acute triangle; at the vertex of the right angle in a right-angled triangle; outside an obtuse triangle.
For an obtuse triangle, the opposite sides must be extended to draw two of the altitudes.
A median (मध्यगा) is the segment joining a vertex to the midpoint of the opposite side.
Every triangle has three medians, and they are concurrent. Their point of concurrence is the centroid, usually shown by G.
The centroid always lies inside the triangle.
The centroid divides each median in the ratio 2 : 1 – the part from vertex to centroid is twice the part from centroid to midpoint.
In an equilateral triangle, the altitude and median from a vertex are the same segment, so the orthocentre and centroid coincide.
Important terms
Altitude – perpendicular segment from a vertex to the opposite side.
Orthocentre – point where the three altitudes meet.
Median – segment joining a vertex to the midpoint of the opposite side.
Centroid – point where the three medians meet.
Concurrent lines – three or more lines passing through one point.
Formulas
Centroid G on median AD: AG : GD = 2 : 1, so AG = (2/3) AD and GD = (1/3) AD.
5 important questions
Q1. In △ABC, G is the centroid and median AD = 9 cm. Find AG and GD.
AG : GD = 2 : 1. AG = (2/3) × 9 = 6 cm; GD = (1/3) × 9 = 3 cm.
Q2. G is the centroid of △PQR and the median PM. If GM = 2.5 cm, find PG and PM.
PG = 2 × GM = 5 cm. PM = PG + GM = 5 + 2.5 = 7.5 cm.
Q3. Where does the orthocentre lie in acute, right and obtuse triangles?
Acute – inside; right – at the vertex of the right angle; obtuse – outside the triangle.
Q4. Distinguish between altitude and median.
An altitude is perpendicular from a vertex to the opposite side; a median joins a vertex to the midpoint of the opposite side. Altitudes meet at the orthocentre; medians at the centroid.
Q5. In which triangle do the altitude and median from a vertex coincide?
In an equilateral triangle (and from the apex of an isosceles triangle), the altitude and median from the vertex are the same segment.
Exam tip
Use a set-square to draw altitudes and compass to find midpoints for medians; keep construction marks visible.
This chapter builds new expansion formulae from the ones learnt earlier: the product (x + a)(x + b), the cubes (a + b)³ and (a − b)³, and the square of a trinomial (a + b + c)². These formulae make multiplication of algebraic expressions and some number calculations quick.
Key points
Earlier formulae: (a + b)² = a² + 2ab + b²; (a − b)² = a² − 2ab + b²; (a + b)(a − b) = a² − b².
(x + a)(x + b) is a product of two binomials with one common term x; it can be shown using areas of squares and rectangles.
(a + b)³ is found by multiplying (a + b) by (a + b)².
(a − b)³ is found similarly; signs alternate +, −, +, −.
The cube formulae can be written in short form using 3ab(a + b) and 3ab(a − b).
(a + b + c)² has six terms: three squares and three double products.
These formulae help find cubes of numbers near round numbers, e.g. 99³ = (100 − 1)³, 101³ = (100 + 1)³.
Expansion (विस्तार) means writing a product as a sum of terms.
Important terms
Binomial – algebraic expression with two terms.
Trinomial – algebraic expression with three terms.
Expansion – writing a product of expressions as a sum of terms.
Coefficient – number multiplied with a variable.
Formulas
(x + a)(x + b) = x² + (a + b)x + ab
(a + b)³ = a³ + 3a²b + 3ab² + b³ = a³ + b³ + 3ab(a + b)
(a − b)³ = a³ − 3a²b + 3ab² − b³ = a³ − b³ − 3ab(a − b)
Factorisation means writing an algebraic expression as a product of its factors. This chapter teaches how to factorise quadratic trinomials by splitting the middle term, how to factorise a³ + b³ and a³ − b³, and how to simplify rational algebraic expressions using factors.
Key points
Recall: common factor method, e.g. 4xy + 8xy² = 4xy(1 + 2y); difference of squares, p² − 9q² = (p + 3q)(p − 3q).
An expression of the form ax² + bx + c (a ≠ 0) is a quadratic trinomial.
To factorise ax² + bx + c, find two numbers whose product is a × c and whose sum is b. Split the middle term using them and take common factors in pairs.
Signs: if the constant term is positive, both numbers have the same sign as b; if it is negative, the numbers have opposite signs.
From (a + b)³ = a³ + b³ + 3ab(a + b) we get the factors of a³ + b³.
From (a − b)³ = a³ − b³ − 3ab(a − b) we get the factors of a³ − b³.
If A and B are algebraic expressions (B ≠ 0), A/B is a rational algebraic expression. It is simplified by factorising the numerator and denominator and cancelling common factors.
Rational algebraic expressions are added, subtracted, multiplied and divided like fractions.
Factorisation (अवयव पाडणे) is the reverse of expansion.
Important terms
Factor – an expression that divides another exactly.
Quadratic trinomial – expression ax² + bx + c with three terms and highest power 2.
Splitting the middle term – writing bx as the sum of two terms to factorise.
Rational algebraic expression – expression of the form A/B where B ≠ 0.
Variation tells us how one quantity changes when another changes. This chapter explains direct variation (both increase together in the same proportion) and inverse variation (one increases when the other decreases), the symbol ∝, the constant of variation, and their use in problems on time, work and speed.
Key points
"x and y are in direct variation" means x ∝ y, i.e. x = ky, where k is a non-zero constant called the constant of variation.
In direct variation, x/y is always constant. Examples: cost and number of articles; distance and time at constant speed; wages and days worked.
"x is inversely proportional to y" means x ∝ 1/y, i.e. x × y = k (constant).
In inverse variation, the product x × y is always constant. Examples: number of workers and days to finish a job; speed and time for a fixed distance.
The symbol ∝ is read "varies as" or "is proportional to".
Steps to solve: (1) write the variation, (2) change ∝ to = with k, (3) find k from given values, (4) use k to find the unknown.
Time, work, speed problems are mostly inverse variation: more workers → fewer days; more speed → less time.
Variation (चलन) problems can also be solved by ratio and proportion.
Important terms
Direct variation – two quantities increase or decrease in the same ratio; x/y = k.
Inverse variation – one quantity increases as the other decreases; xy = k.
Constant of variation – the fixed number k in a variation equation.
Formulas
Direct: x ∝ y ⇒ x = ky ⇒ x₁/y₁ = x₂/y₂
Inverse: x ∝ 1/y ⇒ xy = k ⇒ x₁y₁ = x₂y₂
Distance = speed × time
5 important questions
Q1. x varies directly as y. When x = 5, y = 30. Find the constant of variation and the equation.
x = ky → 5 = k × 30 → k = 1/6. Equation: x = y/6 (or y = 6x).
Q2. a varies inversely as b. When a = 10, b = 8. Find a when b = 20.
ab = k → k = 10 × 8 = 80. When b = 20: a = 80/20 = 4.
Q3. 15 workers build a wall in 48 hours. How many workers will build it in 30 hours?
Workers and time are in inverse variation: 15 × 48 = n × 30 → n = 720/30 = 24 workers.
Q4. The cost of 12 notebooks is ₹ 240. Find the cost of 20 notebooks.
Direct variation: cost/notebooks = 240/12 = 20. Cost of 20 = 20 × 20 = ₹ 400.
Q5. A car at 60 km/hr takes 8 hours for a journey. How long will it take at 80 km/hr?
Speed and time vary inversely: 60 × 8 = 80 × t → t = 480/80 = 6 hours.
Exam tip
First decide: if one quantity goes up and the other also goes up, it is direct; if the other goes down, it is inverse.
This chapter teaches how to construct a quadrilateral when five suitable measures (sides, angles or diagonals) are given, and then studies the properties of special quadrilaterals – rectangle, square, rhombus, parallelogram, trapezium and kite – especially the properties of their sides, angles and diagonals.
Key points
A quadrilateral (चौकोन) has 4 sides, 4 angles and 2 diagonals. The sum of its angles is 360°.
A unique quadrilateral can be constructed if 5 independent measures are known, e.g. 4 sides and 1 diagonal; 3 sides and 2 diagonals; 4 sides and 1 angle; 3 sides and 2 included angles.
Always draw a rough figure first and mark the given measures on it.
Rectangle: opposite sides equal, all angles 90°; diagonals are congruent and bisect each other.
Square: all sides equal, all angles 90°; diagonals are equal, bisect each other at right angles and bisect the opposite angles.
Rhombus: all sides equal; opposite angles equal; diagonals bisect each other at right angles and bisect the opposite angles.
Parallelogram: both pairs of opposite sides parallel and equal; opposite angles equal; adjacent angles supplementary; diagonals bisect each other.
Trapezium: only one pair of opposite sides is parallel. Adjacent angles between the parallel sides are supplementary. In an isosceles trapezium the non-parallel sides are equal.
Kite: two pairs of adjacent sides are equal; the diagonals are perpendicular, and the diagonal joining the vertices where equal sides meet bisects the other diagonal.
Every square is a rectangle and a rhombus; every rectangle and rhombus is a parallelogram.
Important terms
Diagonal – segment joining opposite vertices of a quadrilateral.
Adjacent sides – sides with a common vertex.
Opposite angles – angles that do not share a side.
Bisect – divide into two equal parts.
Formulas
Sum of angles of a quadrilateral = 360°
Parallelogram: adjacent angles sum to 180°
Diagonals of a rhombus/square are perpendicular bisectors of each other
5 important questions
Q1. In parallelogram ABCD, ∠A = 70°. Find ∠B, ∠C and ∠D.
Q2. The diagonals of a rhombus are 24 cm and 10 cm. Find its side.
Diagonals bisect at right angles, so half-diagonals are 12 cm and 5 cm. Side = √(12² + 5²) = √169 = 13 cm.
Q3. The diagonals of rectangle PQRS meet at T. If PR = 10 cm, find QS and PT.
Diagonals of a rectangle are equal and bisect each other: QS = 10 cm and PT = 5 cm.
Q4. Write the properties of diagonals of a square.
They are equal, bisect each other, are perpendicular to each other and bisect the opposite angles.
Q5. Construct quadrilateral ABCD with AB = 4.5 cm, BC = 5.2 cm, CD = 4 cm, AD = 3.8 cm and diagonal AC = 6 cm (steps).
Draw AC = 6 cm. With centres A and C, radii 4.5 cm and 5.2 cm, mark B on one side. With radii 3.8 cm and 4 cm, mark D on the other side. Join AB, BC, CD, DA.
Exam tip
Learn a comparison table of diagonal properties of all special quadrilaterals – true/false and reason questions come from it.
Shopkeepers often reduce the price written on an item to increase sales; this reduction is the discount. Agents who help sell goods or property get a payment called commission (brokerage). This chapter shows how to calculate discount, selling price, rate of discount, commission and rebate.
Key points
The price printed on the tag of an item is its marked price (छापील किंमत).
The amount by which the marked price is reduced while selling is the discount (सूट).
Selling price = marked price − discount.
Discount is usually given as a percentage of the marked price. A 20% discount on ₹ 100 means the item is sold for ₹ 80.
Even after a discount, a shopkeeper can make more total profit by selling more items.
Online shopping companies also offer discounts and home delivery.
A person who brings the seller and buyer together is an agent, mediator or commission agent (broker).
Commission is a percentage of the value of the sale; rates differ for different goods. It may be taken from the seller, the buyer, or both.
Foodgrains, vegetables, fruits, flowers, land and houses are often sold through agents.
The discount given on some goods (e.g. Khadi items, during special occasions) is also called a rebate.
Important terms
Marked price – price printed on the item.
Discount – reduction given on the marked price.
Selling price – price actually paid by the customer.
Commission / brokerage – payment to an agent for arranging a sale.
Rebate – discount, usually a special reduction on total purchase.
This chapter defines a polynomial and its degree, and then teaches division: a monomial by a monomial, a polynomial by a monomial, and a polynomial by a binomial using long division. The result is written as dividend = divisor × quotient + remainder.
Key points
An algebraic expression in one variable is a polynomial (बहुपदी) if the index of the variable in every term is a whole number, e.g. x² + 2x + 3, 3y³ + 2y² + y + 5.
Expressions like x + 1/x or √x + 2 are not polynomials (index is negative or a fraction).
The greatest index of the variable is the degree of the polynomial, e.g. degree of 7x³ + 5x + 4x⁵ + 2x² is 5.
Addition, subtraction and multiplication of polynomials are done like other algebraic expressions.
Division is the opposite of multiplication. Monomial ÷ monomial: divide coefficients and subtract indices, e.g. 15p³ ÷ 3p = 5p².
Polynomial ÷ monomial: divide each term separately, e.g. (6x³ + 8x²) ÷ 2x = 3x² + 4x.
Polynomial ÷ binomial: use long division like numbers. Write both in descending powers (put 0 for a missing power), divide the first term, multiply, subtract and bring down, until the degree of the remainder is less than that of the divisor.
If the remainder is 0, the divisor is a factor of the dividend.
Important terms
Polynomial – algebraic expression where every index of the variable is a whole number.
Degree – highest index of the variable in a polynomial.
Monomial / binomial – polynomial with one term / two terms.
Quotient – result of division.
Remainder – what is left after division; its degree is less than the divisor's.
Statistics is about collecting numerical data, organising it and drawing conclusions. This chapter teaches the arithmetic mean of data, the mean from a frequency table, and two new kinds of bar graphs – the subdivided bar graph and the percentage bar graph.
Key points
Information in the form of numbers is numerical data. Each number is a score (observation).
The average of the data is called the arithmetic mean or simply the mean (मध्य).
Mean = sum of all scores ÷ total number of scores. Example: the mean of 60, 50, 54, 46, 50 is 260 ÷ 5 = 52.
When scores repeat, make a frequency table: score (xᵢ), tally marks, frequency (fᵢ) and the product fᵢ × xᵢ.
N = Σfᵢ (total frequency). Mean x̄ = Σfᵢxᵢ ÷ N.
The symbol Σ (sigma) means "sum of".
A subdivided bar graph shows the total as one bar, divided into parts for each component (e.g. boys and girls in each class) using different colours or shades.
A percentage bar graph shows each bar as 100%, divided into parts according to the percentage of each component. It is useful for comparing proportions, e.g. trees survived in two towns.
In bar graphs, bars have equal width and equal gaps; choose a suitable scale on the vertical axis.
Important terms
Data – collected numerical information.
Arithmetic mean – sum of scores divided by number of scores.
Frequency – number of times a score occurs.
Subdivided bar graph – bar graph where each bar is divided to show its components.
Percentage bar graph – bar graph in which each bar represents 100% and is divided into percentages.
Formulas
Mean = (sum of all scores) / (number of scores)
Mean x̄ = Σfᵢxᵢ / N, where N = Σfᵢ
Percentage of a part = (part / total) × 100
5 important questions
Q1. Find the mean of 60, 50, 54, 46 and 50.
Sum = 260; number of scores = 5. Mean = 260/5 = 52.
Q2. Find the mean from the table: marks 2, 3, 4 with frequencies 4, 3, 3.
An equation is a statement that two expressions are equal. This chapter teaches how to solve equations in one variable using the same operation on both sides (including equations with fractions and variables on both sides) and how to turn word problems into equations and solve them.
Key points
An equation in one variable has only one variable whose highest power is 1, e.g. 3x + 5 = 20.
The value of the variable that makes both sides equal is the solution (उकल) of the equation.
Rule: we may add, subtract, multiply or divide both sides by the same number (not dividing by zero); the equation stays true.
Steps: remove brackets, clear fractions by multiplying both sides by the LCM of denominators, bring variable terms to one side and numbers to the other (transposing changes the sign), simplify, and divide by the coefficient.
For an equation like (a/b) = (c/d), cross-multiply: a × d = b × c.
Always check the answer by putting it back in the original equation.
Word problems: read carefully, choose a variable for the unknown, write the given information as an equation, solve and answer in words with units.
Common word problems: ages, numbers, consecutive numbers, perimeter, money, speed–time.
Important terms
Equation – statement of equality containing a variable.
Variable – letter standing for an unknown number.
Solution (root) – value of the variable that satisfies the equation.
Transposing – moving a term to the other side with the opposite sign.
Formulas
If a = b then a + c = b + c, a − c = b − c, ac = bc, a/c = b/c (c ≠ 0)
Consecutive numbers: x, x + 1, x + 2; consecutive even/odd numbers: x, x + 2, x + 4
5 important questions
Q1. Solve: 2(x − 3) = (3/5)(x + 4).
Multiply both sides by 5: 10(x − 3) = 3(x + 4) → 10x − 30 = 3x + 12 → 7x = 42 → x = 6.
Q2. Solve: (x − 1)/(x + 2) = 2/3 ... check if x = 7.
Two triangles are congruent if one fits exactly on the other – all corresponding sides and angles are equal. This chapter shows that we need only three specific parts to prove congruence and gives the five tests: SAS, SSS, ASA, AAS (SAA) and hypotenuse-side, along with writing the correct correspondence of vertices.
Key points
Congruent (एकरूप) triangles have the same shape and size; the symbol is ≅.
If △ABC ≅ △PQR, then A ↔ P, B ↔ Q, C ↔ R, so AB = PQ, BC = QR, CA = RP and ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R.
The order of letters shows the one-to-one correspondence of vertices; it must be written correctly.
Three specific parts that fix a unique triangle give a test of congruence.
SAS test: two sides and the included angle of one triangle equal to those of the other.
SSS test: all three sides equal.
ASA test: two angles and the included side equal.
AAS (SAA) test: two angles and a non-included side equal (it reduces to ASA because angle sum is 180°).
Hypotenuse-side test: in right-angled triangles, hypotenuse and one side equal.
AAA is not a test of congruence – triangles with equal angles may be of different sizes.
After proving triangles congruent, the remaining corresponding parts are equal: "corresponding sides/angles of congruent triangles" (c.s.c.t. / c.a.c.t.).
Important terms
Congruent triangles – triangles that coincide exactly when placed on each other.
Correspondence – matching of vertices of two triangles.
Included angle – angle between two given sides.
Included side – side between two given angles.
Hypotenuse – side opposite the right angle.
Formulas
Tests of congruence: SAS, SSS, ASA, AAS (SAA), Hypotenuse-side
Sum of angles of a triangle = 180°
5 important questions
Q1. In △PQR and △UTS, PQ = UT, QR = TS, RP = SU. Which test shows they are congruent?
SSS test, with correspondence PQR ↔ UTS, so △PQR ≅ △UTS.
Q2. In △XYZ and △DEF, XZ = DF, ∠X = ∠D, ∠Z = ∠F. State the test and correspondence.
ASA test; XYZ ↔ DEF; △XYZ ≅ △DEF.
Q3. Why is AAA not a test of congruence?
Two triangles with all angles equal can have different side lengths (one may be an enlargement of the other), so they need not be congruent.
Q4. If △ABC ≅ △LMN, write the six pairs of congruent parts.
AB ≅ LM, BC ≅ MN, CA ≅ NL, ∠A ≅ ∠L, ∠B ≅ ∠M, ∠C ≅ ∠N.
Q5. In right triangles PQR and LMN, ∠Q = ∠M = 90°, hypotenuse PR = LN and PQ = LM. Are they congruent?
Yes, by the hypotenuse-side test: △PQR ≅ △LMN.
Exam tip
Mark equal parts on the figure with ticks and arcs first; then the test becomes easy to spot.
In compound interest, the interest of each year is added to the principal, and the next year's interest is calculated on this new amount. This chapter derives the formula for the amount by compound interest, covers half-yearly compounding, and applies the formula to population growth and depreciation of machines and vehicles.
Key points
Simple interest is calculated on the same principal every year: I = PNR/100.
In compound interest (चक्रवाढ व्याज), the amount at the end of one year becomes the principal for the next year. So interest grows every year.
Banks use compound interest for fixed deposits and loans.
Amount after N years: A = P(1 + R/100)ᴺ. Compound interest = A − P.
If interest is compounded every six months, take rate R/2 and number of periods 2N.
For the first year, simple interest and compound interest are equal; from the second year compound interest is more.
The same formula is used for increase in population at a fixed rate.
The reduction in the value of a vehicle or machine over time is called depreciation. Value after N years: A = P(1 − R/100)ᴺ.
p.c.p.a. means per cent per annum.
Important terms
Principal (P) – money borrowed or deposited.
Rate (R) – interest per ₹ 100 for one year.
Amount (A) – principal + interest.
Compound interest – interest calculated on principal plus accumulated interest.
Depreciation – decrease in value of an article over time.
Formulas
Simple interest I = PNR/100
A = P(1 + R/100)ᴺ
Compound interest = A − P
Half-yearly: A = P(1 + R/200)²ᴺ
Population after N years = P(1 + R/100)ᴺ
Depreciation: A = P(1 − R/100)ᴺ
5 important questions
Q1. Find the compound interest on ₹ 10,000 at 10% p.a. for 2 years.
This chapter gives formulas for the area of a parallelogram, rhombus and trapezium, introduces Heron's formula to find the area of a triangle when only its three sides are known, and shows how to find the area of irregular plots by dividing them into triangles and trapeziums.
Key points
A parallelogram can be cut and rearranged into a rectangle with the same base and height, so its area = base × height.
The diagonals of a rhombus are perpendicular bisectors of each other, so the rhombus divides into 4 equal right triangles. Area = ½ × product of diagonals.
Area of a trapezium = ½ × (sum of parallel sides) × height, where height is the distance between the parallel sides.
Area of a triangle = ½ × base × height.
When the height is not known but all three sides a, b, c are known, use Heron's formula with the semiperimeter s = (a + b + c)/2.
Irregular fields and plots are divided into triangles, rectangles and trapeziums; the areas are found separately and added.
Pythagoras' theorem helps find a missing side or diagonal (e.g. side of a rhombus from its diagonals).
Area (क्षेत्रफळ) is measured in square units: sq cm, sq m; land is also measured in are and hectare (1 are = 100 sq m, 1 hectare = 10,000 sq m).
Important terms
Height of a parallelogram/trapezium – perpendicular distance between the parallel sides.
Semiperimeter – half the perimeter of a triangle.
Heron's formula – formula for area of a triangle using its three sides.
Formulas
Parallelogram = base × height
Rhombus = ½ × d₁ × d₂
Trapezium = ½ × (a + b) × h
Triangle = ½ × base × height
Heron's formula: s = (a + b + c)/2; Area = √[s(s − a)(s − b)(s − c)]
5 important questions
Q1. The diagonals of a rhombus are 11.2 cm and 7.5 cm. Find its area.
Area = ½ × 11.2 × 7.5 = 5.6 × 7.5 = 42 sq cm.
Q2. The area of a rhombus is 96 sq cm and one diagonal is 12 cm. Find the other diagonal and the side.
96 = ½ × 12 × d₂ → d₂ = 16 cm. Half-diagonals are 6 and 8, so side = √(6² + 8²) = √100 = 10 cm.
Q3. The parallel sides of a trapezium are 7 cm and 8 cm and the distance between them is 6 cm. Find its area.
Area = ½ × (7 + 8) × 6 = 45 sq cm.
Q4. Find the area of a triangle with sides 17 cm, 25 cm and 26 cm.
s = (17 + 25 + 26)/2 = 34. Area = √(34 × 17 × 9 × 8) = √41,616 = 204 sq cm.
Q5. Find the area of a parallelogram with base 12 cm and height 7 cm.
Area = base × height = 12 × 7 = 84 sq cm.
Exam tip
In Heron's formula, factorise the product under the root into primes and pair them – it avoids big multiplications.
This chapter deals with three-dimensional solids – cuboid, cube and cylinder. It explains the volume of a solid as the space it occupies, and the total surface area as the area of all its faces. Formulas for the surface area and volume of a cuboid, cube and cylinder are derived and used in practical problems like tanks and boxes.
Key points
Solid figures like cuboid, cube and cylinder have three dimensions and occupy space.
The measure of space occupied by a solid is its volume (घनफळ). Units: cubic cm, cubic m. Capacity is measured in litres: 1 litre = 1000 cubic cm; 1 cubic metre = 1000 litres.
A cuboid has 6 rectangular faces; a cube has 6 equal square faces.
Total surface area of a cuboid = 2(lb + bh + lh). Volume = l × b × h.
Total surface area of a cube = 6l². Volume = l³.
A cylinder's curved surface opens into a rectangle whose length is the circumference of the base (2πr) and breadth is the height h. So curved surface area = 2πrh.
Total surface area of a closed cylinder = curved surface + two circular bases = 2πr(h + r).
General formula: volume of a prism or cylinder = area of base × height. Volume of cylinder = πr²h.
Use π = 22/7 or 3.14 as told.
Surface area tells the material needed to cover or make a solid; volume tells how much it can hold.
Important terms
Volume – space occupied by a solid.
Total surface area – sum of areas of all faces/surfaces of a solid.
Curved surface area – area of the curved part of a cylinder.
Capacity – amount of liquid a container can hold.
Formulas
Cuboid: TSA = 2(lb + bh + lh); V = lbh
Cube: TSA = 6l²; V = l³
Cylinder: CSA = 2πrh; TSA = 2πr(r + h); V = πr²h
1 litre = 1000 cm³; 1 m³ = 1000 litres
5 important questions
Q1. Find the total surface area and volume of a box 20 cm × 12 cm × 10 cm.
This chapter studies chords and arcs of a circle. It proves two important properties: the perpendicular from the centre to a chord bisects the chord, and the segment joining the centre to the midpoint of a chord is perpendicular to it. It also explains minor and major arcs corresponding to a chord.
Key points
A chord (जीवा) is a segment joining any two points on a circle. The diameter is the longest chord and passes through the centre.
Radius = ½ diameter.
Property 1: the perpendicular drawn from the centre of a circle to its chord bisects the chord. (If OP ⊥ AB, then AP = PB.)
Property 2 (converse): the segment joining the centre of a circle to the midpoint of its chord is perpendicular to the chord.
Both properties are proved using congruent triangles (hypotenuse-side test and SSS test).
The distance of a chord from the centre is the length of the perpendicular from the centre to the chord.