Mathematics Part 2 (Geometry) गणित भाग 2 (भूमिती)

Quick summary

This chapter introduces the basic undefined terms of geometry – point, line and plane – and teaches co-ordinates of points on a number line, distance between points, betweenness, conditional statements and how to write a geometric proof.

Key points

• Point, line and plane are basic (undefined) terms. A point has no length or breadth; a line extends endlessly in both directions; a plane is a flat surface extending endlessly.
• Points on the same line are collinear; lines in the same plane are coplanar.
• On a number line, the number that a point represents is its co-ordinate. If D shows 1, the co-ordinate of D is 1.
• Distance between two points with co-ordinates a and b: if a > b, d = a − b. It is always non-negative and is written as d(A, B) or AB. d(A, B) = d(B, A).
• Betweenness: for three distinct collinear points P, Q, R, if d(P, Q) + d(Q, R) = d(P, R), then Q is between P and R, written P – Q – R.
• Of three distinct collinear points, exactly one lies between the other two.
• Conditional statement: "If p then q", where p is the antecedent (given) and q is the consequent (to prove). Example: if a quadrilateral is a rhombus, then its diagonals are perpendicular bisectors of each other.
• Converse: "If q then p". The converse of a true statement need not be true. (If a number is even it is divisible by 2 – converse true. If a figure is a square it is a quadrilateral – converse false.)
• Proof: a logical argument using definitions, properties and already proved theorems. A direct proof proceeds step by step from the given; an indirect proof assumes the opposite and reaches a contradiction.
• Steps for a theorem: write Given, To prove, figure, Proof with reasons.

Important terms

• Co-ordinate (निर्देशक) – the number associated with a point on a number line.
• Collinear points (एकरेषीय बिंदू) – points on the same line.
• Line segment – the part of a line between two points, including them; seg AB.
• Ray – part of a line with one end point; ray AB.
• Congruent segments – segments of equal length.
• Midpoint – point M on seg AB with AM = MB.

Formulas

• d(A, B) = a − b, when a > b (a, b are co-ordinates)
• If P – Q – R then PQ + QR = PR
• Midpoint co-ordinate of a and b = (a + b)/2

5 important questions

Q1. Points A, B, C have co-ordinates −3, 5 and 2. Find d(A, B), d(B, C), d(A, C). Which point is between the other two?

d(A, B) = 5 − (−3) = 8; d(B, C) = 5 − 2 = 3; d(A, C) = 2 − (−3) = 5. Since d(A, C) + d(C, B) = 5 + 3 = 8 = d(A, B), point C is between A and B (A – C – B).

Q2. The co-ordinate of A is 5. Find the co-ordinates of points 13 units away from A.

To the left: 5 − 13 = −8; to the right: 5 + 13 = 18. The points have co-ordinates −8 and 18.

Q3. Write the converse of: "If a triangle is equilateral, then all its angles are equal." Is it true?

Converse: "If all angles of a triangle are equal, then it is equilateral." Yes, it is true.

Q4. The co-ordinates of P and Q are −7 and 3. Find the co-ordinate of the midpoint of seg PQ.

Midpoint = (−7 + 3)/2 = −4/2 = −2.

Q5. Write the following in "if–then" form: "Opposite angles of a parallelogram are congruent."

If a quadrilateral is a parallelogram, then its opposite angles are congruent.

Exam tip

For betweenness, check which pair of distances adds up to the third; the common point of those two is in the middle.

Quick summary

Parallel lines are coplanar lines that never meet. This chapter studies the angles formed when a transversal cuts two lines, the properties of these angles when the lines are parallel, and the tests used to prove two lines parallel.

Key points

• A transversal is a line that intersects two or more lines at distinct points. It forms 8 angles.
• Pairs of angles formed: corresponding angles (4 pairs), alternate angles (2 pairs of interior and 2 of exterior), interior angles on the same side (2 pairs).
• Properties when lines are parallel:
• Interior angles on the same side are supplementary (sum = 180°).
• Alternate angles are congruent.
• Corresponding angles are congruent.
• Tests for parallel lines (converses): two lines are parallel if, with a transversal, (1) interior angles on the same side are supplementary, or (2) a pair of alternate angles is congruent, or (3) a pair of corresponding angles is congruent.
• Two lines perpendicular to the same line in a plane are parallel to each other.
• If a line is perpendicular to one of two parallel lines, it is perpendicular to the other also.
• Linear pair angles are supplementary; vertically opposite angles are congruent – these are used in the proofs.
• Use: the sum of the angles of a triangle is 180° is proved using a line through a vertex parallel to the opposite side.

Important terms

• Parallel lines (समांतर रेषा) – coplanar lines that do not intersect.
• Transversal (छेदिका) – a line cutting two or more lines at distinct points.
• Corresponding angles (संगत कोन) – angles on the same side of the transversal, one interior and one exterior.
• Alternate angles (व्युत्क्रम कोन) – angles on opposite sides of the transversal, both interior or both exterior.

Formulas

• If l ∥ m: corresponding ∠ = corresponding ∠; alternate ∠ = alternate ∠; interior angles on the same side add to 180°.
• Linear pair: ∠a + ∠b = 180°.

5 important questions

Q1. Line l ∥ line m and a transversal makes an angle of 70° with l. Find the corresponding angle, the alternate angle and the interior angle on the same side.

Corresponding angle = 70°; alternate interior angle = 70°; interior angle on the same side = 180° − 70° = 110°.

Q2. Prove that if two parallel lines are cut by a transversal, the interior angles on the same side are supplementary.

Given l ∥ m, transversal n. ∠1 and ∠2 are interior angles on the same side. ∠1 + ∠3 = 180° (linear pair). ∠3 = ∠2 (alternate angles). So ∠1 + ∠2 = 180°.

Q3. Two angles in a figure are (3x + 10)° and (2x + 20)°. They are interior angles on the same side of a transversal of parallel lines. Find x.

(3x + 10) + (2x + 20) = 180, so 5x + 30 = 180, 5x = 150 and x = 30. The angles are 100° and 80°.

Q4. Prove that two lines perpendicular to the same line are parallel.

Let l ⊥ n and m ⊥ n. Then the corresponding angles formed by transversal n are both 90°, so they are congruent. By the corresponding angles test, l ∥ m.

Q5. In a figure, ∠a = 115° and ∠b = 65° are interior angles on the same side of lines p and q. Are p and q parallel?

115° + 65° = 180°. The interior angles are supplementary, so by the test, p ∥ q.

Exam tip

Always write the name of the property in brackets (alternate angles, corresponding angles) as the reason – it carries the mark.

Quick summary

This chapter proves important properties of triangles: the remote interior angles theorem, congruence tests, the isosceles triangle theorem, properties of 30°–60°–90° and 45°–45°–90° triangles, medians, the perpendicular bisector and angle bisector theorems, inequalities in a triangle and an introduction to similar triangles.

Key points

• Sum of the angles of a triangle = 180°.
• Remote interior angles theorem: an exterior angle of a triangle equals the sum of its two remote interior angles. So an exterior angle is greater than each remote interior angle.
• Congruence tests: SAS, ASA, AAS (SAA), SSS, and hypotenuse–side test for right triangles. Corresponding parts of congruent triangles are congruent (c.s.c.t. / c.a.c.t.).
• Isosceles triangle theorem: if two sides of a triangle are congruent, the angles opposite them are congruent. Converse: if two angles are congruent, the sides opposite them are congruent.
• An equilateral triangle has all angles 60°.
• 30°–60°–90° triangle: the side opposite 30° is half the hypotenuse; the side opposite 60° is (√3/2) × hypotenuse.
• 45°–45°–90° triangle: each side containing the right angle is hypotenuse/√2.
• Median: joins a vertex to the midpoint of the opposite side. The three medians meet at the centroid, which divides each median in the ratio 2 : 1. In a right triangle, the median on the hypotenuse is half the hypotenuse.
• Perpendicular bisector theorem: every point on the perpendicular bisector of a segment is equidistant from its end points; converse is also true.
• Angle bisector theorem: every point on the bisector of an angle is equidistant from its sides; converse is also true.
• Inequalities: in a triangle, the larger side has the larger opposite angle, and the larger angle has the larger opposite side. The sum of any two sides is greater than the third side.
• Similar triangles: triangles with equal corresponding angles have corresponding sides in proportion.

Important terms

• Congruent triangles (एकरूप त्रिकोण) – triangles that match exactly.
• Median (मध्यगा) – segment from a vertex to the midpoint of the opposite side.
• Centroid – point where medians meet.
• Similar triangles (समरूप त्रिकोण) – same shape, proportional sides.

Formulas

• Exterior angle = sum of two remote interior angles
• 30°–60°–90°: side opp. 30° = ½ × hyp; side opp. 60° = (√3/2) × hyp
• 45°–45°–90°: each side = hyp/√2
• Median on hypotenuse = ½ × hypotenuse
• Centroid divides a median in 2 : 1

5 important questions

Q1. In ΔPQR, ∠P = 70° and ∠Q = 65°. Find the exterior angle at R.

Exterior angle at R = ∠P + ∠Q = 70° + 65° = 135°.

Q2. In a right triangle with angles 30°, 60°, 90°, the hypotenuse is 12 cm. Find the other sides.

Side opposite 30° = ½ × 12 = 6 cm. Side opposite 60° = (√3/2) × 12 = 6√3 cm.

Q3. In a right isosceles triangle, the hypotenuse is 8√2 cm. Find each of the other sides.

Each side = hypotenuse/√2 = 8√2/√2 = 8 cm.

Q4. In ΔABC, AB = 5 cm, BC = 7 cm, AC = 4 cm. Which is the largest and smallest angle?

The largest side BC is opposite ∠A, so ∠A is the largest. The smallest side AC is opposite ∠B, so ∠B is the smallest.

Q5. G is the centroid of ΔABC and the median AD = 12 cm. Find AG and GD.

The centroid divides the median in 2 : 1. AG = (2/3) × 12 = 8 cm; GD = 4 cm.

Exam tip

Write the congruence test name and the correspondence (e.g. ΔABC ≅ ΔPQR by SAS) – correspondence order matters.

Quick summary

This chapter teaches how to construct a triangle when some measurements are given: (1) the base, a base angle and the sum of the other two sides; (2) the base, a base angle and the difference of the other two sides; (3) the perimeter and both base angles. It uses the perpendicular bisector and angle bisector properties.

Key points

• Earlier you learnt to construct a triangle given three sides (SSS), two sides and the included angle (SAS), two angles and the included side (ASA), and the hypotenuse and one side of a right triangle.
• Perpendicular bisector property is the key idea: every point on the perpendicular bisector of a segment is equidistant from its end points. This lets us find the third vertex.
• Always draw a rough figure first and mark the given measures on it. Then plan the steps.
• Construction 1: base BC, ∠B and AB + AC given.
• Draw BC. At B draw ray BP making the given ∠B.
• On ray BP mark D such that BD = AB + AC.
• Join DC and draw the perpendicular bisector of DC; it cuts BD at A. Join AC. ΔABC is required (AD = AC, so AB + AC = BD).
• Construction 2: base BC, ∠B and AB − AC given (AB > AC).
• Draw BC and ray BP at ∠B. Mark D on BP with BD = AB − AC.
• Join DC; its perpendicular bisector meets ray BP at A. Join AC.
• If AC > AB, mark D on the opposite ray of BP (below BC) with BD = AC − AB, then do the same.
• Construction 3: perimeter and both base angles given.
• Draw seg PQ equal to the perimeter. At P and Q draw angles equal to half of ∠B and half of ∠C; the rays meet at A.
• Draw perpendicular bisectors of AP and AQ; they cut PQ at B and C. Join AB and AC. ΔABC is required.
• In Construction 3, AB = PB and AC = CQ (perpendicular bisector property), so AB + BC + CA = PQ, and ∠ABC = 2∠APB (exterior angle).
• A triangle is possible only if the sum of two sides is greater than the third side, and the sum of the two base angles is less than 180°.

Important terms

• Perpendicular bisector (लंबदुभाजक) – line perpendicular to a segment through its midpoint.
• Angle bisector (कोनदुभाजक) – ray dividing an angle into two equal angles.
• Perimeter (परिमिती) – sum of all sides.
• Rough figure – a quick sketch used to plan the construction.

Formulas

• In Construction 1: AB + AC = BD
• In Construction 3: PQ = AB + BC + CA and ∠B = 2 × (angle drawn at P)

5 important questions

Q1. Construct ΔXYZ with YZ = 7.5 cm, ∠Y = 45° and XY + XZ = 10.5 cm. Write the steps.

Draw YZ = 7.5 cm. At Y draw ray YP with ∠ZYP = 45°. Mark D on YP with YD = 10.5 cm. Join DZ. Draw the perpendicular bisector of DZ; it meets YD at X. Join XZ. ΔXYZ is the required triangle.

Q2. Why does the vertex A lie on the perpendicular bisector of DC in Construction 1?

Because A must be equidistant from D and C (AD = AC). Points equidistant from D and C lie on the perpendicular bisector of DC.

Q3. Construct ΔABC with BC = 6 cm, ∠B = 50° and AB − AC = 2 cm. Give steps.

Draw BC = 6 cm. Draw ray BP at 50°. Mark D on BP with BD = 2 cm. Join DC; draw its perpendicular bisector to meet ray BP at A. Join AC. ΔABC is required.

Q4. Construct ΔPQR with perimeter 11.6 cm, ∠Q = 70° and ∠R = 80°. Write the steps.

Draw seg LM = 11.6 cm. At L draw an angle of 35° and at M an angle of 40°; the rays meet at P. Draw perpendicular bisectors of PL and PM, cutting LM at Q and R. Join PQ and PR. ΔPQR is required.

Q5. Can a triangle be made with BC = 8 cm, ∠B = 60° and AB + AC = 7 cm? Why?

No. AB + AC must be greater than BC (triangle inequality). Here 7 cm < 8 cm, so the triangle is not possible.

Exam tip

Keep all construction arcs visible and always draw the rough figure – both carry marks.

Quick summary

This chapter studies the properties and tests of a parallelogram, the special parallelograms – rhombus, rectangle and square – the trapezium, and the mid-point theorem for triangles.

Key points

• Parallelogram: a quadrilateral with both pairs of opposite sides parallel.
• Properties of a parallelogram: opposite sides are congruent; opposite angles are congruent; diagonals bisect each other; adjacent angles are supplementary; a diagonal divides it into two congruent triangles.
• Tests for a parallelogram – a quadrilateral is a parallelogram if: both pairs of opposite sides are congruent; or both pairs of opposite angles are congruent; or diagonals bisect each other; or one pair of opposite sides is parallel and congruent.
• Rectangle: a parallelogram with each angle 90°. Its diagonals are congruent.
• Rhombus: a parallelogram with all sides congruent. Its diagonals are perpendicular bisectors of each other and bisect the opposite angles.
• Square: all sides congruent and all angles 90°. Its diagonals are congruent, are perpendicular bisectors of each other and make 45° with the sides.
• Trapezium: a quadrilateral with only one pair of opposite sides parallel. In an isosceles trapezium, the non-parallel sides are congruent and its base angles are congruent.
• In a trapezium, the angles between a pair of parallel sides on the same non-parallel side are supplementary.
• Mid-point theorem: the segment joining the midpoints of any two sides of a triangle is parallel to the third side and half of it.
• Converse of the mid-point theorem: a line through the midpoint of one side, parallel to another side, bisects the third side.

Important terms

• Parallelogram (समांतरभुज चौकोन) – opposite sides parallel.
• Rhombus (समभुज चौकोन) – all sides equal.
• Trapezium (समलंब चौकोन) – exactly one pair of parallel sides.
• Diagonal (कर्ण) – segment joining opposite vertices.

Formulas

• Sum of angles of a quadrilateral = 360°
• Rhombus with diagonals d₁ and d₂: side² = (d₁/2)² + (d₂/2)²
• Mid-point theorem: PQ = ½ BC and PQ ∥ BC

5 important questions

Q1. In parallelogram ABCD, ∠A = (3x + 12)° and ∠B = (2x − 32)°. Find x and all angles.

Adjacent angles are supplementary: 3x + 12 + 2x − 32 = 180, so 5x = 200, x = 40. ∠A = ∠C = 132°, ∠B = ∠D = 48°.

Q2. The diagonals of a rhombus are 24 cm and 10 cm. Find its side and perimeter.

Diagonals bisect each other at right angles: halves are 12 and 5. Side = √(12² + 5²) = √169 = 13 cm. Perimeter = 4 × 13 = 52 cm.

Q3. Prove that the diagonals of a parallelogram bisect each other.

In parallelogram ABCD, let diagonals meet at O. In ΔAOB and ΔCOD: AB = CD (opposite sides), ∠OAB = ∠OCD and ∠OBA = ∠ODC (alternate angles). So ΔAOB ≅ ΔCOD (ASA), giving AO = CO and BO = DO.

Q4. In ΔPQR, M and N are midpoints of PQ and PR. If QR = 9 cm, find MN.

By the mid-point theorem, MN = ½ × QR = 4.5 cm and MN ∥ QR.

Q5. The diagonal of a square is 13√2 cm. Find its side.

In a square, diagonal = side × √2. So side = 13√2 ÷ √2 = 13 cm.

Exam tip

Learn properties in a table: sides, angles, diagonals for each quadrilateral – "true or false" and short proofs come from it.

Quick summary

This chapter revises the parts of a circle and proves properties of chords: the perpendicular from the centre bisects a chord, and congruent chords are equidistant from the centre. It also explains the incircle and circumcircle of a triangle and how to construct them.

Key points

• A circle is the set of points in a plane at a fixed distance (radius) from a fixed point (centre).
• Chord: a segment joining two points on a circle. Diameter: a chord through the centre; it is the longest chord and diameter = 2 × radius.
• Circles with equal radii are congruent circles; circles with the same centre are concentric circles.
• Theorem 1: a perpendicular drawn from the centre of a circle to a chord bisects the chord.
• Theorem 2 (converse): the segment joining the centre to the midpoint of a chord is perpendicular to the chord.
• Theorem 3: congruent chords of a circle (or of congruent circles) are equidistant from the centre.
• Theorem 4 (converse): chords that are equidistant from the centre are congruent.
• The perpendicular bisector of a chord passes through the centre.
• Incircle: a circle that touches all three sides of a triangle from inside. Its centre (incentre) is the point where the angle bisectors meet. The radius is the perpendicular distance from the incentre to any side.
• Circumcircle: a circle passing through all three vertices of a triangle. Its centre (circumcentre) is where the perpendicular bisectors of the sides meet.
• The circumcentre lies inside an acute triangle, at the midpoint of the hypotenuse in a right triangle, and outside an obtuse triangle. The incentre is always inside.
• In an equilateral triangle, the incentre and circumcentre are the same point.

Important terms

• Chord (जीवा) – segment joining two points of a circle.
• Incircle (अंतर्वर्तुळ) – circle touching all sides of a triangle from inside.
• Circumcircle (परिवर्तुळ) – circle passing through all vertices of a triangle.
• Concurrent lines – three or more lines passing through one point.

Formulas

• Diameter = 2r
• If a chord of length 2x is at distance d from the centre: r² = x² + d² (Pythagoras)

5 important questions

Q1. A chord of a circle is 24 cm long and is 5 cm from the centre. Find the radius.

The perpendicular from the centre bisects the chord, so half-chord = 12 cm. r² = 12² + 5² = 144 + 25 = 169, r = 13 cm.

Q2. The radius of a circle is 10 cm and a chord is 6 cm from the centre. Find the length of the chord.

Half-chord = √(10² − 6²) = √64 = 8 cm. Chord = 2 × 8 = 16 cm.

Q3. Prove that a perpendicular from the centre to a chord bisects the chord.

Let O be the centre, AB a chord and OP ⊥ AB. Join OA and OB. In right triangles OPA and OPB, OA = OB (radii) and OP = OP. By hypotenuse–side test, ΔOPA ≅ ΔOPB, so AP = BP.

Q4. Where does the circumcentre of a right-angled triangle lie? Why?

At the midpoint of the hypotenuse, because the midpoint of the hypotenuse is equidistant from all three vertices (the median on the hypotenuse is half the hypotenuse).

Q5. How do you construct the incircle of ΔABC?

Draw bisectors of any two angles; they meet at I (incentre). Draw IM ⊥ to a side. With centre I and radius IM, draw the circle; it touches all three sides.

Exam tip

In chord problems, always draw the perpendicular from the centre and use Pythagoras with half the chord.

Quick summary

Co-ordinate geometry tells the exact position of a point in a plane using two numbers. This chapter explains the X-axis, Y-axis, origin and quadrants, how to read and plot co-ordinates of points, and the equations of lines parallel to the axes.

Key points

• Two perpendicular number lines are drawn: the horizontal X-axis and the vertical Y-axis. They meet at the origin O (0, 0).
• The position of a point P is written as an ordered pair (x, y): x is the x co-ordinate (distance from the Y-axis) and y is the y co-ordinate (distance from the X-axis).
• The order matters: (2, 3) and (3, 2) are different points.
• The axes divide the plane into four quadrants, counted anticlockwise:
• Quadrant I: (+, +); Quadrant II: (−, +); Quadrant III: (−, −); Quadrant IV: (+, −).
• A point on an axis does not belong to any quadrant.
• Every point on the X-axis has y co-ordinate 0, like (5, 0). Every point on the Y-axis has x co-ordinate 0, like (0, −3).
• Line parallel to the X-axis: all its points have the same y co-ordinate. Its equation is y = b. For example, y = 3 is a line 3 units above the X-axis.
• Line parallel to the Y-axis: all its points have the same x co-ordinate. Its equation is x = a. For example, x = −2.
• The equation of the X-axis is y = 0 and of the Y-axis is x = 0.
• Lines y = b and x = a are perpendicular to each other and meet at the point (a, b).
• To plot a point (a, b): start at O, move a units along the X-axis (right if positive, left if negative), then b units parallel to the Y-axis (up if positive, down if negative).

Important terms

• Origin (आरंभबिंदू) – point where the axes meet, (0, 0).
• Abscissa – x co-ordinate; ordinate – y co-ordinate.
• Quadrant (चरण) – one of the four parts of the plane made by the axes.
• Ordered pair – a pair (x, y) in which order matters.

Formulas

5 important questions

Q1. In which quadrant or on which axis do these points lie: A(−3, 2), B(5, 0), C(−1, −4), D(0, 7), E(4, −6)?

A – Quadrant II; B – on the X-axis; C – Quadrant III; D – on the Y-axis; E – Quadrant IV.

Q2. Write the equation of the line parallel to the Y-axis passing through (−4, 6).

All points on it have x = −4. Equation: x = −4.

Q3. Write the equation of the line parallel to the X-axis at a distance of 5 units below it.

All points have y = −5. Equation: y = −5.

Q4. Find the point of intersection of the lines x = 3 and y = −2.

The lines meet at (3, −2), which lies in Quadrant IV.

Q5. A point has its x co-ordinate 0 and y co-ordinate −3. Where is it? What are the co-ordinates of the origin?

The point (0, −3) lies on the negative side of the Y-axis, 3 units below the origin. The origin is (0, 0).

Exam tip

Remember the sign pattern of quadrants (+,+), (−,+), (−,−), (+,−) going anticlockwise; it answers many one-mark questions.

Quick summary

Trigonometry means "measurement of triangles". This chapter defines the trigonometric ratios sine, cosine and tangent for an acute angle of a right triangle, shows the relations between them, and gives their values for the angles 0°, 30°, 45°, 60° and 90°.

Key points

• In right ΔABC with ∠B = 90° and acute angle θ = ∠C: side opposite θ = AB, side adjacent to θ = BC, hypotenuse = AC.
• sin θ = opposite side ÷ hypotenuse
• cos θ = adjacent side ÷ hypotenuse
• tan θ = opposite side ÷ adjacent side
• Reciprocal ratios: cosec θ = 1/sin θ, sec θ = 1/cos θ, cot θ = 1/tan θ.
• tan θ = sin θ / cos θ.
• The ratios depend only on the angle, not on the size of the triangle (similar triangles give the same ratios).
• For an acute angle, sin θ and cos θ are always between 0 and 1.
• Complementary angles: sin θ = cos (90° − θ) and cos θ = sin (90° − θ).
• Identity: sin²θ + cos²θ = 1.
• Values come from the 30°–60°–90° and 45°–45°–90° triangles.

Important terms

• Trigonometric ratio (त्रिकोणमितीय गुणोत्तर) – ratio of two sides of a right triangle related to an acute angle.
• Hypotenuse (कर्ण) – the side opposite the right angle.
• Complementary angles – angles whose sum is 90°.

Formulas

• sin²θ + cos²θ = 1
• tan θ = sin θ / cos θ
• sin θ = cos (90° − θ)

5 important questions

Q1. In right ΔPQR, ∠Q = 90°, PQ = 12 and QR = 5. Find sin R, cos R and tan R.

PR = √(12² + 5²) = 13. For ∠R: opposite = PQ = 12, adjacent = QR = 5. sin R = 12/13, cos R = 5/13, tan R = 12/5.

Q2. If sin θ = 3/5, find cos θ and tan θ.

cos²θ = 1 − 9/25 = 16/25, so cos θ = 4/5. tan θ = (3/5)/(4/5) = 3/4.

Q3. Find the value of 2 sin 30° + 3 tan 45° − cos 60°.

2 × ½ + 3 × 1 − ½ = 1 + 3 − ½ = 3.5 (or 7/2).

Q4. Evaluate: cos 48° / sin 42°.

sin 42° = cos (90° − 42°) = cos 48°. So the value is 1.

Q5. Find sin²60° + cos²60° and verify the identity.

sin²60° = (√3/2)² = 3/4; cos²60° = (1/2)² = 1/4. Sum = 3/4 + 1/4 = 1. ✓

Exam tip

Memorise the table by the trick: sin 0°, 30°, 45°, 60°, 90° = √0/2, √1/2, √2/2, √3/2, √4/2; cos is the same list in reverse.

Quick summary

This chapter revises the surface area and volume of a cuboid, cube and cylinder, and teaches the surface area and volume of a right circular cone, a sphere and a hemisphere, with real-life problems.

Key points

• Surface area is the area of all the outer faces of a solid (unit cm², m²). Volume is the space a solid occupies (unit cm³, m³).
• 1 litre = 1000 cm³; 1 m³ = 1000 litres.
• Cone: has a circular base of radius r, height h (from vertex to centre of base, perpendicular to base) and slant height l. They are related by l² = r² + h².
• A cone is formed by rotating a right triangle about one side containing the right angle.
• The volume of a cone is one-third of the volume of a cylinder with the same base and height (three cone-fulls of water fill the cylinder).
• Sphere: every point of the surface is at distance r from the centre. A hemisphere is half a sphere.
• The curved surface area of a hemisphere = 2πr²; total surface area of a solid hemisphere = 3πr² (curved part + flat circle).
• Use π = 22/7 or 3.14 as the question says.
• In practical problems (tent, ice-cream cone, ball, bowl), decide first whether curved or total surface area is needed.

Important terms

• Slant height (तिरकी उंची) – distance from the vertex of a cone to any point on the edge of its base.
• Curved surface area (वक्रपृष्ठफळ) – area of the curved part only.
• Total surface area (एकूण पृष्ठफळ) – curved area + area of flat faces.

Formulas

5 important questions

Q1. A cone has radius 7 cm and height 24 cm. Find its slant height and curved surface area.

l = √(7² + 24²) = √625 = 25 cm. CSA = πrl = (22/7) × 7 × 25 = 550 cm².

Q2. Find the volume of a cone of radius 6 cm and height 7 cm.

V = ⅓ πr²h = ⅓ × (22/7) × 36 × 7 = 22 × 12 = 264 cm³.

Q3. Find the surface area and volume of a sphere of radius 7 cm.

SA = 4πr² = 4 × (22/7) × 49 = 616 cm². V = (4/3)πr³ = (4/3) × (22/7) × 343 = 4312/3 ≈ 1437.33 cm³.

Q4. A hemispherical bowl has radius 10.5 cm. How much water can it hold?

V = (2/3)πr³ = (2/3) × (22/7) × 10.5 × 10.5 × 10.5 = 2425.5 cm³ ≈ 2.43 litres.

Q5. The total surface area of a cone is 704 cm² and its radius is 7 cm. Find its slant height.

πr(l + r) = 704, so (22/7) × 7 × (l + 7) = 704. 22(l + 7) = 704, l + 7 = 32, l = 25 cm.

Exam tip

Write the formula first, then substitute, then simplify with units – each step earns a mark.