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

Quick summary

Similar figures have the same shape but may have different sizes. This chapter covers ratios of areas, the Basic Proportionality Theorem, similarity tests and areas of similar triangles.

Important terms

• Similar triangles – triangles whose corresponding angles are equal and corresponding sides are in the same ratio.
• Basic Proportionality Theorem (BPT) – a line parallel to one side of a triangle divides the other two sides in the same ratio.
• Angle bisector theorem – the bisector of an angle divides the opposite side in the ratio of the remaining two sides.

Formulas

• Equal heights: A(△1)/A(△2) = base1/base2
• Equal bases: A(△1)/A(△2) = height1/height2
• BPT: In △ABC, if DE ∥ BC, then AD/DB = AE/EC
• Angle bisector: if AD bisects ∠A, then BD/DC = AB/AC
• Similar triangles: A(△ABC)/A(△PQR) = AB²/PQ² = BC²/QR² = AC²/PR²

Solved examples

Example 1. In △ABC, DE ∥ BC, AD = 6, DB = 9, AE = 8. Find EC.

By BPT, AD/DB = AE/EC
6/9 = 8/EC → EC = (8 × 9)/6
EC = 12

Example 2. △ABC ~ △PQR, A(△ABC) = 81, A(△PQR) = 121, BC = 6.3. Find QR.

A(△ABC)/A(△PQR) = BC²/QR²
81/121 = (6.3)²/QR² → 9/11 = 6.3/QR
QR = 7.7

Quick revision

• Similarity tests: AAA (or AA), SAS, SSS.
• Converse of BPT: if a line divides two sides in the same ratio, it is parallel to the third side.
• Three parallel lines cut any two transversals in the same ratio.
• Ratio of areas = square of ratio of sides (not just ratio of sides).

Exam tip

Write the correspondence (ABC ↔ PQR) carefully before writing ratios. A wrong order gives a wrong answer.

Quick summary

This chapter is about right-angled triangles: Pythagoras theorem, its converse, Pythagorean triplets, the geometric mean theorem and Apollonius theorem.

Important terms

• Hypotenuse – side opposite the right angle; the longest side.
• Pythagorean triplet – three natural numbers a, b, c with a² + b² = c², e.g. (3, 4, 5), (5, 12, 13), (8, 15, 17).
• Geometric mean – x is the geometric mean of a and b if x² = ab.

Formulas

• Pythagoras: in △ABC with ∠B = 90°, AC² = AB² + BC²
• Triplet formula: (a² + b², a² − b², 2ab), for natural numbers a > b
• Geometric mean: in △ABC, ∠B = 90°, BD ⊥ AC → BD² = AD × DC
• 30°–60°–90° triangle: sides in ratio 1 : √3 : 2
• 45°–45°–90° triangle: sides in ratio 1 : 1 : √2
• Apollonius: if M is the midpoint of BC, AB² + AC² = 2AM² + 2BM²

Solved examples

Example 1. The two sides of a right-angled triangle are 9 cm and 40 cm. Find the hypotenuse.

Hypotenuse² = 9² + 40² = 81 + 1600 = 1681
Hypotenuse = √1681 = 41 cm

Example 2. In △ABC, M is the midpoint of BC, AB² + AC² = 290 and AM = 8. Find BC.

By Apollonius theorem: 290 = 2(8)² + 2BM²
290 = 128 + 2BM² → BM² = 81 → BM = 9
BC = 2 × 9 = 18

Quick revision

• Converse: if a² + b² = c², the triangle is right-angled (right angle opposite c).
• In a 30°–60°–90° triangle: side opposite 30° = half the hypotenuse.
• In a 45°–45°–90° triangle: each equal side = hypotenuse/√2.
• Diagonal of a square of side a = a√2.

Exam tip

Draw a small figure and mark the right angle first. Then you will never confuse the hypotenuse.

Quick summary

This chapter covers tangents, touching circles, arcs and the angles they make, cyclic quadrilaterals, and theorems on intersecting chords and secants.

Important terms

• Tangent – a line that touches the circle at exactly one point.
• Secant – a line that cuts the circle at two points.
• Inscribed angle – an angle with its vertex on the circle.
• Cyclic quadrilateral – a quadrilateral whose all four vertices lie on a circle.
• Touching circles – circles that meet at exactly one point.

Formulas

• Tangent ⊥ radius at the point of contact.
• Tangent segments from one outside point are equal: PA = PB.
• Touching circles: distance between centres = r₁ + r₂ (outside) or r₁ − r₂ (inside).
• Inscribed angle = ½ × measure of the intercepted arc.
• Angle in a semicircle = 90°.
• Cyclic quadrilateral: opposite angles add to 180°.
• Chords AB and CD meeting at E: AE × EB = CE × ED.
• Tangent PT and secant through A, B: PA × PB = PT².

Solved examples

Example 1. The radius of a circle is 5 cm. A point P is 13 cm from the centre. Find the length of the tangent from P.

Tangent ⊥ radius, so tangent² = 13² − 5² = 169 − 25 = 144
Tangent = 12 cm

Example 2. Chords AB and CD meet at E inside a circle. AE = 4, EB = 6, CE = 3. Find ED.

AE × EB = CE × ED
4 × 6 = 3 × ED
ED = 8

Quick revision

• Angles in the same arc are equal.
• Exterior angle of a cyclic quadrilateral = interior opposite angle.
• Measure of a semicircle arc = 180°; a full circle = 360°.
• Tangent-secant angle = ½ × intercepted arc.

Exam tip

Name the theorem you use in every step ("tangent-radius theorem", "inscribed angle theorem"). Reasons get marks in geometry.

Quick summary

In this chapter you construct a triangle similar to a given triangle, and tangents to a circle — at a point on it and from a point outside it.

Important terms

• Scale factor – the ratio of sides of the new triangle to the given triangle, e.g. 3 : 4.
• Tangent – a line touching the circle at one point; it is perpendicular to the radius there.

Formulas

• Tangent from outside point P to a circle (centre O, radius r): length = √(OP² − r²)
• Angle in a semicircle = 90° (this is why the midpoint method for tangents works)

Solved examples

Example 1. Construct △ABC ~ △LMN with AB : LM = 3 : 4 (steps)

1. Draw △LMN.
2. From M, draw a ray making an acute angle below MN; mark 4 equal parts M₁, M₂, M₃, M₄ (4 is the bigger number).
3. Join M₄ to N.
4. From M₃, draw a line parallel to M₄N, meeting MN at C.
5. From C, draw a line parallel to NL, meeting ML at A.

△ABC is the required triangle (B is the same point as M).

Example 2. Draw tangents from P to a circle of radius 3 cm, with OP = 7 cm (steps)

1. Draw the circle with centre O, radius 3 cm. Mark P with OP = 7 cm.
2. Find M, the midpoint of OP (perpendicular bisector).
3. With centre M and radius MO, draw an arc cutting the circle at A and B.
4. Join PA and PB.

PA and PB are the tangents. Each is √(49 − 9) = √40 ≈ 6.3 cm.

Quick revision

• Tangent at a point A on the circle: draw radius OA, then a line through A perpendicular to OA.
• Without the centre: use the tangent-secant / alternate segment method.
• Keep all construction lines (arcs) on the figure; do not erase them.
• Write the steps (analysis) in short only if the question asks.

Exam tip

Use a sharp pencil and a compass. Neat arcs and labels get full marks even if the steps are short.

Quick summary

Co-ordinate geometry uses points (x, y) to find distances, the point that divides a segment, and the slope of a line.

Important terms

• Distance formula – length between two points.
• Section formula – co-ordinates of a point dividing a segment in a ratio m : n.
• Slope (m) – steepness of a line; m = tan θ.
• Collinear points – points on one straight line.

Formulas

• Distance: d(A, B) = √[(x₂ − x₁)² + (y₂ − y₁)²]
• Internal division (m : n): ( (mx₂ + nx₁)/(m + n), (my₂ + ny₁)/(m + n) )
• External division (m : n): ( (mx₂ − nx₁)/(m − n), (my₂ − ny₁)/(m − n) )
• Midpoint: ( (x₁ + x₂)/2, (y₁ + y₂)/2 )
• Centroid: ( (x₁ + x₂ + x₃)/3, (y₁ + y₂ + y₃)/3 )
• Slope: m = (y₂ − y₁)/(x₂ − x₁)

Solved examples

Example 1. Find the distance between A(3, 8) and B(−9, 3).

d = √[(−9 − 3)² + (3 − 8)²] = √(144 + 25) = √169
d(A, B) = 13

Example 2. Find the point dividing A(−3, 5) and B(4, −9) in the ratio 2 : 5 internally.

x = (2 × 4 + 5 × (−3))/7 = (8 − 15)/7 = −1
y = (2 × (−9) + 5 × 5)/7 = (−18 + 25)/7 = 1
Point = (−1, 1)

Quick revision

• Parallel lines have equal slopes.
• Slope of the X-axis = 0; slope of the Y-axis is not defined.
• Three points are collinear if slope AB = slope BC.
• Distance of point (x, y) from origin = √(x² + y²).

Exam tip

Write (x₁, y₁) and (x₂, y₂) above the points before using any formula. It avoids sign mistakes.

Quick summary

Trigonometry links the angles and sides of a right-angled triangle. This chapter covers trigonometric identities and heights and distances (angle of elevation and depression).

Important terms

• Angle of elevation – angle between the horizontal and the line of sight when looking up.
• Angle of depression – angle between the horizontal and the line of sight when looking down.
• Line of sight – the line from the eye to the object.

Formulas

• sin θ = opposite/hypotenuse, cos θ = adjacent/hypotenuse, tan θ = opposite/adjacent
• cosec θ = 1/sin θ, sec θ = 1/cos θ, cot θ = 1/tan θ
• sin²θ + cos²θ = 1
• 1 + tan²θ = sec²θ
• 1 + cot²θ = cosec²θ
• tan 30° = 1/√3, tan 45° = 1, tan 60° = √3
• sin 30° = cos 60° = 1/2, sin 60° = cos 30° = √3/2

Solved examples

Example 1. If sin θ = 7/25, find cos θ and tan θ.

cos²θ = 1 − sin²θ = 1 − 49/625 = 576/625
cos θ = 24/25, tan θ = sin θ/cos θ = 7/24
cos θ = 24/25, tan θ = 7/24

Example 2. From a point 50 m away from the foot of a tower, the angle of elevation of its top is 60°. Find the height.

tan 60° = height/50 → √3 = h/50
h = 50√3 = 50 × 1.73
Height ≈ 86.5 m

Quick revision

• Angle of elevation from A to B = angle of depression from B to A.
• Draw the figure: tower vertical, ground horizontal, right angle at the foot.
• Use tan when height and distance are involved.
• To prove identities, change everything to sin and cos.

Exam tip

Always draw and label the figure in heights and distance sums. Use √3 = 1.73 only at the last step.

Quick summary

Mensuration means measuring surface area and volume of solids like cylinder, cone, sphere and frustum, and the area of a sector and segment of a circle.

Important terms

• Slant height (l) – length from the top of a cone to the edge of its base.
• Frustum – the part of a cone left after cutting off its top with a parallel cut.
• Sector – the part of a circle between two radii and an arc.
• Segment – the part of a circle between a chord and its arc.

Formulas

• Cuboid: V = l × b × h, total surface = 2(lb + bh + hl); Cube: V = a³, total surface = 6a²
• Cylinder: V = πr²h, curved surface = 2πrh, total surface = 2πr(r + h)
• Cone: l = √(r² + h²), V = ⅓πr²h, curved surface = πrl, total surface = πr(l + r)
• Sphere: V = (4/3)πr³, surface = 4πr²
• Hemisphere: V = (2/3)πr³, curved surface = 2πr², total surface = 3πr²
• Frustum: V = ⅓πh(r₁² + r₂² + r₁r₂), curved surface = πl(r₁ + r₂), l = √[h² + (r₁ − r₂)²]
• Sector area = (θ/360) × πr², arc length = (θ/360) × 2πr
• Segment area = sector area − triangle area

Solved examples

Example 1. A cone has r = 7 cm and h = 24 cm. Find its curved surface area and volume.

l = √(49 + 576) = √625 = 25 cm
Curved surface = πrl = (22/7) × 7 × 25 = 550 cm²
Volume = ⅓ × (22/7) × 49 × 24 = 1232 cm³
CSA = 550 cm², V = 1232 cm³

Example 2. Find the area and arc length of a sector with r = 21 cm and θ = 60°.

Area = (60/360) × (22/7) × 21 × 21 = (1/6) × 1386 = 231 cm²
Arc = (60/360) × 2 × (22/7) × 21 = (1/6) × 132 = 22 cm
Area = 231 cm², arc = 22 cm

Quick revision

• Volume is in cm³ (or m³); area in cm² (or m²).
• 1 litre = 1000 cm³; 1 m³ = 1000 litres.
• Melting and recasting: volume stays the same.
• Use π = 22/7 when the radius is a multiple of 7, else 3.14.

Exam tip

Write the formula first, then the values, then the unit. Missing units can cost marks.