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State Board · Class 10 · Mathematics Part 2

Lesson 4: Geometric Constructions

5 concepts · 40 practice questions with answers · Free to read

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Similar triangle when vertices are distinct

From daily life

A mapmaker redraws a big triangular plot at two-thirds of its size on a small sheet. Every side shrinks by the same ratio and the angles stay the same. That is a similar triangle construction.

Simple definition

To construct a triangle similar (समरूप) to a given triangle, first find the new side lengths by multiplying each given side by the ratio, then draw the new triangle. Corresponding angles stay equal.

Rule 1: Finding the new sides — ΔABC ~ ΔPQR with AB = 6, BC = 4.5, AC = 7.5 and AB : PQ = 3 : 2. Then PQ = 6 × 23 = 4, QR = 4.5 × 23 = 3, PR = 7.5 × 23 = 5. If the ratio is 2 : 5 (new is bigger): AB = 4.8, BC = 3.6, AC = 6 become 12, 9, 15.

  • Easy : Easy: Ratio 1 : 2 and side 3 cm gives a new side 6 cm.
  • Medium : Medium: AB : PQ = 7 : 5 means ΔABC is the bigger one.
  • Exam-type : Exam: A side of 11.63 cm is hard to measure; divide a 11.6 cm segment into 3 equal parts and take one part.

Rule 2: Which way the ratio goes — New side = given side × (new part ÷ given part). If the given triangle's term is the larger one, the new triangle is smaller. With ratio 5 : 4 for BC : MN and BC = 6, MN = 6 × 45 = 4.8. Angles do not change: if ∠RST = 40°, then ∠XYZ = 40°.

  • Easy : Easy: Ratio 5 : 4, side 10 gives 8.
  • Medium : Medium: Ratio 3 : 5, side 4.5 gives 7.5.
  • Exam-type : Exam: Sides 5.5, 6, 4.5 with BC : MN = 5 : 4: MN = 4.8, LM = 4.4, LN = 3.6.
Remember

New side = old side × new ÷ old. Angles stay the same.

Common mistake

Using the ratio upside down. Check: is the new triangle bigger or smaller?

Shortcut

Memory clue: "नवीन ÷ जुना = गुणाकार".

Spot the difference

Ratio old : newSide 6 becomes
3 : 24
1 : 212
5 : 44.8

Practice questions

Easy
  1. 1What stays the same in similar triangles?
  2. 2How do we find a new side?
  3. 3Ratio 3 : 2, side 9. New side?
Medium
  1. 4Ratio 2 : 3, side 8. New side?
  2. 5Is the new triangle bigger if AB : PQ = 7 : 5?
  3. 6Sides 4.5, 5.7 with included angle 40°, ratio 3 : 5. Find the new sides and angle.
Challenging
  1. 7Sides 5.1 and 4.8 with ratio 4 : 7. Find the new sides.
  2. 8How do you handle a length like 73 cm?
Show answers (try them yourself first)
  1. 1Corresponding angles.
  2. 2Multiply the given side by the ratio (new ÷ given).
  3. 36.
  4. 412.
  5. 5No, smaller.
  6. 67.5, 9.5 and 40°.
  7. 78.925 and 8.4.
  8. 8Divide a 7 cm segment into 3 equal parts and take one part.

Similar triangle with a common vertex

From daily life

If the old and new triangles share a corner, you can draw the new one inside or around the old one. Think of a photo and its enlargement stuck at the same corner of a frame.

Simple definition

With a common vertex B (समाईक शिरोबिंदू), divide BC into equal parts, mark C′ on BC (smaller) or on ray BC (bigger), then draw a line through C′ parallel to AC.

Rule 1: Smaller triangle — For ratio AB : A′B = 4 : 3, ΔABC is bigger and BC′ = 34 BC. Steps: (1) draw ΔABC; (2) divide BC into 4 equal parts; (3) mark C′ at the third part from B; (4) draw a line through C′ parallel to AC meeting AB at A′; (5) ΔA′BC′ is the required triangle.

A small triangle A'BC' inside ABC sharing vertex B with sides in the ratio 3 to 5ABCA'C'ΔA'BC' ~ ΔABCAB : A'B = 5 : 3BC' = 3/5 × BCCommon vertex B: B, A', A and B, C', C are in a line

In the figure : In the figure a similar triangle with a common vertex: the smaller triangle sits inside the bigger one.

  • Easy : Easy: Ratio 4 : 3 with BC = 8: BC′ = 6.
  • Medium : Medium: Divide BC into 4 parts and take 3 parts from B.
  • Exam-type : Exam: By BPT, a line through C′ parallel to AC gives BA′A′A = BC′C′C, so ΔA′BC′ ~ ΔABC.

Rule 2: Dividing a segment and the bigger triangle — To divide BC into equal parts, draw a ray from B on the side opposite to A, mark equal steps on it, join the last mark to C and draw parallels to it through the other marks. For a bigger triangle with ratio AB : A′B = 3 : 5: divide BC into 3 equal parts and mark C′ on ray BC at 5 such parts, so BC′ = 53 BC. For BC = 9, BC′ = 15.

A segment BC divided in the ratio 3 to 2 by drawing a ray with 5 equal marks and lines parallel to T5CDividing BC in the ratio 3 : 2 (5 equal parts)T1T2T3T4T5BCC'BC' : C'C = 3 : 2parallel lines cutequal parts

In the figure : In the figure dividing a segment into equal parts: a ray with equal marks and parallel lines.

  • Easy : Easy: Ratio 3 : 5 and BC = 6: BC′ = 10.
  • Medium : Medium: Mark 5 equal steps on the ray for a division into 5 parts.
  • Exam-type : Exam: ΔPQR ~ ΔLTR with PQ = 4.2, QR = 5.4, PR = 4.8 and PQ : LT = 3 : 4 (common vertex R): LT = 5.6, TR = 7.2, LR = 6.4.
Remember

Smaller: C′ lies on BC. Bigger: C′ lies on ray BC beyond C. The new line is parallel to AC.

Common mistake

Dividing into the wrong number of parts. Divide by the smaller term of the ratio, then take the larger number of parts.

Shortcut

Memory clue: "जुन्याने भाग, नवीनाइतके घे".

Spot the difference

Ratio old : newDivide BC intoTake
4 : 343
3 : 535
5 : 353

Practice questions

Easy
  1. 1What does a common vertex do for the construction?
  2. 2Which theorem proves the construction?
  3. 3Ratio 4 : 3, BC = 12. Find BC′.
Medium
  1. 4Ratio 3 : 5, BC = 6. Find BC′.
  2. 5Ratio 5 : 7: into how many parts do we divide BC?
  3. 6ΔAMT ~ ΔAHE, AM = 6.3, AT = 5.6, AM : AH = 7 : 5. Find AH and AE.
Challenging
  1. 7YZ : YQ = 6 : 5, YQ = 7.2. Find YZ.
  2. 8Why do we draw the ray on the side where A does not lie?
Show answers (try them yourself first)
  1. 1The new triangle shares corner B and its side is parallel to AC.
  2. 2Basic proportionality theorem.
  3. 39.
  4. 410.
  5. 55 parts; take 7 for the bigger triangle.
  6. 64.5 and 4.0.
  7. 78.64.
  8. 8So the marks and parallels do not cross the triangle.

Tangent at a point using the centre

From daily life

A wheel on the ground touches the road at one point, and the road is at a right angle to the wheel's spoke at that point. To draw such a straight edge on a circle, draw the radius and then a perpendicular.

Simple definition

The tangent at a point P of a circle is the line through P perpendicular to the radius CP.

Rule 1: Steps with the centre — (1) Draw a circle with centre C and take a point P on it. (2) Draw ray CP, extended beyond P. (3) Through P draw line l perpendicular to ray CP. Line l is the tangent. Only one tangent exists at a point of a circle.

Tangent at point P of a circle with centre C: draw the line l through P perpendicular to the radius CPCPl1. Draw ray CP2. Draw l ⊥ CP at Pl is the tangent

In the figure : In the figure tangent at a point of a circle using the centre.

  • Easy : Easy: Radius 3 cm, point P on the circle: draw CP = 3 cm, then the perpendicular at P.
  • Medium : Medium: A perpendicular at the end of a radius meets the circle only at that end.
  • Exam-type : Exam: Why can only one tangent be drawn at P? Only one perpendicular to CP passes through P.

Rule 2: Parallel tangents — Tangents at the ends of a diameter are both perpendicular to it, so they are parallel and the distance between them equals the diameter. In a circle of radius 3.3, a chord PQ = 6.6 is a diameter: tangents at P and Q are parallel. A shorter chord MN = 5.7 in radius 3.4: tangents at M and N meet. For an arc of 120° the tangents at its ends meet at T with ∠T = 360° − (90° + 90° + 120°) = 60°.

  • Easy : Easy: Radius 4 cm gives parallel tangents 8 cm apart.
  • Medium : Medium: Chord 5 cm in a circle of radius 3 cm (diameter 6 cm): the tangents at its ends meet.
  • Exam-type : Exam: For arc AB = 100° the tangents at A and B meet at T with ∠T = 80°.
Remember

Tangent ⊥ radius at P. A diameter gives parallel tangents.

Common mistake

Drawing the perpendicular at the wrong point. It must pass through P itself.

Shortcut

Memory clue: "त्रिज्या काढ, तिला लंब काढ".

Spot the difference

ChordTangents at its ends
DiameterParallel
Shorter than diameterMeet outside
Distance if parallelDiameter

Practice questions

Easy
  1. 1How is a tangent drawn using the centre?
  2. 2How many tangents at a point on a circle?
  3. 3Tangents at the ends of a diameter: relation?
Medium
  1. 4Radius 2.7 cm: draw a tangent at a point. Which angle is used?
  2. 5Distance between parallel tangents of a radius 4.5 cm circle?
  3. 6Chord MN = 5.7 cm in a circle of radius 3.4 cm. Do the tangents at M and N meet?
Challenging
  1. 7Arc AB = 80°, tangents meet at T. Find ∠T.
  2. 8Why does the perpendicular at the end of a radius touch the circle only once?
Show answers (try them yourself first)
  1. 1Draw the radius to P and the perpendicular to it at P.
  2. 2One.
  3. 3Parallel.
  4. 490° with the radius.
  5. 59 cm.
  6. 6Yes, 5.7 < 6.8 (diameter).
  7. 7360° − 260° = 100°.
  8. 8Every other point of the line is farther from the centre than the radius.

Tangent at a point without using the centre

From daily life

Suppose someone has lost the centre of a coin-shaped plate but must still draw a tangent at a point. We use the angle between the tangent and a chord, which equals the angle in the opposite arc.

Simple definition

Tangent–chord angle (स्पर्शिका-जीवा कोन): the angle between a tangent and a chord equals the inscribed angle in the opposite arc. We copy that angle at C to get the tangent.

Rule 1: Steps without the centre — (1) Draw a circle and take C on it. (2) Draw chord CB and an inscribed ∠CAB. (3) With centre A, draw an arc cutting the sides of ∠BAC at M and N. (4) With the same radius and centre C, draw an arc cutting CB at R. (5) With radius MN and centre R, draw an arc cutting the previous arc at D. Draw line CD, the tangent.

Tangent at C without the centre: the angle between the tangent and chord CB is made equal to the inscribed angle CABxxCBADInscribed ∠CAB = xCopy it at C:∠BCD = xCD is thetangent at C

In the figure : In the figure tangent at a point without using the centre.

  • Easy : Easy: ∠CAB = 48° means ∠BCD = 48°.
  • Medium : Medium: ∠BCD = 35° means the intercepted arc CB = 70°.
  • Exam-type : Exam: Why it works: ΔMAN ≅ ΔRCD (SSS), so ∠BCD = ∠CAB, and by the converse of the tangent–chord angle theorem CD is a tangent.

Rule 2: Angle facts — The tangent–chord angle = 12 × the intercepted arc = the inscribed angle in the opposite arc. Arc 76° gives angle 38°. Angle 54° gives arc 108°.

  • Easy : Easy: Arc 90° gives angle 45°.
  • Medium : Medium: Angle 62° gives arc 124°.
  • Exam-type : Exam: If the inscribed angle in the opposite arc is 55°, the tangent–chord angle is 55° and the arc is 110°.
Remember

Copy the inscribed angle at C. The new arm is the tangent.

Common mistake

Copying the angle on the wrong side of the chord. It must be on the other side from A.

Shortcut

Memory clue: "जीवा-स्पर्शिका कोन = समोरचा कोन".

Spot the difference

QuantityValue
Angle in opposite arc54°
Tangent–chord angle54°
Intercepted arc108°

Practice questions

Easy
  1. 1Can a tangent be drawn without the centre?
  2. 2What equals the tangent–chord angle?
  3. 3Arc 70°: tangent–chord angle?
Medium
  1. 4∠CAB = 62°. Find ∠BCD.
  2. 5Angle 41°: intercepted arc?
  3. 6Which congruence test proves ∠MAN = ∠RCD?
Challenging
  1. 7Which angle is copied to C in the construction?
  2. 8The tangent at C makes ∠BCD = 38° with chord CB. What is the arc CB, and the angle at the centre?
Show answers (try them yourself first)
  1. 1Yes, by copying an inscribed angle.
  2. 2The inscribed angle in the opposite arc.
  3. 335°.
  4. 462°.
  5. 582°.
  6. 6SSS.
  7. 7The inscribed angle ∠CAB.
  8. 8Arc 76°, central angle 76°.

Tangents from a point outside the circle

From daily life

A guard at a gate points at a round fountain. From outside, you can draw exactly two straight lines from your spot that just touch the fountain. We build them using a right angle in a semicircle.

Simple definition

From an outside point P to a circle with centre O, exactly two tangents can be drawn, with equal lengths PA = PB = √(OP2 − r2).

Rule 1: Steps — (1) Draw a circle with centre O and take an outside point P. (2) Draw seg OP and its perpendicular bisector to get the midpoint M. (3) Draw a circle with centre M and radius OM. (4) Where the two circles meet, name the points A and B. (5) Draw lines PA and PB. They are tangents because ∠OAP = 90° (angle in a semicircle).

Tangents from an outside point P: the circle on diameter OP cuts the given circle at A and BOPMABCircle on OP as diameter meets the circle at A, B∠OAP = 90° (angle in a semicircle), so PA and PB are tangents

In the figure : In the figure tangents from an outside point using the circle on diameter OP.

  • Easy : Easy: Only one circle can have diameter OP, and its centre is M.
  • Medium : Medium: Radius of the second circle = OP2.
  • Exam-type : Exam: Tangents can be drawn only if OP > r.

Rule 2: Length of a tangent — PA = √(OP2 − r2) by Pythagoras. r = 5, OP = 13: PA = 12. r = 6, OP = 10: PA = 8. r = 7, OP = 25: PA = 24. If OP = 2r (for example r = 4, OP = 8), sin∠OPA = rOP = 12, so ∠OPA = 30° and the angle between the tangents is 60°.

  • Easy : Easy: r = 3, OP = 5 gives PA = 4.
  • Medium : Medium: r = 9, OP = 15 gives PA = 12.
  • Exam-type : Exam: r = 6, OP = 12: PA = √(144 − 36) = 6√3 and the angle between tangents is 60°.
Remember

Two tangents from an outside point. Draw the circle on diameter OP.

Common mistake

Trying to draw tangents from a point inside the circle. There are none.

Shortcut

Memory clue: "OP चा मध्य → वर्तुळ → A, B".

Spot the difference

Position of PTangents
Outside2
On the circle1
Inside0

Practice questions

Easy
  1. 1How many tangents from an outside point?
  2. 2Which circle is used in the construction?
  3. 3r = 3, OP = 5: PA?
Medium
  1. 4r = 4.1, OP = 7.3: PA (to one decimal)?
  2. 5Why is ∠OAP = 90°?
  3. 6Circle of diameter 6.4 cm, OP = 6.4 cm. Find the angle between tangents.
Challenging
  1. 7r = 3.4 cm and OP = 5.5 cm. Can we draw tangents? Why?
  2. 8Find the tangent length for r = 3.5 and OP = 5.7.
Show answers (try them yourself first)
  1. 1Two.
  2. 2The circle with diameter OP.
  3. 34.
  4. 4√36.48 ≈ 6.04, so about 6.0.
  5. 5It is an angle in a semicircle (diameter OP).
  6. 6sin = 3.26.4 = 12, angle 30° each side, so 60°.
  7. 7Yes, 5.5 > 3.4, so P is outside.
  8. 8√(32.49 − 12.25) = √20.24 ≈ 4.5.

Quick revision

  • Corresponding sides of similar triangles are proportional and corresponding angles are equal. To construct a similar triangle with a given ratio, multiply the given sides by the ratio of the new part to the given part. With a common vertex B, divide BC into equal parts, mark C′ on BC (smaller triangle) or on ray BC (bigger triangle) and draw a line through C′ parallel to AC to get A′.
  • The tangent at a point P of a circle with centre C is the line through P perpendicular to ray CP; without the centre, draw a chord CB and an inscribed ∠CAB and copy it at C as ∠BCD ≅ ∠CAB (tangent-secant angle theorem and its converse). Only one tangent can be drawn at a point of a circle.
  • For tangents from an exterior point P, draw the circle with diameter OP (centre M, the midpoint of OP); it cuts the given circle at A and B because an angle in a semicircle is 90°. PA = PB = √(OP² − r²), and two tangents exist only when OP is greater than the radius.
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