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State Board · Class 10 · Mathematics Part 2

Lesson 1: Similarity

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

This lesson is also in: मराठी

Ratio of areas of two triangles

From daily life

Two plots of land: one is 12 m wide and 5 m deep, the other 8 m wide and 9 m deep. Which is bigger, and by how much? Comparing areas by base and height is the first step in similarity.

Simple definition

Area of a triangle = 12 × base × height. So the ratio of areas of two triangles = ratio of (base × height).

Rule 1: Same height or same base — Bases 12, 8 and heights 5, 9: areas in the ratio 60 : 72 = 5 : 6. Same height: areas are in the ratio of the bases (bases 7 and 21 give 1 : 3). Same base: areas are in the ratio of the heights. Triangles on the same base between the same parallel lines have equal areas.

Triangles ABD and ADC with the same height from A have areas in the ratio of their bases 4 to 5ABCDcommon height hBD = 4DC = 5Same height ⇒ areas are in the ratio of the basesA(ΔABD) : A(ΔADC) = 4 : 5

In the figure : In the figure triangles with the same height: areas are in the ratio of the bases.

  • Easy : Easy: Heights equal, bases 7 and 21: areas in the ratio 1 : 3.
  • Medium : Medium: D on side BC of ΔABC with BC = 20, BD = 8, DC = 12: A(ΔABD) : A(ΔADC) = 8 : 12 = 2 : 3 and A(ΔABD) : A(ΔABC) = 8 : 20 = 2 : 5.
  • Exam-type : Exam: In ΔPQR with PS ⊥ RQ, QT ⊥ PR, RQ = 10, PS = 8 and PR = 16: area = 12 × 10 × 8 = 40 = 12 × 16 × QT, so QT = 5.
Remember

Same height → bases decide. Same base → heights decide. A median halves the area.

Common mistake

Forgetting the 12 cancels in a ratio. It cancels, so do not worry about it.

Shortcut

Memory clue: "क्षेत्रफळ ∝ पाया × उंची".

Spot the difference

ConditionRatio of areas
Same heightRatio of bases
Same baseRatio of heights
Different bothRatio of base × height

Practice questions

Easy
  1. 1Write the area of a triangle.
  2. 2Same heights, bases 5 and 15: ratio of areas?
  3. 3Same base, heights 6 and 9: ratio of areas?
Medium
  1. 4Bases 9, 10 and heights 5, 6: ratio of areas?
  2. 5BC = 15, BD = 6 on side BC. Find A(ΔABD) : A(ΔADC).
  3. 6What does a median do to the area of a triangle?
Challenging
  1. 7AE = 4 and DF = 6 are heights on the common base BC. Find A(ΔABC) : A(ΔDBC).
  2. 8Why do ΔABC and ΔBDC on the same base BC of a parallelogram have equal areas?
Show answers (try them yourself first)
  1. 112 × base × height.
  2. 21 : 3.
  3. 32 : 3.
  4. 445 : 60 = 3 : 4.
  5. 5DC = 9, ratio 6 : 9 = 2 : 3.
  6. 6Divides it into two equal areas.
  7. 74 : 6 = 2 : 3.
  8. 8Same base and same height (between the parallel sides).

Basic proportionality theorem and its converse

From daily life

A shopkeeper cuts a triangular banner with a straight line parallel to its bottom edge. The cut divides both slanting edges in the same proportion. That is the Basic Proportionality Theorem.

Simple definition

Basic Proportionality Theorem (मूलभूत समानुपात प्रमेय, BPT): a line parallel to one side of a triangle divides the other two sides in the same ratio: if PQ ∥ BC then APPB = AQQC.

Rule 1: Using BPT — In ΔABC, DE ∥ BC, AD = 3, DB = 6, AE = 4: 36 = 4EC, so EC = 8. If AD : DB = 3 : 5 and AC = 24, then AE = 9 and EC = 15. In a trapezium with AB ∥ PQ ∥ DC, AP = 10, PD = 15, QC = 18: BQ = 10 × 1815 = 12.

A line DE parallel to BC divides AB and AC in the same ratio 3 to 2ABCDEAD = 3DB = 2AE = 4.5EC = 3AD/DB = AE/EC = 3/2DE ∥ BC divides both sides in the same ratio

In the figure : In the figure the basic proportionality theorem: a line parallel to a side divides the other two sides in the same ratio.

  • Easy : Easy: AP = 2, PB = 4, AQ = 3: QC = 6.
  • Medium : Medium: AD = 4, DB = 6, AE = x, EC = x + 3 gives 6x = 4x + 12, x = 6.
  • Exam-type : Exam: Proof idea: ΔAPQ and ΔPQB share a height, ΔPQB and ΔPQC have equal areas (same base PQ, between parallels PQ and BC).

Rule 2: Converse of BPT — If a line divides two sides in the same ratio, it is parallel to the third side. In ΔPQR, PM = 6, MQ = 9, PN = 8, NR = 12: 69 = 812 = 23, so MN ∥ QR. If PM = 6, MQ = 9, PN = 8, NR = 14: not equal, so not parallel.

  • Easy : Easy: 12 = 24, so the line is parallel.
  • Medium : Medium: In ΔABC with AB = AC, the bisectors of ∠B and ∠C meet AC and AB at D and E. By symmetry AE = AD, so ED ∥ BC.
  • Exam-type : Exam: If X is inside ΔDEF and PQ ∥ DE, QR ∥ EF (P, Q, R on XD, XE, XF), then XP : PD = XQ : QE = XR : RF, so PR ∥ DF.
Remember

Parallel line → equal ratios on the other two sides. Equal ratios → parallel.

Common mistake

Comparing part with whole on one side and part with part on the other. Use the same kind of ratio on both sides.

Shortcut

Memory clue: "समांतर ⇔ समान गुणोत्तर".

Spot the difference

StatementMeaning
BPTPQ ∥ BC ⇒ AP/PB = AQ/QC
ConverseAP/PB = AQ/QC ⇒ PQ ∥ BC
Test6/9 = 8/12 ⇒ parallel

Practice questions

Easy
  1. 1State BPT.
  2. 2State its converse.
  3. 3DE ∥ BC, AD = 2, DB = 6, AE = 3. Find EC.
Medium
  1. 4AD : DB = 3 : 5 and AC = 24. Find AE.
  2. 5Is PQ ∥ BC if AP = 4, PB = 6, AQ = 6, QC = 9?
  3. 6Trapezium AB ∥ PQ ∥ DC: AP = 15, PD = 12, QC = 14. Find BQ.
Challenging
  1. 7Why can we say the converse by contradiction?
  2. 8In ΔABC, DE ∥ BC, AD = x, DB = 2x and AE = 5. Find EC.
Show answers (try them yourself first)
  1. 1A line parallel to a side divides the other two sides in the same ratio.
  2. 2If a line divides two sides in the same ratio, it is parallel to the third side.
  3. 39.
  4. 49.
  5. 546 = 69, yes.
  6. 617.5.
  7. 7Assume it is not parallel; then a different parallel line gives the same ratio, impossible.
  8. 8x2x = 5EC, EC = 10.

Angle bisector theorem and three parallel lines

From daily life

When you fold a paper triangle so that one corner angle is cut exactly in half, the fold line splits the opposite edge in the same ratio as the two edges that make the corner.

Simple definition

Angle bisector theorem (कोन दुभाजक प्रमेय): the bisector of an angle of a triangle divides the opposite side in the ratio of the remaining two sides: BDDC = ABAC.

Rule 1: Angle bisector — AB = 9, AC = 6, BC = 10 and AD bisects ∠A: BD : DC = 9 : 6 = 3 : 2, so BD = 6 and DC = 4. In ΔLMN, LM = 8, MN = 12, LN = 15, MT bisects ∠M: LT : TN = 8 : 12, so LT = 6 and TN = 9. Converse: if ABAC = BDDC, AD bisects ∠A.

The bisector AD of angle A divides BC into 4 and 6, the same ratio as AB to AC, 6 to 9ABCDAB = 6AC = 9BD = 4DC = 6AD bisects ∠A ⇒ BD : DC = AB : AC4 : 6 = 6 : 9 = 2 : 3

In the figure : In the figure the angle bisector theorem: BD : DC = AB : AC.

  • Easy : Easy: AB = 6, AC = 3, BC = 12: BD = 8, DC = 4.
  • Medium : Medium: AB = 12, AC = 8, BC = 15: BD = 9, DC = 6.
  • Exam-type : Exam: BD = 3, DC = 5, AB = 6 and AD bisects ∠A: AC = 10.

Rule 2: Three parallel lines — Three parallel lines cut two transversals in the same ratio: ABBC = PQQR. If AB = 4, BC = 6, PQ = 5, then QR = 7.5. If AC = 5.4, CE = 9, BD = 7.5 then DF = 7.5 × 95.4 = 12.5.

  • Easy : Easy: AB = 2, BC = 3, PQ = 4: QR = 6.
  • Medium : Medium: AB = 5, BC = 10, PR = 36: PQ : QR = 1 : 2, PQ = 12.
  • Exam-type : Exam: Proof: draw PC meeting line m at D. In ΔACP, BD ∥ AP gives ABBC = PDDC, and in ΔCPR, DQ ∥ CR gives PDDC = PQQR.
Remember

Bisector → ratio of the two sides at that corner. Parallel lines cut transversals in equal ratios.

Common mistake

Putting the opposite side in the ratio. The ratio uses the two sides at the angle.

Shortcut

Memory clue: "दुभाजक = बाजू : बाजू".

Spot the difference

TheoremFormula
Angle bisectorBD/DC = AB/AC
Three parallel linesAB/BC = PQ/QR
ConverseAB/AC = BD/DC ⇒ bisector

Practice questions

Easy
  1. 1State the angle bisector theorem.
  2. 2AB = 6, AC = 9, BC = 10. Find BD.
  3. 3AB = 15, AC = 10, BD = 6. Find DC.
Medium
  1. 4PR = 15, RQ = 20, PS = 12, RS bisects ∠R. Find SQ.
  2. 5AB = 4, BC = 6, PQ = 6 for three parallel lines. Find QR.
  3. 6In ΔMNP, NQ bisects ∠N, MN = 5, PN = 7, MQ = 2.5. Find QP.
Challenging
  1. 7AB = x, BC = x + 5, AD = x − 2, DC = x + 2 and BD bisects ∠B. Find x.
  2. 8A triangle has sides 6, 8, 12... why does the bisector of the angle between 6 and 8 never cut the 12-side in equal parts?
Show answers (try them yourself first)
  1. 1The bisector divides the opposite side in the ratio of the other two sides.
  2. 2BD : DC = 2 : 3, BD = 4.
  3. 34.
  4. 416.
  5. 59.
  6. 63.5.
  7. 7x(x + 2) = (x − 2)(x + 5), x = 10.
  8. 8It splits in the ratio 6 : 8 = 3 : 4, not 1 : 1.

Tests of similarity of triangles

From daily life

A 6 m pole casts a 4 m shadow. At the same time a tower casts a 20 m shadow. How tall is the tower? Same-shape triangles, differently sized, make this easy.

Simple definition

Two triangles are similar (समरूप) if corresponding angles are equal and corresponding sides are in proportion. The tests are AA, SAS and SSS.

Rule 1: The three tests — AA: two angles equal (∠X = 70°, ∠Y = 50° and ∠P = 70°, ∠Q = 50°). SAS: two sides proportional with the included angles equal (4 : 6 = 6 : 9 = 2 : 3). SSS: three sides proportional (3, 4, 6 and 9, 12, 18 with ratio 1 : 3). Similarity is reflexive, symmetric and transitive.

A 3 m pole with a 2 m shadow and a tree with a 6 m shadow form similar triangles, so the tree is 9 m tall3 m2 mh = ?6 mSun rays are parallel,so the triangles are similarh / 3 = 6 / 2h = 9 m

In the figure : In the figure an application of similar triangles: finding a height from shadows.

Live figure

Tests of similarity of triangles

In the app you can move this figure with your finger.

Open in the app
  • Easy : Easy: Angles 40°, 60°, 80° and 80°, 40°, 60° are similar by AA.
  • Medium : Medium: Pole 6 m, shadow 4 m; tower shadow 20 m: 64 = h20, h = 30 m.
  • Exam-type : Exam: Trapezium AB ∥ DC, AB = 12, DC = 4, OB = 9: ΔAOB ~ ΔCOD gives OD = 9 × 412 = 3.

Rule 2: Choosing the right test — Angle given: use AA. Two sides and the angle between: use SAS (the angle must be between the proportional sides). Only sides: use SSS. If BP ⊥ AC and CQ ⊥ AB, ΔAPB ~ ΔAQC by AA (common ∠A, and 90° each).

  • Easy : Easy: 3, 4, 6 and 9, 12, 18: SSS with ratio 1 : 3.
  • Medium : Medium: If the equal angles are not between the proportional sides, SAS does not apply.
  • Exam-type : Exam: Diagonals of ABCD meet at Q with 2QA = QC and 2QB = QD: ΔAQB ~ ΔCQD by SAS, so DC = 2AB.
Remember

Similar triangles have equal angles and proportional sides. Write the corresponding vertices in the same order.

Common mistake

Writing vertices in the wrong order. ΔABC ~ ΔPQR means A matches P, B matches Q.

Shortcut

Memory clue: "AA, SAS, SSS".

Spot the difference

TestNeeds
AA2 equal angles
SAS2 proportional sides + included angle
SSS3 proportional sides

Practice questions

Easy
  1. 1Name the three similarity tests.
  2. 2When are two triangles similar?
  3. 3∠X = 100°, ∠Y = 30° and ∠M = 100°, ∠N = 30°. Which test?
Medium
  1. 4Sides 2, 3, 4 and 6, 9, 12: similar? Ratio?
  2. 5Poles 8 m and 4 m, shadow of the small one 6 m. Shadow of the big one?
  3. 6AP ⊥ BC, BQ ⊥ AC, AP = 7, BQ = 8, BC = 12. Find AC.
Challenging
  1. 7In a trapezium AB ∥ DC, AB = 20, DC = 6, OB = 15. Find OD.
  2. 8Sides 4, 6 and 6, 9 but the equal angle is not between them. Can we say similar?
Show answers (try them yourself first)
  1. 1AA, SAS, SSS.
  2. 2Equal corresponding angles and proportional sides.
  3. 3AA.
  4. 4Yes, SSS, ratio 1 : 3.
  5. 512 m.
  6. 6ΔCPA ~ ΔCQB, AC = 12 × 78 = 10.5.
  7. 74.5.
  8. 8No, SAS needs the included angle.

Areas of similar triangles

From daily life

A photograph and its enlargement have the same shape. If every length is doubled, the paper needed becomes four times, not two times. Area grows as the square.

Simple definition

For similar triangles, the ratio of areas = the ratio of the squares of corresponding sides: A(ΔABC)A(ΔPQR) = AB2PQ2 = BC2QR2.

Rule 1: Using the theorem — Sides 2 : 3 give areas 4 : 9. Areas 49 and 100 give sides 7 : 10. Sides in the ratio 3 : 5 and the smaller area 36: bigger area = 36 × 259 = 100. Areas 18 and 50 with a side 6 of the smaller: the bigger side is 6 × 53 = 10. Equilateral triangles of sides 6 and 12 have areas 1 : 4.

Two similar triangles with sides 2 and 3 and unit squares showing the areas are in the ratio 4 to 923Similar triangles: side ratio 2 : 32 × 2 = 43 × 3 = 9Areas are in the ratio2² : 3² = 4 : 9Area ratio = (ratio of corresponding sides)²

In the figure : In the figure areas of similar triangles are in the ratio of the squares of the sides.

  • Easy : Easy: Sides 1 : 4 give areas 1 : 16.
  • Medium : Medium: Areas 16 and 25 give sides 4 : 5.
  • Exam-type : Exam: PQ ∥ DE, FP : PD = 1 : 2. So FP : FD = 1 : 3 and areas 1 : 9. If A(ΔPQF) = 8, A(ΔFDE) = 72 and A(quadrilateral DPQE) = 72 − 8 = 64.
Remember

Ratio of areas = (ratio of sides)2. Take the square root to go back.

Common mistake

Using the plain ratio of sides for the ratio of areas. Always square it.

Shortcut

Memory clue: "क्षेत्रफळ = बाजू²".

Spot the difference

Sides ratioAreas ratio
2 : 34 : 9
3 : 59 : 25
1 : 41 : 16

Practice questions

Easy
  1. 1State the theorem on areas of similar triangles.
  2. 2Sides 3 : 5: ratio of areas?
  3. 3Areas 4 : 9: ratio of sides?
Medium
  1. 4Areas 80 and 125: AB/PQ?
  2. 5Sides 2 : 5, smaller area 64: bigger area?
  3. 6Areas 225 and 81, a side 12 of the smaller. Find the corresponding side of the bigger.
Challenging
  1. 79 × A(ΔPQR) = 16 × A(ΔLMN), QR = 20. Find MN.
  2. 8Why do the diagonals of a trapezium make triangles with areas in the ratio AB² : CD²?
Show answers (try them yourself first)
  1. 1Ratio of areas = ratio of squares of corresponding sides.
  2. 29 : 25.
  3. 32 : 3.
  4. 445.
  5. 564 × 254 = 400.
  6. 612 × 159 = 20.
  7. 7Areas 16 : 9, sides 4 : 3, MN = 15.
  8. 8ΔAPB ~ ΔCPD by AA, so areas go as the squares of the parallel sides.

Quick revision

  • Triangles with equal heights have areas proportional to their bases, and triangles with equal bases have areas proportional to their heights.
  • BPT: a line parallel to a side divides the other two sides in the same ratio (and conversely); the bisector of an angle of a triangle divides the opposite side in the ratio of the remaining sides.
  • Tests of similarity: AA (AAA), SAS, SSS; the ratio of the areas of similar triangles = the ratio of the squares of their corresponding sides.