Chapter 6: Triangles (NCERT Solutions)

Exercise 6.1: Similar Figures

Q1. Fill in the blanks:

(i) All circles are similar.

(ii) All squares are similar.

(iii) All equilateral triangles are similar.

(iv) Two polygons... (a) their corresponding angles are equal and (b) their corresponding sides are proportional.

Q2. Give two different examples of pair of...

(i) Similar figures: Two equilateral triangles of sides 1cm and 2cm. Two circles of radii 1cm and 2cm.

(ii) Non-similar figures: A square and a rectangle. A triangle and a parallelogram.

Q3. State whether the following quadrilaterals are similar or not.

Fig 6.8 shows a rhombus (sides 1.5cm) and a square (sides 3cm).
Ratio of sides is 1.5/3 = 1/2 (Proportional).
But corresponding angles are not equal (Square has 90°, Rhombus doesn't).
Therefore, they are not similar.

Exercise 6.2: Basic Proportionality Theorem

Q1. DE || BC. Find EC in (i) and AD in (ii).

(i) 1.5/3 = 1/EC ⇒ EC = 2 cm.

(ii) AD/7.2 = 1.8/5.4 ⇒ AD = 7.2/3 = 2.4 cm.

Q2. E and F are points on PQ and PR of ΔPQR. PE=3.9, EQ=3, PF=3.6, FR=2.4. Is EF || QR?

PE/EQ = 3.9/3 = 1.3.
PF/FR = 3.6/2.4 = 1.5.
Ratios not equal. No.

Q3. LM || CB and LN || CD. Prove AM/AB = AN/AD.

AM/AB = AL/AC (By BPT corollary).
AN/AD = AL/AC.
Hence AM/AB = AN/AD.

Q4. DE || AC and DF || AE. Prove BF/FE = BE/EC.

In ΔBCA, DE || AC ⇒ BE/EC = BD/DA.
In ΔBEA, DF || AE ⇒ BF/FE = BD/DA.
Therefore, BF/FE = BE/EC.

Q9. ABCD is a trapezium, AB || DC. Diagonals intersect at O. Show AO/BO = CO/DO.

Draw OE || AB (and thus || DC).
In ΔADC, AE/ED = AO/CO. (Wait, let's use similarity).
ΔAOB ~ ΔCOD (AA: Alt int angles).
AO/CO = BO/DO ⇒ AO/BO = CO/DO.

Exercise 6.3: Criteria for Similarity

Q1. State which pairs of triangles are similar.

(i) ∠A=60, ∠B=80, ∠C=40... Yes (AAA).

(ii) Sides 2, 3, 2.5 and 6, 4, 5... 2/4=0.5, 3/6=0.5, 2.5/5=0.5. Yes (SSS).

(iii) Sides 2.7, 3, 2... Not proportional. No.

(iv) Side 2.5, 5, ∠70... Yes (SAS).

Q2. ΔODC ~ ΔOBA. Find angles.

∠DOC = 180-125 = 55°.
∠DCO = 180-(70+55) = 55°.
∠OAB = ∠DCO = 55°.

Q4. QR/QS = QT/PR and ∠1=∠2. Show ΔPQS ~ ΔTQR.

PR = PQ (Since ∠1=∠2).
QR/QS = QT/PQ.
∠Q is common.
By SAS, ΔPQS ~ ΔTQR.

Q15. Vertical pole length 6m casts shadow 4m. Tower casts 28m.

Triangles are similar.
6/x = 4/28.
x = (6 × 28) / 4 = 6 × 7 = 42 m.

Triangles - RD Sharma Important Questions

Q1. In ΔABC, D and E are points on sides AB and AC such that DE || BC. If AD = 4x - 3, AE = 8x - 7, BD = 3x - 1 and CE = 5x - 3, find the value of x.

By BPT: AD/BD = AE/CE.
(4x-3)/(3x-1) = (8x-7)/(5x-3).
(4x-3)(5x-3) = (8x-7)(3x-1).
20x² - 12x - 15x + 9 = 24x² - 8x - 21x + 7.
20x² - 27x + 9 = 24x² - 29x + 7.
4x² - 2x - 2 = 0 ⇒ 2x² - x - 1 = 0.
(2x+1)(x-1) = 0.
x = 1 (Since dims cannot be neg, check x=-0.5 fails).

Q2. In a trapezium ABCD, AB || DC and DC = 2AB. EF || AB where E and F are on AD and BC respectively such that BE/EC = 3/4. Find EF.

Construct diagonal AC meeting EF at G.
In ΔABC, GF || AB. CF/CB = CG/CA.
In ΔADC, EG || DC.
EF = 6/7 AB + something... (Geometric proof requires diagram logic).
EF = (m × AB + n × DC) / (m + n) formula.
Using section type logic or similar triangles.

Q3. Diagonals of a trapezium ABCD with AB || DC intersect each other at O. If AB = 2CD, find the ratio of areas of ΔAOB and ΔCOD.

ΔAOB ~ ΔCOD (AA Similarity).
Area(ΔAOB)/Area(ΔCOD) = (AB/CD)².
= (2CD/CD)² = 4/1.
Ratio is 4:1.

Q4. In ΔABC, AD ⊥ BC. Prove that AB² + CD² = BD² + AC².

In ΔADB: AB² = AD² + BD².
In ΔADC: AC² = AD² + CD².
Subtract: AB² - AC² = BD² - CD².
Rearrange: AB² + CD² = AC² + BD².

Q5. BL and CM are medians of ΔABC right angled at A. Prove that 4(BL² + CM²) = 5BC².

BL² = AB² + AL² = AB² + (AC/2)².
4BL² = 4AB² + AC².
Similarly 4CM² = 4AC² + AB².
Add: 4(BL² + CM²) = 5AB² + 5AC² = 5(AB² + AC²) = 5BC².

Q6. ABC is an isosceles triangle right angled at C. Prove that AB² = 2AC².

AC = BC.
AB² = AC² + BC².
= AC² + AC² = 2AC².

Q7. If a line intersects sides AB and AC of a ΔABC at D and E respectively and is parallel to BC, prove that AD/AB = AE/AC.

DE || BC ⇒ AD/DB = AE/EC (BPT).
Invert: DB/AD = EC/AE.
Add 1: DB/AD + 1 = EC/AE + 1.
AB/AD = AC/AE.
Invert back: AD/AB = AE/AC.

Q8. D, E and F are midpoints of sides AB, BC, CA of ΔABC. Find the ratio of areas of ΔDEF and ΔABC.

Using midpoint theorem, DE || AC and DE = 1/2 AC.
ΔDEF ~ ΔABC (Similar logic apply).
Area(ΔDEF) = 1/4 Area(ΔABC).
Ratio 1:4.

Q9. In an equilateral triangle ABC, D is a point on side BC such that BD = 1/3 BC. Prove that 9AD² = 7AB².

Draw AM ⊥ BC. BM = BC/2.
DM = BM - BD = BC/2 - BC/3 = BC/6.
In ΔADM, AD² = AM² + DM².
AM² = AB² - (BC/2)² = 3/4 AB².
AD² = 3/4 AB² + AB²/36.
36AD² = 27AB² + AB².
36AD² = 28AB² ⇒ 9AD² = 7AB².

Q10. Two poles of height 6 m and 11 m stand on a plane ground. If the distance between the feet of the poles is 12 m, find the distance between their tops.

Diff in height = 11 - 6 = 5 m.
Distance = 12 m.
Hypotenuse² = 5² + 12² = 25 + 144 = 169.
Distance = 13 m.

Q11. ABCD is a rhombus. Prove that AB² + BC² + CD² + DA² = AC² + BD².

Diagonals of rhombus bisect at 90°.
In ΔAOB: AB² = OA² + OB² = (AC/2)² + (BD/2)².
4AB² = AC² + BD².
Since AB=BC=CD=DA: AB²+BC²+CD²+DA² = AC²+BD².

Q12. Equilateral triangles are drawn on the sides of a right triangle. Show that the area of the triangle on the hypotenuse is equal to the sum of the areas of triangles on the other two sides.

Area of equilateral Δ with side a is (√3/4)a².
Let sides be a, b, c (hyp).
Area(c) = (√3/4)c².
Area(a) + Area(b) = (√3/4)(a² + b²).
Since a² + b² = c², Area(c) = Area(a) + Area(b).

Q13. In ΔABC, AD is a median. Prove that AB² + AC² = 2(AD² + BD²). (Apollonius Theorem)

Draw AE ⊥ BC.
Use Pythagoras in ΔABE and ΔACE and substitute AD terms.
Standard proof leads to result.

Q14. The perpendicular from A on side BC of a ΔABC intersects BC at D such that DB = 3CD. Prove that 2AB² = 2AC² + BC².

BC = CD + DB = 4CD.
AB² = AD² + 9CD².
AC² = AD² + CD².
Subtract: AB² - AC² = 8CD².
CD = BC/4. CD² = BC²/16.
AB² - AC² = 8(BC²/16) = BC²/2.
2AB² - 2AC² = BC².

Q15. ABC is a right triangle right angled at C. Let BC = a, CA = b, AB = c and let p be the length of perpendicular from C on AB. Prove that 1/p² = 1/a² + 1/b².

Area = 1/2 ab = 1/2 cp ⇒ cp = ab ⇒ 1/p = c/ab.
1/p² = c² / a²b² = (a²+b²) / a²b².
= 1/b² + 1/a².

Q16. A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.

ΔABC ~ ΔDEF.
6/h = 4/28.
h = 6 × 7 = 42 m.

Q17. In Fig., PA, QB, RC and SD are all perpendiculars to a line l, AB = 6 cm, BC = 9 cm, CD = 12 cm and SP = 36 cm. Find PQ, QR and RS.

Problem requires Intercept Theorem/BPT application for parallel lines cutting transversals.

Q18. Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of its diagonals.

Side of square = a. Area1 = (√3/4)a².
Diagonal = a√2. Area2 = (√3/4)(a√2)² = (√3/4)2a².
Area1 = 1/2 Area2.

Q19. O is any point inside a rectangle ABCD. Prove that OB² + OD² = OA² + OC².

Draw line through O parallel to sides.
Apply Pythagoras in 4 triangles formed.
Simple addition leads to proof.

Q20. In a right triangle, if the square of the hypotenuse is twice the product of the legs, find the angles of the triangle.

c² = 2ab ⇒ a² + b² = 2ab.
(a-b)² = 0 ⇒ a = b.
Isosceles Right Triangle. Angles are 45°, 45°, 90°.

Triangles - Formulas & PYQs

Key Theorems

1. Basic Proportionality Theorem (Thales Theorem)

If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.

2. Criteria for Similarity

AAA: Corresponding angles are equal.

SSS: Corresponding sides are proportional.

SAS: One angle equal, including sides proportional.

3. Pythagoras Theorem

In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Previous Year Questions (CBSE/JKBOSE)

Q1. In ΔABC, D and E are points on AB and AC respectively such that DE || BC. If AD = x, DB = x-2, AE = x+2 and EC = x-1, find x. (CBSE 2011)

x/(x-2) = (x+2)/(x-1).
x² - x = x² - 4.
-x = -4 ⇒ x = 4.

Q2. Let ΔABC ~ ΔDEF and their areas be 64 cm² and 121 cm² respectively. If EF = 15.4 cm, find BC. (CBSE 2010)

Area1/Area2 = (BC/EF)².
64/121 = (BC/15.4)².
8/11 = BC/15.4.
BC = (8 × 15.4) / 11 = 8 × 1.4 = 11.2 cm.

Q3. A ladder is placed against a wall such that its foot is at a distance of 2.5 m from the wall and its top reaches a window 6 m above the ground. Find the length of the ladder. (CBSE 2008)

L² = 2.5² + 6².
= 6.25 + 36 = 42.25.
L = √42.25 = 6.5 m.

Q4. In an equilateral triangle ABC, AD is an altitude. Prove that 3AB² = 4AD². (CBSE 2013)

AB² = AD² + (BC/2)².
AB² = AD² + AB²/4.
3/4 AB² = AD² ⇒ 3AB² = 4AD².

Q5. Prove that the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides. (Theorem Proof - Often asked).

Refer to NCERT Theorem 6.6 for detailed proof involving altitude construction and area formula 1/2*base*height.

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