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NCERT Solutions · Class 10 · Chapter 6: Triangles

NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.3

Exercise 6.3: Criteria for similarity of triangles. AAA (or AA): two angles equal. SSS: all three sides in the same ratio. SAS: one angle equal and the sides including it in the same ratio. Always write the vertices in corresponding order.

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Exercise 6.3 questions and solutions

Exercise 6.3, Question 1

State which pairs of triangles in the textbook figure are similar, the criterion used, and the correspondence.
(i) \(\triangle ABC\): \(\angle A = 60^\circ, \angle B = 80^\circ, \angle C = 40^\circ\); \(\triangle PQR\): \(\angle P = 60^\circ, \angle Q = 80^\circ, \angle R = 40^\circ\)
Show solution
  1. All three angles match in order.
Answer: \(\triangle ABC \sim \triangle PQR\) (AAA)
(ii) \(\triangle ABC\): \(AB = 2, BC = 2.5, CA = 3\); \(\triangle PQR\): \(PQ = 6, QR = 4, RP = 5\)
Show solution
  1. Match smallest with smallest: \[\begin{gathered}\dfrac{AB}{QR} = \dfrac24, \\ \dfrac{BC}{RP} = \dfrac{2.5}{5}, \\ \dfrac{CA}{PQ} = \dfrac36\end{gathered}\], all \(\tfrac12\).
Answer: \(\triangle ABC \sim \triangle QRP\) (SSS)
(iii) \(\triangle LMP\): \(LM = 2.7, MP = 2, LP = 3\); \(\triangle DEF\): \(DE = 4, EF = 5, DF = 6\)
Show solution
  1. \(\dfrac{MP}{DE} = \dfrac12\) and \(\dfrac{LP}{DF} = \dfrac12\), but \(\dfrac{LM}{EF} = \dfrac{2.7}{5} = 0.54\).
Answer: Not similar
(iv) \(\triangle MNL\): \(\angle M = 70^\circ,\ MN = 2.5,\ ML = 5\); \(\triangle QPR\): \(\angle Q = 70^\circ,\ QP = 5,\ QR = 10\)
Show solution
  1. The equal angles are included between the given sides.
  2. \[\begin{aligned}\dfrac{MN}{QP} &= \dfrac{2.5}{5} \\ &= \dfrac12\end{aligned}\] and \[\begin{aligned}\dfrac{ML}{QR} &= \dfrac{5}{10} \\ &= \dfrac12\end{aligned}\]
Answer: \(\triangle MNL \sim \triangle QPR\) (SAS)
(v) \(\triangle ABC\): \(\angle A = 80^\circ,\ AB = 2.5,\ BC = 3\); \(\triangle DEF\): \(\angle F = 80^\circ,\ DF = 5,\ EF = 6\)
Show solution
  1. The \(80^\circ\) angle in \(\triangle ABC\) is not between AB and BC, so SAS cannot be used, and no other criterion applies.
Answer: Not similar (no criterion applies)
(vi) \(\triangle DEF\): \(\angle D = 70^\circ, \angle E = 80^\circ\); \(\triangle PQR\): \(\angle Q = 80^\circ, \angle R = 30^\circ\)
Show solution
  1. \[\begin{aligned}\angle F &= 180^\circ - 150^\circ \\ &= 30^\circ\end{aligned}\] and \[\begin{aligned}\angle P &= 180^\circ - 110^\circ \\ &= 70^\circ\end{aligned}\]
  2. So \[\begin{gathered}\angle D = \angle P, \\ \angle E = \angle Q, \\ \angle F = \angle R\end{gathered}\]
Answer: \(\triangle DEF \sim \triangle PQR\) (AAA)

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Exercise 6.3, Question 2

\(\triangle ODC \sim \triangle OBA\) (O is where AC and BD cross), \(\angle BOC = 125^\circ\) and \(\angle CDO = 70^\circ\). Find \(\angle DOC\), \(\angle DCO\) and \(\angle OAB\).
Show solution
  1. DOB is a straight line: \[\begin{aligned}\angle DOC &= 180^\circ - 125^\circ \\ &= 55^\circ\end{aligned}\]
  2. In \(\triangle ODC\): \[\begin{aligned}\angle DCO &= 180^\circ - 70^\circ - 55^\circ \\ &= 55^\circ\end{aligned}\]
  3. Corresponding angles of similar triangles: \(\angle OAB = \angle OCD = 55^\circ\).
Answer: \(55^\circ,\ 55^\circ,\ 55^\circ\)

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Exercise 6.3, Question 3

The diagonals AC and BD of trapezium ABCD (\(AB \parallel DC\)) meet at O. Using a similarity criterion, show \(\dfrac{OA}{OC} = \dfrac{OB}{OD}\).
Show solution
  1. In \(\triangle OAB\) and \(\triangle OCD\): \(\angle OAB = \angle OCD\) and \(\angle OBA = \angle ODC\) (alternate angles, \(AB \parallel DC\)).
  2. So \(\triangle OAB \sim \triangle OCD\) (AA).
  3. Corresponding sides: \(\dfrac{OA}{OC} = \dfrac{OB}{OD}\).
Answer: Proved.

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Exercise 6.3, Question 4

In \(\triangle PQR\), T is on QR and S is on QP produced so that \(\dfrac{QR}{QS} = \dfrac{QT}{PR}\) and \(\angle 1 = \angle 2\) (\(\angle PQR = \angle PRQ\)). Show \(\triangle PQS \sim \triangle TQR\).
Show solution
  1. \(\angle PQR = \angle PRQ\), so \(PQ = PR\).
  2. Then \(\dfrac{QR}{QS} = \dfrac{QT}{PQ}\), that is \(\dfrac{QS}{QR} = \dfrac{QP}{QT}\).
  3. The angle at Q is common to \(\triangle PQS\) and \(\triangle TQR\), and the sides including it are proportional.
  4. So \(\triangle PQS \sim \triangle TQR\) (SAS).
Answer: Proved (SAS).

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Exercise 6.3, Question 5

S and T are points on PR and QR of \(\triangle PQR\) with \(\angle P = \angle RTS\). Show \(\triangle RPQ \sim \triangle RTS\).
Show solution
  1. \(\angle RPQ = \angle RTS\) (given) and \(\angle R\) is common.
  2. Two angles equal, so \(\triangle RPQ \sim \triangle RTS\) (AA).
Answer: Proved (AA).

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Exercise 6.3, Question 6

If \(\triangle ABE \cong \triangle ACD\) (D on AB, E on AC), show \(\triangle ADE \sim \triangle ABC\).
Show solution
  1. Congruence gives \(AB = AC\) and \(AE = AD\).
  2. So \(\dfrac{AD}{AB} = \dfrac{AE}{AC}\), and \(\angle A\) is common.
  3. \(\triangle ADE \sim \triangle ABC\) (SAS).
Answer: Proved (SAS).

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Exercise 6.3, Question 7

In \(\triangle ABC\), altitudes AD and CE meet at P. Show that:
(i) \(\triangle AEP \sim \triangle CDP\)
Show solution
  1. \(\angle AEP = \angle CDP = 90^\circ\) and \(\angle APE = \angle CPD\) (vertically opposite).
  2. AA similarity.
Answer: Proved (AA).
(ii) \(\triangle ABD \sim \triangle CBE\)
Show solution
  1. \(\angle ADB = \angle CEB = 90^\circ\) and \(\angle B\) is common: AA.
Answer: Proved (AA).
(iii) \(\triangle AEP \sim \triangle ADB\)
Show solution
  1. \(\angle AEP = \angle ADB = 90^\circ\) and \(\angle PAE = \angle BAD\) (the same angle): AA.
Answer: Proved (AA).
(iv) \(\triangle PDC \sim \triangle BEC\)
Show solution
  1. \(\angle PDC = \angle BEC = 90^\circ\) and \(\angle PCD = \angle BCE\) (the same angle): AA.
Answer: Proved (AA).

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Exercise 6.3, Question 8

E is a point on side AD produced of parallelogram ABCD, and BE meets CD at F. Show \(\triangle ABE \sim \triangle CFB\).
Show solution
  1. \(\angle A = \angle C\) (opposite angles of a parallelogram).
  2. \(\angle AEB = \angle CBF\) (alternate angles, \(AE \parallel BC\)).
  3. So \(\triangle ABE \sim \triangle CFB\) (AA).
Answer: Proved (AA).

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Exercise 6.3, Question 9

\(\triangle ABC\) and \(\triangle AMP\) are right-angled at B and M, and share the angle at A. Prove that:
(i) \(\triangle ABC \sim \triangle AMP\)
Show solution
  1. \(\angle ABC = \angle AMP = 90^\circ\) and \(\angle A\) is common: AA.
Answer: Proved (AA).
(ii) \(\dfrac{CA}{PA} = \dfrac{BC}{MP}\)
Show solution
  1. From (i), corresponding sides are proportional: \(\dfrac{CA}{PA} = \dfrac{BC}{MP}\).
Answer: Proved.

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Exercise 6.3, Question 10

CD and GH bisect \(\angle ACB\) and \(\angle EGF\), with D on AB and H on FE, and \(\triangle ABC \sim \triangle FEG\). Show that:
(i) \(\dfrac{CD}{GH} = \dfrac{AC}{FG}\)
Show solution
  1. \(\angle A = \angle F\) and \(\angle C = \angle G\), so \(\tfrac12\angle C = \tfrac12\angle G\): \(\angle ACD = \angle FGH\).
  2. \(\triangle ACD \sim \triangle FGH\) (AA), so \(\dfrac{CD}{GH} = \dfrac{AC}{FG}\).
Answer: Proved.
(ii) \(\triangle DCB \sim \triangle HGE\)
Show solution
  1. \(\angle B = \angle E\) and \[\begin{aligned}\angle DCB &= \tfrac12\angle C \\ &= \tfrac12\angle G \\ &= \angle HGE\end{aligned}\]: AA.
Answer: Proved (AA).
(iii) \(\triangle DCA \sim \triangle HGF\)
Show solution
  1. \(\angle A = \angle F\) and \(\angle DCA = \angle HGF\) (halves of equal angles): AA.
Answer: Proved (AA).

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Exercise 6.3, Question 11

E is on side CB produced of isosceles \(\triangle ABC\) (\(AB = AC\)). \(AD \perp BC\) and \(EF \perp AC\). Prove \(\triangle ABD \sim \triangle ECF\).
Show solution
  1. \(AB = AC\) gives \(\angle ABD = \angle ACB = \angle ECF\).
  2. \(\angle ADB = \angle EFC = 90^\circ\).
  3. So \(\triangle ABD \sim \triangle ECF\) (AA).
Answer: Proved (AA).

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Exercise 6.3, Question 12

Sides AB, BC and median AD of \(\triangle ABC\) are proportional to sides PQ, QR and median PM of \(\triangle PQR\). Show \(\triangle ABC \sim \triangle PQR\).
Show solution
  1. \[\begin{aligned}\dfrac{AB}{PQ} &= \dfrac{BC}{QR} \\ &= \dfrac{AD}{PM}\end{aligned}\]; since BD \(= \tfrac12\)BC and QM \(= \tfrac12\)QR, also \(\dfrac{BD}{QM} = \dfrac{BC}{QR}\).
  2. So \(\triangle ABD \sim \triangle PQM\) (SSS), giving \(\angle B = \angle Q\).
  3. Then \(\dfrac{AB}{PQ} = \dfrac{BC}{QR}\) with \(\angle B = \angle Q\): \(\triangle ABC \sim \triangle PQR\) (SAS).
Answer: Proved.

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Exercise 6.3, Question 13

D is a point on side BC of \(\triangle ABC\) with \(\angle ADC = \angle BAC\). Show \(CA^2 = CB \cdot CD\).
Show solution
  1. In \(\triangle ADC\) and \(\triangle BAC\): \(\angle ADC = \angle BAC\) and \(\angle C\) is common.
  2. So \(\triangle ADC \sim \triangle BAC\) (AA), giving \(\dfrac{CA}{CB} = \dfrac{CD}{CA}\).
  3. Cross-multiply: \(CA^2 = CB \cdot CD\).
Answer: Proved.

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Exercise 6.3, Question 14

Sides AB, AC and median AD of \(\triangle ABC\) are proportional to sides PQ, PR and median PM of \(\triangle PQR\). Show \(\triangle ABC \sim \triangle PQR\).
Show solution
  1. Extend AD to E with \(DE = AD\), and PM to L with \(ML = PM\). ABEC and PQLR are parallelograms, so \(BE = AC\) and \(QL = PR\).
  2. Then \[\begin{aligned}\dfrac{AB}{PQ} &= \dfrac{BE}{QL} \\ &= \dfrac{AE}{PL}\end{aligned}\]: \(\triangle ABE \sim \triangle PQL\) (SSS), so \(\angle BAE = \angle QPL\).
  3. Similarly \(\triangle AEC \sim \triangle PLR\), so \(\angle CAE = \angle RPL\). Adding, \(\angle BAC = \angle QPR\).
  4. With \(\dfrac{AB}{PQ} = \dfrac{AC}{PR}\): \(\triangle ABC \sim \triangle PQR\) (SAS).
Answer: Proved.

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Exercise 6.3, Question 15

A 6 m pole casts a 4 m shadow when a tower casts a 28 m shadow. Find the tower's height.
Show solution
  1. Sun's rays make the same angle, and both stand vertically: the two right triangles are similar (AA).
  2. \[\begin{aligned}&\dfrac{h}{28} = \dfrac{6}{4} \\ \Rightarrow\ &h = 42\end{aligned}\]
Answer: \(42\) m

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Exercise 6.3, Question 16

AD and PM are medians of \(\triangle ABC \sim \triangle PQR\). Prove \(\dfrac{AB}{PQ} = \dfrac{AD}{PM}\).
Show solution
  1. Similarity gives \(\angle B = \angle Q\) and \[\begin{aligned}\dfrac{AB}{PQ} &= \dfrac{BC}{QR} \\ &= \dfrac{2BD}{2QM} \\ &= \dfrac{BD}{QM}\end{aligned}\]
  2. So \(\triangle ABD \sim \triangle PQM\) (SAS).
  3. Hence \(\dfrac{AB}{PQ} = \dfrac{AD}{PM}\).
Answer: Proved.

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Triangles in our sample papers: Sample paper 1 (questions 8, 23, 28, 33) · Sample paper 2 (questions 9, 23, 29, 33) · Sample paper 3 (questions 10, 12, 19, 28, 33) · Sample paper 4 (questions 9, 29, 33) · Sample paper 5 (questions 9, 10, 28, 33).

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Textbook: NCERT Mathematics Class 10 (rationalised edition, 2023-24 reprint onward), free from ncert.nic.in. Question statements are shortened to the minimum needed; the solutions and tips are our own. CBSE Math Revision is independent and not affiliated with NCERT or CBSE. Spotted a slip? Tell us and it goes in the corrections log.