Class 10 Mathematics: Triangles important questions

Triangles is a chapter of Class 10 Mathematics for the Andhra Pradesh State Board. These 15 questions are marked Very Important, Important or Practice, and each answer has been checked by our team. Use them to revise, not as a prediction of what will come in the exam.

1. Important · 4 marks

   1. ABCD is a trapezium with AB ∥ DC, and its diagonals intersect at O. Show that AO/BO = CO/DO.

   In ΔAOB and ΔCOD: ∠OAB = ∠OCD (alternate angles, AB ∥ DC) ∠OBA = ∠ODC (alternate angles) So ΔAOB ~ ΔCOD (AA similarity). Therefore AO/CO = BO/DO, which rearranges to AO/BO = CO/DO.

2. Practice · 1 marks

   2. Statement I: All similar triangles are congruent. Statement II: All right-angled isosceles triangles are similar. A) Both true B) I true, II false C) I false, II true D) Both false

   C) Statement I is false: similar triangles have the same shape but can differ in size. Statement II is true: every right-angled isosceles triangle has angles 90°, 45°, 45°, so they are similar by AA.

3. Important · 2 marks

   3. Give an example of i) similar figures ii) non-similar figures

   i) Any two circles, any two squares, or any two equilateral triangles. ii) A square and a rectangle that is not a square, or a circle and a triangle.

4. Practice · 1 marks

   4. In ΔABC, D is on AB and E is on AC with DE ∥ BC. AD = 2 cm, BD = 4 cm and AE = 5 cm. Find EC. A) 7 B) 10 C) 14 D) 20

   B) 10 cm By the Basic Proportionality Theorem, AD/DB = AE/EC, so 2/4 = 5/EC and EC = 10 cm.

5. Important · 2 marks

   5. State the SAS criterion for similarity of triangles.

   If one angle of a triangle is equal to one angle of another triangle and the sides including these angles are proportional, then the two triangles are similar.

6. Very Important · 8 marks

   6. Sides AB and AC and median AD of ΔABC are respectively proportional to sides PQ and PR and median PM of ΔPQR. Show that ΔABC ~ ΔPQR.

   Given: AB/PQ = AC/PR = AD/PM. Construction: Extend AD to E so that DE = AD, and extend PM to N so that MN = PM. Join BE, CE, QN, RN. ABEC is a parallelogram (diagonals AE and BC bisect each other), so BE = AC. Similarly QN = PR. So AB/PQ = BE/QN = AE/PN (AE = 2AD, PN = 2PM), and ΔABE ~ ΔPQN (SSS similarity). Hence ∠BAE = ∠QPN … (1) In the same way ΔACE ~ ΔPRN, so ∠CAE = ∠RPN … (2) Adding (1) and (2): ∠BAC = ∠QPR. Now AB/PQ = AC/PR and the included angles are equal, so ΔABC ~ ΔPQR (SAS similarity).

7. Practice · 1 marks

   7. If ΔABC ~ ΔXYZ, which of the following may be true? A) ∠A = ∠Z B) AB = XY C) AC/BC ≠ XZ/YZ D) ∠B = ∠Z

   B) AB = XY Similar triangles have ∠A = ∠X, ∠B = ∠Y, ∠C = ∠Z and proportional sides (so AC/BC = XZ/YZ). Corresponding sides can be equal when the scale factor is 1, i.e. when the triangles are congruent.

8. Important · 2 marks

   8. State the Basic Proportionality Theorem (Thales' theorem).

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

9. Very Important · 8 marks

   9. In ΔABC, D is on AB, and E and F are on BC (F between B and E), with DE ∥ AC and DF ∥ AE. Prove that BF/FE = BE/EC.

   In ΔABC, DE ∥ AC, so by the Basic Proportionality Theorem, BD/DA = BE/EC … (1) In ΔABE, DF ∥ AE, so BD/DA = BF/FE … (2) From (1) and (2): BF/FE = BE/EC. Hence proved.

10. Practice · 1 marks

    10. In triangles PQR and TSM, ∠P = 55°, ∠Q = 25°, ∠M = 100° and ∠S = 25°. Which statement is correct? A) ΔPQR ~ ΔTSM B) ΔQPR ~ ΔSMT C) ΔPRQ ~ ΔTSM D) ΔQPR ~ ΔTSM

    A) ΔPQR ~ ΔTSM In ΔPQR, ∠R = 180° − 55° − 25° = 100°. In ΔTSM, ∠T = 180° − 25° − 100° = 55°. So ∠P = ∠T, ∠Q = ∠S and ∠R = ∠M, and by AA similarity ΔPQR ~ ΔTSM (vertices in matching order).

11. Important · 2 marks

    11. State the AA criterion for similarity of two triangles.

    If two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar.

12. Very Important · 8 marks

    12. State and prove the Basic Proportionality (Thales) Theorem.

    Statement: If a line is drawn parallel to one side of a triangle to meet the other two sides at distinct points, the other two sides are divided in the same ratio. Given: In ΔABC, DE ∥ BC, with D on AB and E on AC. To prove: AD/DB = AE/EC. Construction: Join BE and CD. Draw EN ⟂ AB and DM ⟂ AC. Proof: ar(ADE) = ½ × AD × EN and ar(BDE) = ½ × DB × EN, so ar(ADE)/ar(BDE) = AD/DB … (1) ar(ADE) = ½ × AE × DM and ar(DEC) = ½ × EC × DM, so ar(ADE)/ar(DEC) = AE/EC … (2) ΔBDE and ΔDEC are on the same base DE and between the same parallels DE and BC, so ar(BDE) = ar(DEC) … (3) From (1), (2) and (3): AD/DB = AE/EC. Hence proved.

13. Practice · 1 marks

    13. In ΔABC, D is on AB and E is on AC with DE ∥ BC. AD = 3 cm, BD = 4 cm and BC = 14 cm. Find DE. A) 7 cm B) 6 cm C) 4 cm D) 3 cm

    B) 6 cm Since DE ∥ BC, ΔADE ~ ΔABC, so DE/BC = AD/AB = 3/(3 + 4) = 3/7. DE = (3/7) × 14 = 6 cm.

14. Important · 2 marks

    14. State the converse of the Basic Proportionality Theorem.

    If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

15. Very Important · 8 marks

    15. DC ∥ AB, and the diagonals AC and BD meet at O, with ∠CDO = 70° and ∠BOC = 125°. i) Name the pair of similar triangles. ii) Which similarity criterion proves it? State the criterion. iii) Write the proportionality relation from the similar triangles.

    i) ΔODC ~ ΔOBA. ii) AA criterion: ∠ODC = ∠OBA (alternate angles, since DC ∥ AB) and ∠DOC = ∠BOA (vertically opposite angles). If two angles of one triangle are equal to two angles of another, the triangles are similar. iii) OD/OB = OC/OA = DC/BA. (Also: ∠DOC = 180° − 125° = 55°, and ∠DCO = 180° − 70° − 55° = 55°.)