CBSE Class 10 Mathematics Triangles Assignment Set 09

Exercise 3: Criteria for Similarity of Triangles

Question 1. State which pairs of triangles in the given figures are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form.

Answer:
(i) ∠A = ∠P = 60°, ∠B = ∠Q = 80°, ∠C = ∠R = 40°. By AAA (AA) similarity, ΔABC ~ ΔPQR.
(ii) \( \frac{AB}{QR}=\frac{2}{4}=\frac{1}{2},\ \frac{BC}{RP}=\frac{2.5}{5}=\frac{1}{2},\ \frac{CA}{PQ}=\frac{3}{6}=\frac{1}{2} \). By SSS similarity, ΔABC ~ ΔQRP.
(iii) \( \frac{LM}{EF}=\frac{2.7}{5},\ \frac{LP}{DF}=\frac{3}{6}=\frac{1}{2},\ \frac{MP}{DE}=\frac{2}{4}=\frac{1}{2} \). The ratios are not all equal, so the triangles are not similar.
(iv) ∠M = ∠Q = 70° and \( \frac{MN}{QP}=\frac{2.5}{5}=\frac{1}{2},\ \frac{ML}{QR}=\frac{5}{10}=\frac{1}{2} \). By SAS similarity, ΔNML ~ ΔPQR.
(v) In ΔABC the angle 80° is at A, but the sides AB = 2.5 and BC = 3 do not include it, while in ΔDEF the angle 80° is the included angle of DF = 5 and EF = 6. The angle between the proportional sides is not the same, so the triangles are not similar.
(vi) In ΔDEF, ∠F = 180° − 70° − 80° = 30°. In ΔPQR, ∠P = 180° − 80° − 30° = 70°. So ∠D = ∠P = 70°, ∠E = ∠Q = 80° and ∠F = ∠R = 30°. By AA similarity, ΔDEF ~ ΔPQR.
In simple words: Match the corresponding vertices and state the similarity criterion.

Teacher's Note:
1. For SSS compare the ratios of the sides in the same order (smallest with smallest, and so on).
2. For SAS the angle must lie between the two proportional sides.

Question 2. In the figure, ΔODC ~ ΔOBA, ∠BOC = 125° and ∠CDO = 70°. Find ∠DOC, ∠DCO and ∠OAB.

Answer:
DOB is a straight line, so ∠DOC + ∠COB = 180° ⇒ ∠DOC = 180° − 125° = 55°.
In ΔDOC, ∠DCO = 180° − ∠CDO − ∠DOC = 180° − 70° − 55° = 55°.
Since ΔODC ~ ΔOBA, the corresponding angles are equal: ∠OAB = ∠OCD = 55°.
In simple words: Use the linear pair, the angle sum, and then the equal corresponding angles of similar triangles.

Teacher's Note:
1. Name the corresponding vertices in the same order when you write a similarity statement.
2. State the criterion (AA, SAS or SSS) every time you claim that two triangles are similar.

Question 3. Diagonals AC and BD of a trapezium ABCD with AB ∥ DC intersect each other at the point O. Using a similarity criterion for two triangles, show that \( \frac{OA}{OC}=\frac{OB}{OD} \).

Answer:
In ΔAOB and ΔCOD: ∠1 = ∠2 and ∠3 = ∠4 (alternate angles, as AB ∥ DC).
So ΔAOB ~ ΔCOD (AA). Corresponding sides of similar triangles are proportional, so \( \frac{OA}{OC}=\frac{OB}{OD} \). Hence proved.
In simple words: Match the corresponding vertices and state the similarity criterion.

Teacher's Note:
1. Name the corresponding vertices in the same order when you write a similarity statement.
2. State the criterion (AA, SAS or SSS) every time you claim that two triangles are similar.

Question 4. In the given figure, \( \frac{QR}{QS}=\frac{QT}{PR} \) and ∠1 = ∠2. Show that ΔPQS ~ ΔTQR.

Answer:
In ΔPQR, ∠PQR = ∠PRQ (∠1 = ∠2), so PQ = PR ...(i)
Given \( \frac{QR}{QS}=\frac{QT}{PR} \) and by (i) \( \frac{QR}{QS}=\frac{QT}{QP} \) ...(ii)
In ΔPQS and ΔTQR: \( \frac{QR}{QS}=\frac{QT}{QP} \) and ∠Q is common, so ΔPQS ~ ΔTQR (SAS).
In simple words: Match the corresponding vertices and state the similarity criterion.

Teacher's Note:
1. Name the corresponding vertices in the same order when you write a similarity statement.
2. State the criterion (AA, SAS or SSS) every time you claim that two triangles are similar.

Question 5. S and T are points on sides PR and QR of ΔPQR such that ∠P = ∠RTS. Show that ΔRPQ ~ ΔRTS.

Answer:
In ΔRPQ and ΔRTS: ∠RPQ = ∠RTS (given) and ∠PRQ = ∠TRS (common angle).
Hence ΔRPQ ~ ΔRTS (AA).
In simple words: Match the corresponding vertices and state the similarity criterion.

Teacher's Note:
1. Name the corresponding vertices in the same order when you write a similarity statement.
2. State the criterion (AA, SAS or SSS) every time you claim that two triangles are similar.

Question 6. In the figure, if ΔABE ≅ ΔACD, show that ΔADE ~ ΔABC.

Answer:
Since ΔABE ≅ ΔACD, AE = AD and AB = AC (CPCT).
So \( \frac{AD}{AB}=\frac{AE}{AC} \) and ∠DAE = ∠BAC (common angle). Hence ΔADE ~ ΔABC (SAS).
In simple words: Match the corresponding vertices and state the similarity criterion.

Teacher's Note:
1. Name the corresponding vertices in the same order when you write a similarity statement.
2. State the criterion (AA, SAS or SSS) every time you claim that two triangles are similar.

Question 7. In the figure, altitudes AD and CE of ΔABC intersect each other at the point P. Show that:
(i) ΔAEP ~ ΔCDP   (ii) ΔABD ~ ΔCBE   (iii) ΔAEP ~ ΔADB   (iv) ΔPDC ~ ΔBEC

Answer:
(i) In ΔAEP and ΔCDP: ∠AEP = ∠CDP = 90° and ∠APE = ∠CPD (vertically opposite). So ΔAEP ~ ΔCDP (AA).
(ii) In ΔABD and ΔCBE: ∠ADB = ∠CEB = 90° and ∠ABD = ∠CBE (common). So ΔABD ~ ΔCBE (AA).
(iii) In ΔAEP and ΔADB: ∠AEP = ∠ADB = 90° and ∠PAE = ∠DAB (common). So ΔAEP ~ ΔADB (AA).
(iv) In ΔPDC and ΔBEC: ∠PDC = ∠BEC = 90° and ∠PCD = ∠BCE (common). So ΔPDC ~ ΔBEC (AA).
In simple words: Match the corresponding vertices and state the similarity criterion.

Teacher's Note:
1. Name the corresponding vertices in the same order when you write a similarity statement.
2. State the criterion (AA, SAS or SSS) every time you claim that two triangles are similar.

Question 8. E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ΔABE ~ ΔCFB.

Answer:
In ΔABE and ΔCFB: ∠A = ∠C (opposite angles of a parallelogram) and ∠AEB = ∠CBF (alternate angles, as AE ∥ BC).
Hence ΔABE ~ ΔCFB (AA).
In simple words: Match the corresponding vertices and state the similarity criterion.

Teacher's Note:
1. Name the corresponding vertices in the same order when you write a similarity statement.
2. State the criterion (AA, SAS or SSS) every time you claim that two triangles are similar.

Question 9. In the figure, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that (i) ΔABC ~ ΔAMP (ii) \( \frac{CA}{PA}=\frac{BC}{MP} \).

Answer:
(i) In ΔABC and ΔAMP: ∠CAB = ∠MAP (common) and ∠ABC = ∠AMP = 90°. So ΔABC ~ ΔAMP (AA).
(ii) Corresponding sides of similar triangles are proportional, so \( \frac{CA}{PA}=\frac{BC}{MP} \).
In simple words: Match the corresponding vertices and state the similarity criterion.

Teacher's Note:
1. Name the corresponding vertices in the same order when you write a similarity statement.
2. State the criterion (AA, SAS or SSS) every time you claim that two triangles are similar.

Question 10. CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of ΔABC and ΔEFG respectively. If ΔABC ~ ΔFEG, show that (i) \( \frac{CD}{GH}=\frac{AC}{FG} \) (ii) ΔDCB ~ ΔHGE (iii) ΔDCA ~ ΔHGF.

Answer:
Since ΔABC ~ ΔFEG, ∠A = ∠F, ∠B = ∠E and ∠ACB = ∠FGE. Halving the last equality: ∠ACD = ∠DCB = ∠FGH = ∠HGE.
(ii) In ΔDCB and ΔHGE: ∠DCB = ∠HGE and ∠B = ∠E, so ΔDCB ~ ΔHGE (AA).
(iii) In ΔDCA and ΔHGF: ∠ACD = ∠FGH and ∠A = ∠F, so ΔDCA ~ ΔHGF (AA).
(i) From (iii), corresponding sides are proportional: \( \frac{CD}{GH}=\frac{AC}{FG} \).
In simple words: Match the corresponding vertices and state the similarity criterion.

Teacher's Note:
1. Name the corresponding vertices in the same order when you write a similarity statement.
2. State the criterion (AA, SAS or SSS) every time you claim that two triangles are similar.

Question 11. In the figure, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD ⊥ BC and EF ⊥ AC, prove that ΔABD ~ ΔECF.

Answer:
Since AB = AC, ∠ABC = ∠ACB, that is ∠ABD = ∠ECF.
In ΔABD and ΔECF: ∠ADB = ∠EFC = 90° and ∠ABD = ∠ECF. Hence ΔABD ~ ΔECF (AA).
In simple words: Match the corresponding vertices and state the similarity criterion.

Teacher's Note:
1. Name the corresponding vertices in the same order when you write a similarity statement.
2. State the criterion (AA, SAS or SSS) every time you claim that two triangles are similar.

Question 12. Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of ΔPQR. Show that ΔABC ~ ΔPQR.

Answer:
Given \( \frac{AB}{PQ}=\frac{BC}{QR}=\frac{AD}{PM} \). Since AD and PM are medians, BD = \( \frac{1}{2}BC \) and QM = \( \frac{1}{2}QR \), so \( \frac{AB}{PQ}=\frac{BD}{QM}=\frac{AD}{PM} \).
By SSS, ΔABD ~ ΔPQM, so ∠ABD = ∠PQM, that is ∠ABC = ∠PQR.
In ΔABC and ΔPQR: \( \frac{AB}{PQ}=\frac{BC}{QR} \) and ∠ABC = ∠PQR. Hence ΔABC ~ ΔPQR (SAS).
In simple words: Match the corresponding vertices and state the similarity criterion.

Teacher's Note:
1. Name the corresponding vertices in the same order when you write a similarity statement.
2. State the criterion (AA, SAS or SSS) every time you claim that two triangles are similar.

Question 13. D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC. Show that CA² = CB · CD.

Answer:
In ΔADC and ΔBAC: ∠ADC = ∠BAC (given) and ∠ACD = ∠BCA (common). So ΔADC ~ ΔBAC (AA).
Hence \( \frac{CA}{CB}=\frac{CD}{CA} \) ⇒ CA² = CB · CD.
In simple words: Match the corresponding vertices and state the similarity criterion.

Teacher's Note:
1. Name the corresponding vertices in the same order when you write a similarity statement.
2. State the criterion (AA, SAS or SSS) every time you claim that two triangles are similar.

Question 14. Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that ΔABC ~ ΔPQR.

Answer:
Given \( \frac{AB}{PQ}=\frac{AC}{PR}=\frac{AD}{PM} \). Produce AD to E so that DE = AD and produce PM to N so that MN = PM. Join BE, CE, QN and RN.
In ΔABD and ΔCED: BD = DC, AD = DE and ∠ADB = ∠CDE (vertically opposite), so ΔABD ≅ ΔCED and AB = CE. Similarly, in the second figure, PQ = RN.
Then \( \frac{CE}{RN}=\frac{AC}{PR}=\frac{2AD}{2PM}=\frac{AE}{PN} \). By SSS, ΔACE ~ ΔPRN, so ∠CAE = ∠RPN, that is ∠CAD = ∠RPM. Similarly ∠BAD = ∠QPM.
Adding, ∠A = ∠P. With \( \frac{AB}{PQ}=\frac{AC}{PR} \) and ∠A = ∠P, ΔABC ~ ΔPQR (SAS).
In simple words: Match the corresponding vertices and state the similarity criterion.

Teacher's Note:
1. Name the corresponding vertices in the same order when you write a similarity statement.
2. State the criterion (AA, SAS or SSS) every time you claim that two triangles are similar.

Question 15. 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.

Answer:
At the same time the sun's angle is the same, so the two right triangles formed by the objects and their shadows are similar (AA).
Let the height of the tower be h metres: \( \frac{6}{h}=\frac{4}{28} \) ⇒ h = (6 × 28) ÷ 4 = 42 m.
In simple words: Objects and their shadows at the same time form similar right triangles.

Teacher's Note:
1. Name the corresponding vertices in the same order when you write a similarity statement.
2. State the criterion (AA, SAS or SSS) every time you claim that two triangles are similar.

Question 16. If AD and PM are medians of triangles ABC and PQR respectively, where ΔABC ~ ΔPQR, prove that \( \frac{AB}{PQ}=\frac{AD}{PM} \).

Answer:
Since ΔABC ~ ΔPQR, \( \frac{AB}{PQ}=\frac{BC}{QR}=\frac{AC}{PR} \) ...(i) and ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R ...(ii).
AD and PM are medians, so BD = \( \frac{BC}{2} \) and QM = \( \frac{QR}{2} \) ...(iii). From (i) and (iii), \( \frac{AB}{PQ}=\frac{BD}{QM} \) ...(iv).
In ΔABD and ΔPQM: ∠B = ∠Q and \( \frac{AB}{PQ}=\frac{BD}{QM} \), so ΔABD ~ ΔPQM (SAS). Hence \( \frac{AB}{PQ}=\frac{BD}{QM}=\frac{AD}{PM} \). Proved.
In simple words: Match the corresponding vertices and state the similarity criterion.

Teacher's Note:
1. Name the corresponding vertices in the same order when you write a similarity statement.
2. State the criterion (AA, SAS or SSS) every time you claim that two triangles are similar.