Class 10 - Mathematics
Triangles - Exercise 6.3
Top Block 1
Exercise 6.3
Question : 1:State which pairs of triangles in Fig. 6.34 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°
Therefore, Δ PQS ~ Δ TQR [AAA similarity]
(ii) AB/QR = BC/RP = CA/PQ
Therefore, Δ ABC ~ Δ QRP [SSS similarity]
(iii) Triangles are not similar because the corresponding sides are not equal.
(iv) Triangles are not similar because the corresponding sides are not equal.
(v) Triangles are not similar because the corresponding sides are not equal.
(vi) In Δ DEF,
∠D + ∠E + ∠F = 180°
⇒ 70° + 80° + ∠F = 180°
⇒ 150° + ∠F = 180°
⇒ ∠F = 180°– 150°
⇒ ∠F = 30°
Similarly, in Δ PQR,
∠D + ∠E + ∠F = 180°
⇒ ∠P + 80° + 30° = 180°
⇒ ∠P + 110° = 180°
⇒ ∠P = 180°– 110°
⇒ ∠P = 70°
Now, in Δ DEF and Δ PQR,
∠D = ∠P [Each 70°]
∠E = ∠Q [Each 70°]
∠F = ∠R [Each 70°]
So, Δ DEF ~ Δ PQR [AAA similarity]
Question : 2: In Fig. 6.35, Δ ODC ~ Δ OBA, ∠ BOC = 125°
and ∠ CDO = 70°. Find ∠ DOC, ∠ DCO and ∠ OAB.
Mddle block 1
Answer :
Since DOB is a straight line
So, ∠ DOC + ∠ COB = 180°
⇒ ∠ DOC + 125° = 180°
⇒ ∠ DOC = 180°– 125°
⇒ ∠ DOC = 55°
In Δ DOC,
∠ DCO + ∠ DCO + ∠ DOC = 180°
⇒ ∠ DCO + 70° + 55° = 180°
⇒ ∠ DCO + 125° = 180°
⇒ ∠ DCO = 180°– 125°
⇒ ∠ DCO = 55°
Given that, Δ ODC ~ Δ OBA
So, ∠ OAB = ∠ DCO [Corresponding angles of similar triangles]
⇒ ∠ OAB = 55°
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 OA/OC = OB/OD.
Answer :
∠ DOC = ∠ BOA [Vertically opposite angles]
∠ CDO = ∠ ABO [Alternate angles]
∠ DCO = ∠ BAO [Alternate angles]
So, Δ DOC ~ Δ BOA [AA similarity]
⇒ DO/BO = CO/AO
⇒ BO/DO = AO/CO
Question : 4:In Fig. 6.36, QR/QS = QT/PR and ∠ 1 = ∠ 2. Show that Δ PQS ~ Δ TQR.
Answer :
In Δ PQR, ∠ PQR = ∠ PRQ
⇒ PQ = PR …………..1
Given that QR/QS = QT/PR
⇒ QR/QS = QT/PQ ………2 [From equation 2]
In Δ PQS and Δ TQR,
QR/QS = QT/PQ [From equation 2]
∠Q = ∠Q [Common]
So, Δ PQS ~ Δ TQR [SAS similarity]
Question : 5: S and T are points on sides PR and QR of Δ PQR such that ∠ P = ∠ RTS.
Show that: Δ RPQ ~ Δ RTS.
Answer :
Given, S and T are points on sides PR and QR of Δ PQR such that ∠ P = ∠ RTS
Now, in Δ RPQ ~ Δ RTS
∠ RTS= ∠ QPS [Given]
∠ R = ∠ R [Common]
Hence, Δ RPQ ~ Δ RTS [By AA similarity]
Question : 6: In Fig. 6.37, if Δ ABE ≅Δ ACD, show that Δ ADE ~ Δ ABC.
Answer :
Given, Δ ABE ≅Δ ACD
So, AB = AC ………1 [By CPCT]
and AD = AE ………2 [By CPCT]
Now in Δ ADE and Δ ABC,
AD/AB = AE/AC [From equation 1 and 2]
∠ A = ∠ A [Common]
Hence, Δ ADE ~ Δ ABC [By SAS similarity]
Question : 7:In Fig. 6.38, altitudes AD and CE of Δ ABC intersect each other at the point P.
(i) Δ AEP ~ Δ CDP
(ii) Δ ABD ~ Δ CBE
(iii) Δ AEP ~ Δ ADB
(iv) Δ PDC ~ Δ BEC
Answer :
∠ APE = ∠ CPD [Vertically opposite angles]
∠ AEP = ∠ CDP [Common]
So, Δ AEP ~ Δ CDP [AA similarity]
(ii) Δ ABD ~ Δ CBE
∠ ADB = ∠ CEB [Each 90°]
∠ ABD = ∠ CBE [Common]
So, Δ ABD ~ Δ CBE [AA similarity]
∠ AEP = ∠ ADB [Each 90°]
∠ PAE = ∠ DAP [Common]
So, Δ AEP ~ Δ ADB [AA similarity]
(iv) Δ PDC ~ Δ BEC
∠ PDC = ∠ BEC [Each 90°]
∠ PCD = ∠ BCE [Common]
So, Δ PDC ~ Δ BEC [AA similarity]
Question : 8:E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that Δ ABE ~ Δ CFB.
Answer :
∠ A = ∠ C [Opposite angles of parallelogram]
∠ AEB = ∠ CBF [Alternate angles as AE || BC]
So, Δ ABE ~ Δ CFB [AA similarity]
Question : 9:
In Fig. 6.39, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that:
(i) Δ ABC ~ Δ AMP
(ii) CA/PA = BC/MP
Answer :
(i) In Δ ABC and Δ AMP,
∠ ABC = ∠ AMP [Each 90°]
∠ A = ∠ A [Common]
So, Δ ABC ~ Δ AMP [AA similarity]
(ii) Since Δ ABC ~ Δ AMP [Proved]
So, the corresponding sides of the similar triangles are proportional
⇒ AB/AM = BC/PM = AC/AP
⇒ CA/PA = BC/MP
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 Δ ABC and Δ EFG respectively.
If Δ ABC ~ Δ FEG, show that:
(i) CD/GH = AC/FG
(ii) Δ DCB ~ Δ HGE
(iii) Δ DCA ~ Δ HGF
Answer :
So, ∠ A = ∠ F, ∠ B = ∠ E and ∠ ACB = ∠ FGE
⇒ ∠ ACD = ∠ FGH [CD and GH are the bisectors of equal angles]
and ∠ DCB = ∠ HGE [CD and GH are the bisectors of equal angles]
In Δ ACD and Δ FGH,
∠ A = ∠ F and ∠ ACD = ∠ FGH [Proved]
So, Δ ACD ~ Δ FGH [AA similarity]
⇒ CD/GH = AC/FG
(ii) In Δ DCB and Δ HGE,
∠ B = ∠ E [Proved]
∠ DCB = ∠ HGE [Proved]
So, Δ DCB ~ Δ HGE [AA similarity]
(iii) In Δ DCA and Δ HGF,
∠ A = ∠ F [Proved]
∠ ACD = ∠ HGF [Proved]
So, Δ DCA ~ Δ HGF [AA similarity]
Question : 11:In Fig. 6.40, E is a point on side CB produced of an isosceles triangle ABC with AB = AC.
If AD ⊥ BC and EF ⊥ AC, prove that Δ ABD ~ Δ ECF.
Answer :
Given, ABC is an isosceles triangle.
So, AB = BC
⇒ ∠ ABD = ∠ ECF
In Δ ABD and Δ ECF,
∠ ADB = ∠ EFC [Each 90]
∠ ABD = ∠ ECF [Proved]
So, Δ ABD ~ Δ ECF [AA similarity]
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 Δ PQR (see Fig. 6.41). Show that: Δ ABC ~ Δ PQR.
Answer :
Given, AD and Pm are the median of triangle, Therefore,
BD = BC/2 and QM = QR/2
Given that
AB/PQ = BC/QR = AD/PM
⇒ AB/PQ = (BC/2)/(QR/2) = AD/PM
⇒ AB/PQ = BD/QM = AD/PM
In Δ ABD and Δ PQM,
AB/PQ = BD/QM = AD/PM [Proved]
So, Δ ABD ~ Δ PQM [SSS similarity]
Hence, ∠ ABD = ∠ PQM [Corresponding angles of similar triangles]
In Δ ABC and Δ PQR,
∠ ABD = ∠ PQM [Proved]
AB/PQ = BC/QR
So, Δ ABC ~ Δ PQR [SAS similarity]
Question : 13:D is a point on the side BC of a triangle ABC such that ∠ ADC = ∠ BAC. Show that CA2 = CB * CD.
Answer :
∠ ADC = ∠ BAC [Proved]
∠ ACD = ∠ BCA [Proved]
So, Δ ADC ~ Δ BAC [AA similarity]
We know that the corresponding sides of similar
triangles are proportional. Therefore,
CA/CB = CD/CA
⇒ CA2 = CB * CD
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: Δ ABC ~ Δ PQR.
Answer :
AB/PQ = AC/PR = AD/PM
Produce AD and PM to E and L such that AD = DE and PM = DE
Now, join B to E, C to E, Q to L and R to L.
AD and Pm are medians of triangles, therefore
BD and Dc and QM = MR
AD = DE and PM = ML [By construction]
So, the diagonals of ABEC bisecting each other at D,
Therefore, ABEC is a parallelogram.
So, AC = BE, AB = EC, PR = QL and PQ = LR
Given that, AB/PQ = AC/PR = AD/PM
⇒ AB/PQ = BE/QL = 2AD/2PM
⇒ AB/PQ = BE/QL = AE/PL
Hence, Δ ABE ~ Δ PQL [SSS similarity]
We know that the corresponding angles of similar triangles are equal.
Therefore, ∠ BAE = ∠ QPL ……….1
Similarly, Δ AEC ~ Δ PLR
and ∠ CAE = ∠ RPL ……….2
Add equation 1 and 2, we get
∠ BAE + ∠ CAE = ∠ QPL + ∠ RPL
⇒ ∠ CAB = ∠ RPQ ……….3
In Δ ABC and Δ PQR,
AB/PQ = AC/PR [Given]
∠ CAB = ∠ RPQ [Proved]
So, Δ ABC ~ Δ PQR [SAS similarity]
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 :
Therefore, DE and BE are the shadows of pole and tower.
In Δ ABE and Δ CDF,
∠ DFC = ∠ BAE [Angle of sun at same place]
∠ CDF = ∠ ABE [Each 90]
So, Δ ABE ~ Δ CDF [AA similarity]
⇒ AB/CD = BE/QL
⇒ AB/6 = 28/4
⇒ AB/6 = 7
⇒ AB = 6 * 7 = 42
Hence, the height of tower is 42 m.
Question : 16:If AD and PM are medians of triangles ABC and PQR, respectively where Δ ABC ~ Δ PQR,
prove that AB/PQ = AD/PM.
Answer :
We know that the corresponding sides of similar triangles are proportional.
Therefore, AB/PQ = AC/PR = BC/QR …………1
and ∠ A = ∠ P, ∠ B = ∠ Q, ∠ C = ∠ R …………2
AD and PM are medians of triangles. Therefore,
BD = BC/2 and QM = QR/2 ………3
From equation 1 and 3, we get
AB/PQ = BD/QM ……….4
In Δ ABD and Δ PQM,
∠ B = ∠ Q [From equation 2]
AB/PQ = BD/QM [From equation 4]
So, Δ ABD ~ Δ PQM [SAS similarity]
⇒ AB/PQ = BD/QM = AD/PM
