NCERT solutions for class 10 Maths
Chapter 6
Triangles
Exercise 6.3
Ex 6.3 Question 1.
State which pairs of triangles in Figure, 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:
Solution:
(i) ΔABC and ΔPQR,
∠A = ∠P = 60°
∠B = ∠Q = 80°
∠C = ∠R = 40°
∴ ΔABC ~ ΔPQR [AAA similarity]
(ii) ΔABC and ΔPQR,
AB/QR = BC/RP = CA/PQ
ΔABC ~ ΔQRP [SSS similarity]
(iii) In ΔLMP and ΔDEF,
The corresponding sides are not proportional.Hence triangles are not similar.
(iv) In ΔMNL and ΔQPR
The corresponding sides are not proportional.Hence triangles are not similar.
(v) In ΔABC and ΔDEF, given that,
The corresponding sides are not proportional.Hence triangles are not similar.
(vi) In ΔDEF,
∠D + ∠E + ∠F = 180°
⇒ 70° + 80° + ∠F = 180°
⇒ ∠F = 30°
Similarly, in ΔPQR,
∠P + ∠Q + ∠R = 180
⇒ ∠P + 80° + 30° = 180°
⇒ ∠P = 70°
In, ΔDEF and ΔPQR,
∠D = ∠P = 70°
∠F = ∠Q = 80°
∠F = ∠R = 30°
Hence, ΔDEF ~ ΔPQR [AAA similarity]
Ex 6.3 Question 2.
In the figure, ΔODC ∝ ¼ ΔOBA, ∠ BOC = 125° and ∠ CDO = 70°. Find ∠ DOC, ∠ DCO and ∠ OAB.
Solution:
DOB is a straight line.
∴ ∠DOC + ∠ COB = 180°
⇒ ∠DOC = 180° – 125° = 55°
In ΔDOC,
∴ ∠DCO + ∠ CDO + ∠ DOC = 180°
⇒ ∠DCO + 70º + 55º = 180°
⇒ ∠DCO = 55°
Given that, ΔODC ~ ΔOBA.
∠OAB = ∠OCD [Corresponding angles are equal in similar triangles]
⇒ ∠ OAB = 55°
Ex 6.3 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 AO/OC = OB/OD
Solution:
In ΔDOC and ΔBOA,
∠DOC = ∠BOA [vertically opposite angles]
∠CDO = ∠ABO [Alternate angles]
∠DCO = ∠BAO [Alternate angles]
ΔDOC ~ ΔBOA [AAA similarity]
DO/BO = CO/AO ⇒ BO/DO = AO/CO
Ex 6.3 Question 4.
In the fig.6.36, QR/QS = QT/PR and ∠1 = ∠2. Show that ΔPQS ~ ΔTQR.
Solution:
In ΔPQR,
∠PQR = ∠PRQ
∴ PQ = PR ………(i)
Given that,
QR/QS = QT/PR
QR/QS = QT/QP [From equation (i)] …..(ii)
In ΔPQS and ΔTQR
QR/QS = QT/QP [From equation (ii)]
∠Q = ∠Q [Common]
∴ ΔPQS ~ ΔTQR [SAS similarity]
Ex 6.3 Question 5.
S and T are point on sides PR and QR of ΔPQR such that ∠P = ∠RTS. Show that ΔRPQ ~ ΔRTS.
Solution:
In ΔRPQ and ΔRTS,
∠RTS = ∠QPS [Given]
∠R = ∠R [Common]
∴ ΔRPQ ~ ΔRTS [AA similarity criterion]
Ex 6.3 Question 6.
In the figure, if ΔABE ≅ ΔACD, show that ΔADE ~ ΔABC.
Solution:
Given that, ΔABE ≅ ΔACD.
∴ AB = AC [By CPCT] ……….(i)
& AD = AE [By CPCT] …………(ii)
In ΔADE and ΔABC,
By dividing equation (ii) by equation(i),
AD/AB = AE/AC
∠A = ∠A [Common]
∴ ΔADE ~ ΔABC [SAS similarity]
Ex 6.3 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
Solution:
(i) In ΔAEP and ΔCDP,
∠AEP = ∠CDP [Each 90°]
∠APE = ∠CPD [Vertically opposite angles]
ΔAEP ∼ ΔCDP [AA similarity]
(ii) In ΔABD and ΔCBE,
∠ADB = ∠CEB [ Each 90°]
∠ABD = ∠CBE [Common]
ΔABD ~ ΔCBE [AA similarity]
(iii) In ΔAEP and ΔADB,
∠AEP = ∠ADB [Each 90°]
∠PAE = ∠DAB [Common]
ΔAEP ~ ΔADB [AA similarity]
(iv) In ΔPDC and ΔBEC,
∠PDC = ∠BEC [Each 90°]
∠PCD = ∠BCE [Common]
ΔPDC ~ ΔBEC [AA similarity]
Ex 6.3 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.
Solution:
In ΔABE and ΔCFB,
∠A = ∠C [Opposite angles of a parallelogram]
∠AEB = ∠CBF [Alternate angles as AE || BC]
∴ ΔABE ~ ΔCFB [AA similarity]
Ex 6.3 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) CA/PA = BC/MP
Solution:
(i) In ΔABC and ΔAMP,
∠CAB = ∠MAP [common]
∠ABC = ∠AMP [Each 90°]
∴ ΔABC ~ ΔAMP [AA similarity]
(ii) ΔABC ~ ΔAMP [Proved above]
CA/PA = BC/MP [Corresponding parts of similar triangles] are always equal]
Ex 6.3 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) CD/GH = AC/FG
(ii) ΔDCB ~ ΔHGE
(iii) ΔDCA ~ ΔHGF
Solution:
(i) Given that,
ΔABC ~ ΔFEG.
∴ ∠A = ∠F, ∠B = ∠E, and ∠ACB = ∠FGE
Since, ∠ACB = ∠FGE
∴ ∠ACD = ∠FGH [CD & GH are the bisector of equal angles]
& ∠DCB = ∠HGE [CD & GH are the bisector of equal angles]
In ΔACD and ΔFGH,
∠A = ∠F & ∠ACD = ∠FGH [Proved above]
∴ ΔACD ~ ΔFGH [AA similarity]
⇒CD/GH = AC/FG
(ii) In ΔDCB and ΔHGE,
∠DCB = ∠HGE [Proved above]
∠B = ∠E [Proved above]
∴ ΔDCB ~ ΔHGE [AA similarity]
(iii) In ΔDCA and ΔHGF,
∠ACD = ∠FGH [Proved above]
∠A = ∠F [Proved above]
∴ ΔDCA ~ ΔHGF [AA similarity]
Ex 6.3 Question 11.
In the following 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.
Solution:
ABC is an isosceles triangle.
∴ AB = AC
⇒ ∠ABD = ∠ECF
In ΔABD and ΔECF,
∠ADB = ∠EFC [Each 90°]
∠BAD = ∠CEF [Proved above]
∴ ΔABD ~ ΔECF [AA similarity ]
Ex 6.3 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 (see Fig 6.41). Show that ΔABC ~ ΔPQR.
Solution:
Given, ΔABC and ΔPQR, AB, BC and median AD of ΔABC are proportional to sides PQ, QR and median PM of ΔPQR
AD & PM are the median of triangles.
BD = BC/2 and QM = QR/2
Given that, AB/PQ = BC/QR = AD/PM
AB / PQ = ½BC/½QR = AD/ PM
⇒AB/PQ = BC/QR = AD/PM
In ΔABD and ΔPQM
AB/PQ = BC/QR = AD/PM [Proved above]
⇒ ΔABD ~ ΔPQM [SSS similarity]
∴ ∠ABD = ∠PQM [Corresponding angles of two similar triangles]
In ΔABC and ΔPQR
∠ABD = ∠PQM [Proved above]
AB/PQ = BC/QR
ΔABC ~ ΔPQR [SAS similarity]
Ex 6.3 Question 13.
D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC. Show that CA2 = CB.CD
Solution:
In ΔADC and ΔBAC,
∠ADC = ∠BAC [Given]
∠ACD = ∠BCA [Common]
∴ ΔADC ~ ΔBAC [AA similarity]
[The corresponding sides of similar triangles are in proportion.]
∴ CA/CB = CD/CA
⇒ CA2 = CB.CD.
Ex 6.3 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.
Solution:
Given that;
AB/PQ = AC/PR = AD/PM
We have to prove, ΔABC ~ ΔPQR
Contruction : Produce AD to E so that AD = DE. Join CE, Similarly produce PM to N such that PM = MN, also Join RN.
In ΔABD and ΔCDE,
AD = DE [By Construction.]
BD = DC [Since, AP is the median]
and, ∠ADB = ∠CDE [Vertically opposite angles]
∴ ΔABD ≅ ΔCDE [SAS criterion of congruence]
⇒ AB = CE [By CPCT] …………..(i)
Also, in ΔPQM and ΔMNR,
PM = MN [By Construction.]
QM = MR [Since, PM is the median]
and, ∠PMQ = ∠NMR [Vertically opposite angles]
∴ ΔPQM = ΔMNR [SAS criterion of congruence]
⇒ PQ = RN [CPCT] ……………(ii)
Now, AB/PQ = AC/PR = AD/PM
From equation (i) and (ii),
⇒CE/RN = AC/PR = AD/PM
⇒ CE/RN = AC/PR = 2AD/2PM
⇒ CE/RN = AC/PR = AE/PN [Since 2AD = AE and 2PM = PN]
∴ ΔACE ~ ΔPRN [SSS similarity]
Therefore, ∠2 = ∠4
Similarly, ∠1 = ∠3
∴ ∠1 + ∠2 = ∠3 + ∠4
⇒ ∠A = ∠P ………….(iii)
Now, in ΔABC and ΔPQR, we have
AB/PQ = AC/PR [Given]
From equation (iii),
∠A = ∠P
∴ ΔABC ~ ΔPQR [ SAS similarity]
Ex 6.3 Question 15.
A vertical pole of a length 6 m casts a shadow 4m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.
Solution:
Vertical pole Length = 6m
Shadow = 4 m
Let ,tower height = h m
Length of shadow of the tower = 28 m
In ΔABC and ΔDEF,
∠C = ∠E [angular elevation of sum]
∠B = ∠F = [Each 90°]
∴ ΔABC ~ ΔDEF [AA similarity]
∴ AB/DF = BC/EF [Two triangles are similar corresponding sides are proportional]
∴ 6/h = 4/28
⇒h = (6×28)/4
⇒ h = 6 × 7
⇒ h = 42 m
Ex 6.3 Question 16.
If AD and PM are medians of triangles ABC and PQR, respectively where ΔABC ~ ΔPQR prove that AB/PQ = AD/PM.
Solution:
Given that, ΔABC ~ ΔPQR
The corresponding sides of similar triangles are in proportion.
∴AB/PQ = AC/PR = BC/QR………………(i)
Also, ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R ……..(ii)
AD and PM are medians of triangle
∴ BD = BC/2 and QM = QR/2 ………….(iii)
From equations (i) and (iii),
AB/PQ = BD/QM …………….(iv)
In ΔABD and ΔPQM,
From equation (ii),
∠B = ∠Q
From equation (iv)
AB/PQ = BD/QM
∴ ΔABD ~ ΔPQM [SAS similarity]
⇒AB/PQ = BD/QM = AD/PM