ICSE Solutions for Selina Concise Chapter 15 Similarity (with Applications to Maps and Models) Class 10 Maths
Exercise 15(A)
1. In the figure, given below, straight lines AB and CD intersect at P; and AC || BD. Prove that:
(i) ∆APC and ∆BPD are similar.
(ii) If BD = 2.4 cm, AC = 3.6 cm, PD = 4.0 cm and PB = 3.2 cm; find the lengths of PA and PC.
Solution
(i) In ∆APC and ∆BPD, we have
∠APC = ∠BPD [Vertically opposite angles]
∠ACP = ∠BDP [Alternate angles as, AC || BD]
Thus, ∆APC ~ ∆BPD by AA similarity criterion
(ii) So, by corresponding parts of similar triangles, we have
PA/PB = PC/PD = AC/BD
Given, BD = 2.4 cm, AC = 3.6 cm, PD = 4.0 cm and PB = 3.2 cm
PA/(3.2) = PC/4 = 3.6/2.4
PA/3.2 = 3.6/2.4 and PC/4 = 3.6/2.4
Thus,
PA = (3.6×3.2)/2.4 = 4.8 cm and
PC = (3.6×4)/2.4 = 6 cm
2. In a trapezium ABCD, side AB is parallel to side DC; and the diagonals AC and BD intersect each other at point P. Prove that:
(i) Δ APB is similar to Δ CPD.
(ii) PA x PD = PB×PC.
Solution
(i) In ∆APB and ∆CPD, we have
∠APB = ∠CPD [Vertically opposite angles]
∠ABP = ∠CDP [Alternate angles as, AB||DC]
Thus, ∆APB ~ ∆CPD by AA similarity criterion.
(ii) As ∆APB ~ ∆CPD
Since the corresponding sides of similar triangles are proportional, we have
PA/PC = PB/PD
Thus,
PA×PD = PB×PC
3. P is a point on side BC of a parallelogram ABCD. If DP produced meets AB produced at point L, prove that:
(i) DP : PL = DC : BL.
(ii) DL : DP = AL : DC.
Solution
(i) As AD||BC, we have AD|| BP also.
So, by BPT
DP/PL = AB/BL
And, since ABCD is a parallelogram, AB = DC
Hence,
DP/PL = DC/BL
i.e., DP : PL = DC : BL
(ii) As AD||BC, we have AD|| BP also.
So, by BPT
DL/DP = AL/AB
And, since ABCD is a parallelogram, AB = DC
Hence,
DL/DP = AL/AB
i.e., DL : DP = AL : DC
4. In quadrilateral ABCD, the diagonals AC and BD intersect each other at point O. If AO = 2CO and BO = 2DO; show that:
(i) Δ AOB is similar to Δ COD.
(ii) OA x OD = OB×OC.
Solution
(i) Given,
AO = 2CO and BO = 2DO,
⇒ AO/CO = 2/1 = BO/DO
And,
∠AOB = ∠DOC [Vertically opposite angles]
Hence, ∆AOB ~ ∆COD [SAS criterion for similarity]
(ii) As, AO/CO = 2/1 = BO/DO [Given]
Thus, OA×OD = OB×OC
5. In Δ ABC, angle ABC is equal to twice the angle ACB, and bisector of angle ABC meets the opposite side at point P. Show that :
(i) CB : BA = CP : PA
(ii) AB×BC = BP×CA
Solution
(i) In Δ ABC, we have
∠ABC = 2 ∠ACB [Given]
Now, let ∠ACB = x
So, ∠ABC = 2x
Also given, BP is bisector of ∠ABC
Thus, ∠ABP = ∠PBC = x
By using the angle bisector theorem,
i.e. the bisector of an angle divides the side opposite to it in the ratio of other two sides.
∴ CB: BA = CP: PA.
(ii) In Δ ABC and Δ APB,
∠ABC = ∠APB [Exterior angle property]
∠BCP = ∠ABP [Given]
Thus, ∆ABC ~ ∆APB by AA criterion for similarity
Now, since corresponding sides of similar triangles are proportional we have
CA/AB = BC/BP
∴ AB×BC = BP×CA
6. In Δ ABC, BM ⊥ AC and CN ⊥ AB; show that:
AB/AC = BM/CN = AM/AN
Solution
In Δ ABM and Δ ACN,
∠AMB = ∠ANC [Since, BM ⊥ AC and CN ⊥ AB]
∠BAM = ∠CAN [Common angle]
Hence, ∆ABM ~ ∆ACN by AA criterion for similarity
So, by corresponding sides of similar triangles we have
7. In the given figure, DE ‖ BC, AE = 15 cm, EC = 9 cm, NC = 6 cm and BN = 24 cm.
(i) Write all possible pairs of similar triangles.
(ii) Find the lengths of ME and DM.
Solution
(i) In Δ AME and Δ ANC,
∠AME = ∠ANC [Since DE || BC so, ME || NC]
∠MAE = ∠NAC [Common angle]
Hence, ∆AME ~ ∆ANC by AA criterion for similarity
In Δ ADM and Δ ABN,
∠ADM = ∠ABN [Since DE || BC so, DM || BN]
∠DAM = ∠BAN [Common angle]
Hence, ∆ADM ~ ∆ABN by AA criterion for similarity
In Δ ADE and Δ ABC,
∠ADE = ∠ABC [Since DE || BC so, ME || NC]
∠AED = ∠ACB [Since DE || BC]
Hence, ∆ADE ~ ∆ABC by AA criterion for similarity
(ii) Proved above that, ∆AME ~ ∆ANC
So as corresponding sides of similar triangles are proportional, we have
ME/NC = AE/AC
⇒ ME/ 6 = 15/ 24
⇒ ME = 3.75 cm
And, ∆ADE ~ ∆ABC [Proved above]
So as corresponding sides of similar triangles are proportional, we have
AD/AB = AE/AC = 15/24 …. (1)
Also, ∆ADM ~ ∆ABN [Proved above]
So as corresponding sides of similar triangles are proportional, we have
DM/BN = AD/AB = 15/24 …. From (1)
⇒ DM/ 24 = 15/ 24
⇒ DM = 15 cm
8. In the given figure, AD = AE and AD2 = BD x EC. Prove that: triangles ABD and CAE are similar.
Solution
In Δ ABD and Δ CAE,
∠ADE = ∠AED [Angles opposite to equal sides are equal.]
So, ∠ADB = ∠AEC [As ∠ADB + ∠ADE = 180o and ∠AEC + ∠AED = 180o]
And, AD2 = BD×EC [Given]
AD/BD = EC/AD
AD/BD = EC/AE
Thus, ∆ABD ~ ∆CAE by SAS criterion for similarity.
9. In the given figure, AB ‖ DC, BO = 6 cm and DQ = 8 cm; find: BP×DO.
Solution
In Δ DOQ and Δ BOP,
∠QDO = ∠PBO [As AB || DC so, PB || DQ]
So, ∠DOQ = ∠BOP [Vertically opposite angles]
Hence, ∆DOQ ~ ∆BOP by AA criterion for similarity
Since, corresponding sides of similar triangles are proportional we have
DO/BO = DQ/BP
DO/6 = 8/BP
BP×DO = 48 cm2
10. Angle BAC of triangle ABC is obtuse and AB = AC. P is a point in BC such that PC = 12 cm. PQ and PR are perpendiculars to sides AB and AC respectively. If PQ = 15 cm and PR = 9 cm; find the length of PB.
Solution
In Δ ABC,
AC = AB [Given]
So, ∠ABC = ∠ACB [Angles opposite to equal sides are equal.]
In Δ PRC and Δ PQB,
∠ABC = ∠ACB
∠PRC = ∠PQB [Both are right angles.]
Hence, ∆PRC ~ ∆PQB by AA criterion for similarity
Since, corresponding sides of similar triangles are proportional we have
PR/PQ = RC/QB = PC/PB
⇒ PR/PQ = PC/PB
⇒ 9/15 = 12/PB
Thus,
PB = 20 cm
11. State, true or false:
(i) Two similar polygons are necessarily congruent.
(ii) Two congruent polygons are necessarily similar.
(iii) All equiangular triangles are similar.
(iv) All isosceles triangles are similar.
(v) Two isosceles-right triangles are similar.
(vi) Two isosceles triangles are similar, if an angle of one is congruent to the corresponding angle of the other.
(vii) The diagonals of a trapezium, divide each other into proportional segments.
Solution
(i) False
(ii) True
(iii) True
(iv) False
(v) True
(vi) True
(vii) True
12. Given: ∠GHE = ∠DFE = 90o, DH = 8, DF = 12, DG = 3x – 1 and DE = 4x + 2.
Find: the lengths of segments DG and DE.
Solution
In Δ DHG and Δ DFE,
∠GHD = ∠DFE = 90o
∠D = ∠D [Common]
Thus, ∆DHG ~ ∆DFE by AA criterion for similarity
So, we have
DH/DF = DG/DE
⇒ 8/12 = (3x – 1)/ (4x + 2)
⇒ 32x + 16 = 36x – 12
⇒ 28 = 4x
⇒ x = 7
Hence,
DG = 3×7 – 1 = 20
DE = 4×7 + 2 = 30
13. D is a point on the side BC of triangle ABC such that angle ADC is equal to angle BAC. Prove that: CA2 = CB x CD.
Solution
In Δ ADC and Δ BAC,
∠ADC = ∠BAC [Given]
∠ACD = ∠ACB [Common]
Thus, ∆ADC ~ ∆BAC by AA criterion for similarity
So, we have
CA/CB = CD/CA
∴ CA2 = CB×CD
14. In the given figure, ∆ ABC and ∆ AMP are right angled at B and M respectively.
Given AC = 10 cm, AP = 15 cm and PM = 12 cm.
(i) ∆ ABC ~ ∆ AMP.
(ii) Find AB and BC.
Solution
(i) In ∆ ABC and ∆ AMP, we have
∠BAC = ∠PAM [Common]
∠ABC = ∠PMA [Each = 90o]
Hence, ∆ABC ~ ∆AMP by AA criterion for similarity
(ii) Now, in right triangle AMP
By using Pythagoras theorem, we have
AM = √(AP2 – PM2) = √(152 – 122) = 9
As ∆ABC ~ ∆AMP,
AB/AM = BC/PM = AC/AP
⇒ AB/9 = BC/12 = 10/15
So,
AB/9 = 10/15
⇒ AB = (10×9)/15 = 6 cm
⇒ BC/12 = 10/ 15
⇒ BC = 8 cm
Exercise 15(B)
1. In the following figure, point D divides AB in the ratio 3: 5. Find:
(i) AE/EC
(ii) AD/AB
(iii) AE/AC
Also if,
(iv) DE = 2.4 cm, find the length of BC.
(v) BC = 4.8 cm, find the length of DE.
Solution
(i) Given, AD/DB = 3/5
And DE || BC.
So, by Basic Proportionality theorem, we have
AD/DB = AE/EC
AE/EC = 3/5
(ii) Given, AD/DB = 3/5
So, DB/AD = 5/3
Adding 1 both sides, we get
DB/AD + 1 = 5/3 + 1
⇒ (DB + AD)/ AD = (5 + 3)/3
⇒ AB/AD = 8/3
∴ AD/AB = 3/8
(iii) In ∆ABC, as DE || BC
By BPT, we have
AD/DB = AE/ EC
So, AD/AB = AE/AC
From above, we have AD/AB = 3/8
∴ AE/AC = 3/8
(iv) In ∆ADE and ∆ABC,
∠ADE = ∠ABC [As DE || BC, corresponding angles are equal.]
∠A = ∠A [Common angle]
Hence, ∆ADE ~ ∆ABC by AA criterion for similarity
So, we have
AD/AB = DE/BC
⇒ 3/8 = 2.4/BC
⇒ BC = 6.4 cm
(v) Since, ∆ADE ~ ∆ABC by AA criterion for similarity
So, we have
AD/AB = DE/BC
⇒ 3/8 = DE/4.8
⇒ DE = 1.8 cm
2. In the given figure, PQ ‖ AB; CQ = 4.8 cm QB = 3.6 cm and AB = 6.3 cm. Find:
(i) CP/PA
(ii) PQ
(iii) If AP = x, then the value of AC in terms of x.
Solution
(i) In ∆CPQ and ∆CAB,
∠PCQ = ∠APQ [As PQ || AB, corresponding angles are equal.]
∠C = ∠C [Common angle]
Hence, ∆CPQ ~ ∆CAB by AA criterion for similarity
So, we have
CP/CA = CQ/CB
⇒ CP/CA = 4.8/ 8.4 = 4/7
Thus, CP/PA = 4/3
(ii) As, ∆CPQ ~ ∆CAB by AA criterion for similarity
We have,
PQ/AB = CQ/CB
⇒ PQ/6.3 = 4.8/8.4
⇒ PQ = 3.6 cm
(iii) As, ∆CPQ ~ ∆CAB by AA criterion for similarity
We have,
CP/AC = CQ/CB
⇒ CP/AC = 4.8/8.4 = 4/7
So, if AC is 7 parts and CP is 4 parts, then PA is 3 parts.
Hence, AC = 7/3 ×PA = (7/3)x
3. A line PQ is drawn parallel to the side BC of Δ ABC which cuts side AB at P and side AC at Q. If AB = 9.0 cm, CA = 6.0 cm and AQ = 4.2 cm, find the length of AP.
Solution
In ∆ APQ and ∆ ABC,
∠APQ = ∠ABC [As PQ || BC, corresponding angles are equal.]
∠PAQ = ∠BAC [Common angle]
Hence, ∆APQ ~ ∆ABC by AA criterion for similarity
So, we have
AP/AB = AQ/AC
⇒ AP/9 = 4.2/6
Thus, AP = 6.3 cm
4. In Δ ABC, D and E are the points on sides AB and AC respectively.
Find whether DE ‖ BC, if
(i) AB = 9cm, AD = 4cm, AE = 6cm and EC = 7.5cm.
(ii) AB = 6.3 cm, EC = 11.0 cm, AD =0.8 cm and EA = 1.6 cm.
Solution
(i) In ∆ ADE and ∆ ABC,
AE/EC = 6/7.5 = 4/5
⇒ AD/BD = 4/5 [BD = AB – AD = 9 – 4 = 5 cm]
So, AE/EC = AD/BD
∴ DE || BC by the converse of BPT.
(ii) In ∆ ADE and ∆ ABC,
AE/EC = 1.6/11 = 0.8/5.5
⇒ AD/BD = 0.8/5.5 [BD = AB – AD = 6.3 – 0.8 = 5.5 cm]
So, AE/EC = AD/BD
∴ DE || BC by the converse of BPT.
5. In the given figure, Δ ABC ~ Δ ADE. If AE: EC = 4: 7 and DE = 6.6 cm, find BC. If ‘x’ be the length of the perpendicular from A to DE, find the length of perpendicular from A to BC in terms of ‘x’.
Solution
Given,
Δ ABC ~ Δ ADE
So, we have
AE/AC = DE/BC
⇒ 4/11 = 6.6/BC
⇒ BC = (11×6.6)/4 = 18.15 cm
And, also
As, Δ ABC ~ Δ ADE, we have
∠ABC = ∠ADE and ∠ACB = ∠AED
So, DE || BC
And, AB/AD = AC/AE = 11/4 [Since, AE/EC = 4/7]
In ∆ ADP and ∆ ABQ,
∠ADP = ∠ABQ [As DP || BQ, corresponding angles are equal.]
∠APD = ∠AQB [As DP || BQ, corresponding angles are equal.]
Hence, ∆ADP ~ ∆ABQ by AA criterion for similarity
AD/AB = AP/AQ
⇒ 4/11 = x/AQ
Thus, AQ = (11/4)x
Exercise 15(C)
1. (i) The ratio between the corresponding sides of two similar triangles is 2: 5. Find the ratio between the areas of these triangles.
(ii) Areas of two similar triangles are 98 sq. cm and 128 sq. cm. Find the ratio between the lengths of their corresponding sides.
Solution
We know that,
The ratio of the areas of two similar triangles is equal to the ratio of squares of their corresponding sides.
So,
(i) The required ratio is given by,
(ii) The required ratio is given by,
2. A line PQ is drawn parallel to the base BC of Δ ABC which meets sides AB and AC at points P and Q respectively. If AP = 1/3 PB; find the value of:
(i) Area of Δ ABC/Area of Δ APQ
(ii) Area of Δ APQ/ Area of Trapezium PBCQ
Solution
Given, AP = (1/3) PB
So, AP/PB = 1/3
In ∆ APQ and ∆ ABC,
As PQ || BC, corresponding angles are equal
∠APQ = ∠ABC and ∠AQP = ∠ACB
Hence, ∆APQ ~ ∆ABC by AA criterion for similarity
So,
(i) Area of ∆ABC/ Area of ∆APQ
= AB2/ AP2
= 42/12
= 16: 1
[AP/PB = 1/3 so, AB/AP = 4/1]
(ii) Area of Δ APQ/Area of Trapezium PBCQ
= Area of Δ APQ/(Area of Δ ABC – Area of Δ APQ)
= 1/(16/1)
= 1: 16
3. The perimeters of two similar triangles are 30 cm and 24 cm. If one side of the first triangle is 12 cm, determine the corresponding side of the second triangle.
Solution
Let ∆ABC ~ ∆DEF
So, AB/DE = BC/EF = AC/DF = (AB + BC + AC)/(DE + EF + DF)
= Perimeter of Δ ABC/Perimeter of Δ DEF
Perimeter of Δ ABC/Perimeter of Δ DEF = AB/DE
30/24 = 12/DE
⇒ DE = 9.6 cm
4. In the given figure, AX: XB = 3: 5.
Find:
(i) the length of BC, if the length of XY is 18 cm.
(ii) the ratio between the areas of trapezium XBCY and triangle ABC.
Solution
Given, AX/XB = 3/5 ⇒ AX/AB = 3/8 …. (1)
(i) In Δ AXY and Δ ABC,
As XY || BC, corresponding angles are equal.
∠AXY = ∠ABC and ∠AYX = ∠ACB
Hence, ∆AXY ~ ∆ABC by AA criterion for similarity.
So, we have
AX/AB = XY/BC
⇒ 3/8 = 18/BC
⇒ BC = 48 cm
(ii) Area of Δ AXY/ Area of Δ ABC = AX2/AB2 = 9/64
(Area of Δ ABC – Area of Δ AXY)/Area of Δ ABC = (64 – 9)/64 = 55/64
Area of trapezium XBCY/ Area of Δ ABC = 55/64
5. ABC is a triangle. PQ is a line segment intersecting AB in P and AC in Q such that PQ || BC and divides triangle ABC into two parts equal in area. Find the value of ratio BP: AB.
Solution
It’s given that,
Ar(Δ APQ) = ½ Ar(Δ ABC)
⇒ Ar(Δ APQ)/Ar(Δ ABC) = ½
⇒ AP2/AB2 = ½
⇒ AP/AB = 1/√2
⇒ (AB – BP)/ AB = 1/√2
⇒ 1 – (BP/AB) = 1/√2
⇒ BP/AB = 1 – 1/√2
Thus,
[Multiplying by √2 in both numerator & denominator]6. In the given triangle PQR, LM is parallel to QR and PM: MR = 3: 4.
Calculate the value of ratio:
(i) PL/PQ and then LM/QR
(ii) Area of Δ LMN/ Area of Δ MNR
(iii) Area of Δ LQM/ Area of Δ LQN
Solution
(i) In Δ PLM and Δ PQR,
As LM || QR, corresponding angles are equal.
∠PLM = ∠PQR
∠PML = ∠PRQ
Hence, ∆PLM ~ ∆PQR by AA criterion for similarity.
So, we have
PM/ PR = LM/ QR
3/7 = LM/QR [Since, PM/MR = ¾ ⇒ PM/PR = 3/7]
And, by BPT we have
PL/LQ = PM/MR = ¾
⇒ LQ/PL = 4/3
⇒ 1 + (LQ/PL) = 1 + 4/3
⇒ (PL + LQ)/PL = (3 + 4)/3
⇒ PQ/PL = 7/3
Hence, PL/PQ = 3/7
(ii) As Δ LMN and Δ MNR have common vertex at M and their bases LN and NR are along the same straight line
Hence, Ar (Δ LMN)/ Ar (Δ RNQ) = LN/NR
Now, in Δ LMN and Δ RNQ we have,
∠NLM = ∠NRQ [Alternate angles]
∠LMN = ∠NQR [Alternate angles]
Thus, ∆LNM ~ ∆RNQ by AA criterion for similarity.
So,
MN/QN = LN/NR = LM/QR = 3/7
∴ Ar (Δ LMN)/ Ar (Δ MNR) = LN/NR = 3/7
(iii) As Δ LQM and Δ LQN have common vertex at L and their bases QM and QN are along the same straight line.
Area of Δ LQM/Area of Δ LQN = QM/QN = 10/7
[Since, MN/QN = 3/7 ⇒ QM/QN = 10/7]
Exercise 15(D)
1. A triangle ABC has been enlarged by scale factor m = 2.5 to the triangle A’ B’ C’ Calculate:
(i) the length of AB, if A’ B’ = 6 cm.
(ii) the length of C’ A’ if CA = 4 cm.
Solution
Given that, Δ ABC has been enlarged by scale factor m of 2.5 to Δ A’B’C’.
(i) A’B’ = 6 cm
So,
AB(2.5) = A’B’ = 6 cm
⇒ AB = 2.4 cm
(ii) CA = 4 cm
We know that,
CA(2.5) = C’A’
⇒ C’A’ = 4×2.5 = 10 cm
2. A triangle LMN has been reduced by scale factor 0.8 to the triangle L’ M’ N’. Calculate:
(i) the length of M’ N’, if MN = 8 cm.
(ii) the length of LM, if L’ M’ = 5.4 cm.
Solution
Given, Δ LMN has been reduced by a scale factor m = 0.8 to Δ L’M’N’.
(i) MN = 8 cm
So, MN (0.8) = M’N’
⇒ (8)(0.8) = M’N’
⇒ M’N’ = 6.4 cm
(ii) L’M’ = 5.4 cm
So, LM (0.8) = L’M’
⇒ LM (0.8) = 5.4
⇒ LM = 6.75 cm
3. A triangle ABC is enlarged, about the point 0 as centre of enlargement, and the scale factor is 3. Find:
(i) A’B’, if AB = 4 cm.
(ii) BC, if B’C’ = 15 cm.
(iii) OA, if OA’ = 6 cm
(iv) OC’, if OC = 21 cm
Also, state the value of:
(a) OB’/OB (b) C’A’/CA
Solution
Given that, Δ ABC is enlarged and the scale factor m = 3 to the Δ A’B’C’.
(i) AB = 4 cm
So, AB(3) = A’B’
⇒ (4)(3) = A’B’
⇒ A’B’ = 12 cm
(ii) B’C’ = 15 cm
So, BC(3) = B’C’
⇒ BC(3) = 15
⇒ BC = 5 cm
(iii) OA’ = 6 cm
So, OA (3) = OA’
⇒ OA (3) = 6
⇒ OA = 2 cm
(iv) OC = 21 cm
So, OC(3) = OC’
⇒ 21×3 = OC’
⇒ OC’ = 63 cm
The ratio of the lengths of the two corresponding sides of two triangles.
Δ ABC is enlarged and the scale factor m = 3 to the Δ A’B’C’
Hence,
(a) OB’/OB = 3
(b) C’A’/CA = 3
Exercise 15(E)
1. In the following figure, XY is parallel to BC, AX = 9 cm, XB = 4.5 cm and BC = 18 cm.
Find:
(i) AY/YC
(ii) YC/AC
(iii) XY
Solution
Given, XY || BC.
So, In Δ AXY and Δ ABC
∠AXY = ∠ABC [Corresponding angles]
∠AYX = ∠ACB [Corresponding angles]
Hence, ∆AXY ~ ∆ABC by AA criterion for similarity.
As corresponding sides of similar triangles are proportional, we have
(i) AX/AB = AY/AC
⇒ 9/13.5 = AY/AC
⇒ AY/YC = 9/4.5
⇒ AY/YC = 2
⇒ AY/YC = 2/1
(ii) We have,
AX/AB = AY/AC
⇒ 9/13.5 = AY/ AC
⇒ YC/AC = 4.5/13.5 = 1/3
(iii) As, ∆AXY ~ ∆ABC
AX/AB = XY/BC
⇒ 9/13.5 = XY/18
⇒ XY = (9×18)/13.5 = 12 cm
2. In the following figure, ABCD to a trapezium with AB ‖ DC. If AB = 9 cm, DC = 18 cm, CF= 13.5 cm, AP = 6 cm and BE = 15 cm,
Calculate:
(i) EC
(ii) AF
(iii) PE
Solution
(i) In Δ AEB and Δ FEC,
∠AEB = ∠FEC [Vertically opposite angles]
∠BAE = ∠CFE [Since, AB||DC]
Hence, ∆AEB ~ ∆FEC by AA criterion for similarity.
So, we have
AE/FE = BE/EC = AB/FC
⇒ 15/EC = 9/13.5
⇒ EC = 22.5 cm
(ii) In Δ APB and Δ FPD,
∠APB = ∠FPD [Vertically opposite angles]
∠BAP = ∠DFP [Since, AB||DF]
Hence, ∆APB ~ ∆FPD by AA criterion for similarity.
So, we have
AP/FP = AB/FD
⇒ 6/FP = 9/31.5
⇒ FP = 21 cm
So, AF = AP + PF = 6 + 21 = 27 cm
(iii) We already have, ∆AEB ~ ∆FEC
So,
AE/FE = BE/CE = AB/FC
⇒ AE/FE = 9/13.5
⇒ (AF – EF)/ FE = 9/13.5
⇒ AF/EF – 1 = 9/13.5
⇒ 27/EF = 9/13.5 + 1 = 22.5/ 13.5
⇒ EF = (27×13.5)/22.5 = 16.2 cm
Now, PE = PF – EF = 21 – 16.2 = 4.8 cm
3. In the following figure, AB, CD and EF are perpendicular to the straight line BDF.
If AB = x and; CD = z unit and EF = y unit, prove that:
1/x + 1/y = 1/z
Solution
In Δ FDC and Δ FBA,
∠FDC = ∠FBA [As DC || AB]
∠DFC = ∠BFA [common angle]
Hence, ∆FDC ~ ∆FBA by AA criterion for similarity.
So, we have
DC/AB = DF/BF
⇒ z/x = DF/BF …. (1)
In Δ BDC and Δ BFE,
∠BDC = ∠BFE [As DC || FE]
∠DBC = ∠FBE [Common angle]
Hence, ∆BDC ~ ∆BFE by AA criterion for similarity.
So, we have
BD/BF = z/y ... (2)
Now, adding (1) and (2), we get
BD/BF + DF/BF = z/y + z/x
⇒ 1 = z/y + z/x
Thus,
1/z = 1/x + 1/y
Hence, Proved
4. Triangle ABC is similar to triangle PQR. If AD and PM are corresponding medians of the two triangles, prove that: AB/PQ = AD/PM.
Solution
Given, ∆ABC ~ ∆PQR
AD and PM are the medians, so BD = DC and QM = MR
So, we have
AB/PQ = BC/QR [Corresponding sides of similar triangles are proportional.]
Then,
AB/PQ = (BC/2)/ (QR/2) = BD/QM
And, ∠ABC = ∠PQR i.e. ∠ABD = ∠PQM
Hence, ∆ABD ~ ∆PQM by SAS criterion for similarity.
Thus,
AB/PQ = AD/PM.
5. Triangle ABC is similar to triangle PQR. If AD and PM are altitudes of the two triangles, prove that: AB/PQ = AD/PM.
Solution
Given, ∆ABC ~ ∆PQR
So,
∠ABC = ∠PQR i.e. ∠ABD = ∠PQM
Also, ∠ADB = ∠PMQ [Both are right angles]
Hence, ∆ABD ~ ∆PQM by AA criterion for similarity.
Thus,
AB/PQ = AD/PM
6. Triangle ABC is similar to triangle PQR. If bisector of angle BAC meets BC at point D and bisector of angle QPR meets QR at point M, prove that: AB/PQ = AD/PM
Solution
Given, ∆ABC ~ ∆PQR
And, AD and PM are the angle bisectors.
So,
∠BAD = ∠QPM
Also, ∠ABC = ∠PQR i.e. ∠ABD = ∠PQM
Hence, ∆ABD ~ ∆PQM by AA criterion for similarity.
Thus,
AB/PQ = AD/PM
7. In the following figure, ∠AXY = ∠AYX. If BX/AX = CY/AY, show that triangle ABC is isosceles.
Solution
Given,
∠AXY = ∠AYX
So, AX = AY [Sides opposite to equal angles are equal.]
Also, from BPT we have
BX/AX = CY/AY
Thus,
AX + BX = AY + CY
So, AB = AC
∴ ∆ABC is an isosceles triangle.
8. In the following diagram, lines l, m and n are parallel to each other. Two transversals p and q intersect the parallel lines at points A, B, C and P, Q, R as shown.
Prove that: AB/BC = PQ/QR
Solution
Let join AR such that it intersects BQ at point X.
In ∆ACR, BX || CR. By BPT, we have
AB/BC = AX/XR …(1)
In ∆APR, XQ || AP. By BPT, we have
PQ/QR = AX/XR …(2)
From (1) and (2),
AB/BC = PQ/QR
Hence, Proved
9. In the following figure, DE || AC and DC || AP. Prove that: BE/EC = BC/CP.
Solution
Given, DE || AC
So,
BE/EC = BD/DA [By BPT]
And, DC || AP
So,
BC/CP = BD/DA [By BPT]
∴ BE/EC = BC/CP
10. In the figure given below, AB ‖ EF ‖ CD. If AB = 22.5 cm, EP = 7.5 cm, PC = 15 cm and DC = 27 cm.
Calculate: (i) EF (ii) AC
Solution
(i) In ∆PCD and ∆PEF,
∠CPD = ∠EPF [Vertically opposite angles]
∠DCE = ∠FEP [As DC || EF, alternate angles.]
Hence, ∆PCD ~ ∆PEF by AA criterion for similarity.
So, we have
27/EF = 15/7.5
Thus,
EF = 13.5
(ii) And, as EF || AB
∆CEF ~ ∆CAB by AA criterion for similarity.
EC/AC = EF/AB
22.5/AC = 13.5/22.5
Thus, AC = 37.5 cm.
11. In ΔABC, ∠ABC = ∠DAC, AB = 8 cm, AC = 4 cm and AD = 5 cm.
(i) Prove that ΔACD is similar to ΔBCA.
(ii) Find BC and CD
(iii) Find the area of ΔACD: area of ΔABC
Solution
(i) In ∆ACD and ∆BCA,
∠DAC = ∠ABC [Given]
∠ACD = ∠BCA [Common angles]
Hence, ∆ACD ~ ∆BCA by AA criterion for similarity.
(ii) Since, ∆ACD ~ ∆BCA
We have,
AC/BC = CD/CA = AD/AB
⇒ 4/BC = CD/4 = 5/8
⇒ 4/BC = 5/8
So, BC = 32/5 = 6.4 cm
And,
CD/4 = 5/8
Thus, CD = 20/8 = 2.5 cm
(iii) As, ∆ACD ~ ∆BCA
We have,
Ar(∆ACD)/ Ar(∆BCA) = AD2/ AB2 = 52/82
⇒ Ar(∆ACD)/ Ar(∆BCA) = 25/64
12. In the given triangle P, Q and R are mid-points of sides AB, BC and AC respectively. Prove that triangle QRP is similar to triangle ABC.
Solution
In ∆ABC, as PR || BC by BPT we have
AP/PB = AR/RC
And, in ∆PAR and ∆BAC,
∠PAR = ∠BAC [Common]
∠APR = ∠ABC [Corresponding angles]
Hence, ∆PAR ~ ∆BAC by AA criterion for similarity
So, we have
PR/BC = AP/AB
⇒ PR/BC = ½ [Since, P is the mid-point of AB]
⇒ PR = ½ BC
Similarly,
PQ = ½ AC
RQ = ½ AB
So,
PR/BC = PQ/AC = RQ/AB
∴ ∆QRP ~ ∆ABC by SSS similarity.
13. In the following figure, AD and CE are medians of ∆ABC. DF is drawn parallel to CE. Prove that:
(i) EF = FB,
(ii) AG: GD = 2: 1
Solution
(i) In ∆BFD and ∆BEC,
∠BFD = ∠BEC [Corresponding angles]
∠FBD = ∠EBC [Common]
Hence, ∆BFD ~ ∆BEC by AA criterion for similarity.
So,
BF/BE = BD/BC
⇒ BF/BE = ½ [Since, D is the mid-point of BC]
⇒ BE = 2BF
⇒ BF = FE = 2BF
Thus,
EF = FB
(ii) In ∆AFD, EG || FD and using BPT we have
AE/EF = AG/GD ….(1)
Now, AE = EB [Since, E is the mid-point of AB]
AE = 2EF [As, EF = FB, by (1)]
So, from (1) we have
AG/GD = 2/1
∴ AG: GD = 2:1
14. Two similar triangles are equal in area. Prove that the triangles are congruent.
Solution
Let’s consider two similar triangles as ∆ABC ~ ∆PQR
So,
Ar(∆ABC)/Ar(∆PQR) = (AB/PQ)2 = (BC/QR)2 = (AC/PR)2
Since,
Area of ∆ABC = Area of ∆PQR [Given]
Hence,
AB = PQ
BC = QR
AC = PR
So, as the respective sides of two similar triangles are all of same length.
We can conclude that,
∆ABC ≅ ∆PQR [By SSS rule]
Hence, Proved
15. The ratio between the altitudes of two similar triangles is 3: 5; write the ratio between their:
(i) medians. (ii) perimeters. (iii) areas.
Solution
The ratio between the altitudes of two similar triangles is same as the ratio between their sides.
So,
(i) The ratio between the medians of two similar triangles is same as the ratio between their sides.
Thus, the required ratio = 3: 5
(ii) The ratio between the perimeters of two similar triangles is same as the ratio between their sides.
Thus, the required ratio = 3: 5
(iii) The ratio between the areas of two similar triangles is same as the square of the ratio between their corresponding sides.
Thus, the required ratio = (3)2: (5)2 = 9: 25