ML Aggarwal ICSE Solutions Class 10 Math 13th Chapter Similarity
Class 10 Chapter 13 SimilarityExercise 13.1
(1) State which pairs of triangles in the figure given below are similar. Write the similarity rule used and also write the pairs of similar triangles in symbolic form (all lengths of sides are in cm).
Solution:
In △ABC and △PQR,
AB/PQ = 3.2/4, AC/PR = 3.6/4.5 and BC/QR = 3/5.4
In △DEF and △LMN,
∠E = ∠N = 40°
DE/LN = 4/2 = 2/1 and EF/MN = 4.8/2.4 = 2/1
Thus, △DEF ~ △LMN ∵ By SAS axiom
(2) If in two right triangles, one of the acute angle of are triangle is equal to an acute angle of other triangle, can you say that the two triangle are similar? Why?
Solution :
Given that, In two right triangles, one acute angle of one triangle is equal to one of the acute angle of another triangle.
=> The given two right triangles are similar (By AAA axiom)
(3) It is given that △ABC ~ △EDF such that AB = 5cm, AC = 7cm, DF = 15cm and DE = 12cm. Find the lengths of the remaining sides of the triangles.
Solution :
Given that, △ABC ~ △EDF also, AB = 5cm
AC = 7cm
DF = 15cm and DE = 12cm
As △ABC ~ △EDF
=> AB/ED = AC/EF = BC/DF (∵ by similarity of triangle)
Let us consider, AB/ED = AC/EF
5/12 = 7/EF => EF = 7×12/5 = 16.8cm
Now, let us consider, AB/ED = BC/DF
5/12 = BC/15 => BC = 5×15/12 = 6.25 cm
Thus, the lengths of the remaining sides of a triangle is found to be
EF = 16.8cm & BC = 6.25cm
(4) (a) If △ABC ~ △DEF, AB = 4cm, DE = 6cm, EF = 9cm and FD = 12cm, then find the perimeter of △ABC.
(b) If △ABC ~ △PQR, perimeter of △ABC = 32cm, perimeter of △PQR = 48cm and PR = 6cm, then find the length of AC.
Solution:
(a) △ABC ~ DEF (Given)
And, AB = 4cm, DE = 6cm, EF = 9cm and FD = 12cm
As △ABC ~ △DEF
Let us consider, AB/DE = AC/DE ∵ by similarity of triangles
4/6 = AC/12
=> AC = 4×12/6 = 8cm [AC = 8cm]
Now, consider, AB/DE = BC/EF ∵ By similarity of triangles
4/6 = BC/9
BC = 6cm
BC = 36/6 = 6cm
Thus, perimeter of triangle ABC is found to be P (△ABC) = AB + BC + AC = 4 + 6 + 8 = 18cm.
Thus, the perimeter of required triangles is found to be 18cm.
(b) Given that, △ABC ~ △PQR
Perimeter of △ABC = 32cm, △PQR = 48cm, PR = 6cm
Perimeter of △PQR = 48cm
Then, by similarity of triangles,
AB/PQ = AC/PR = BC/QR
=> Perimeter of △ABC/Perimeter of △PQR = AC/PR
32/48 = AC/6
AC = (32×6)/48
AC = 4cm
Thus, the lengths of side AC is found to be AC = 4cm.
(5) Calculate the other side of a triangle whose shortest side is 6cm and which is similar to a triangle whose sides are 4cm, 7cm and 8cm.
Solution:
Let us consider, △ABC ~ △DEF
And the shortest side of △ABC is found to be BC = 6cm
And, in △DEF
DE = 8cm, EF = 4cm and DF = 7cm
=>△ABC ~ △DEF
Thus, AB/DE = BC/EF = AC/DF ∵ by similarity of triangles
Let us consider, AB/DE = BC/EF
AB/8 = 6/4 => AB = 48/4
AB = 12cm
Let us consider, BC/EF = AC/DF
6/4 = AC/7
=> AC = (6×7)/4 = 42/4 = 21/2
AC = 10.5cm
(7) (a) In the figure.
(i) Given below, ∠P = ∠RTS then prove that △RPQ ~ △RTS.
(b) In the fig,
(ii) Given below, ∠ADC = ∠BAC.
Prove that, CA2 = DC × BC
Solution :
(a) In fig (i), ∠P = ∠RTS
In △RPQ and △RTS,
∠R = ∠R (∵ Common angle)
∠P = ∠RTS (∵ given)
△RPQ ~ △RTS (∵ by AA axiom)
(b) In fig (ii), ∠ADC = ∠BAC
Now, In △ABC and △ADC,
∠C = ∠C (∵ common angle)
∠BAC = ∠ADC (∵ given)
=>△ABC ~ △ADC
Thus, by similarity of two triangles, CA/DC = BC/CA
=> (CA)2 = DC × BC (Hence proved).
(9) In the given fig, ABC is a triangle in which AB = AC. P is a point on the side BC such that PM ⊥ AB and PN ⊥ AC. Prove that, BM × NP = CN × MP.
Solution:
Given that,
In △ABC, AB = AC
P is a point on the side BC so that PM ⊥ or AB and PN ⊥ or AC.
In △ABC, AB = AC (∵ given in question)
∠B = ∠C (∵ angles which are opposite to equal sides)
Now, let us consider △BMP & △CNP, ∠M = ∠N
=>△BMP ~ △CNP
Then, by similarity of triangles BM/CN = MP/NP
=> BM × NP = CN × MP (Hence proved)
(10) Prove that the ratio of the perimeter of two similar triangles is the same as the ratio of their corresponding sides.
Solution:
Given that, △ABC ~ △PQR or △MNO ~ △XYZ
We have, if two triangles are similar then their corresponding angles are equal and their corresponding sides are proportional.
=> MN/XY = NO/YZ = MO/XZ
Perimeter of △MNO = MN + NO + MO and perimeter of △XYZ = XY + YZ + XZ
Thus, MN/XY = NO/YZ = MO/XZ = (MN/XY + NO/YZ + MO/XZ)
= perimeter of △MNO/perimeter of △XYZ
(11) In the fig, ABCD is a trapezium in which AB || DC. The diagonals AC and BD interest at 0. Prove that AO/OC = BO/OD. Using the above result, find the value of X if OA = 3x – 19, OB = x-4, OC = x-3 and OD = 4.
Solution :
Given that, ABCD is a, trapezium with AB || DC and diagonals AE & BD of trapezium ABCD interests each other at point ‘o’.
Now, in △AOB and △COD,
∠AOB = ∠COD (∵ vertically opposite angles are always equal)
∠OAB = ∠OCD
=>△AOB ~ △COD
Then, by similarity of triangles, OA/DC = OB/OD
Given that, OA = 3x – 19, OB = x – 4, OC = x – 3 and BD = 4
Then, OA/OC = OB/OD
(3x – 19)/(x – 3) = (x – 4)/4
=> (x – 3) (x – 4) = 4(3x – 19)
x2 – 7x + 12 – 12x + 76 = 0
x2 – 19x + 88 = 0
x2 – 8x – 11x + 88 = 0
x (x – 8) – 11 (x – 8) = 0
(x – 8) (x – 11) = 0
x – 8 = 0 or x – 11 = 0
x = 8 or x = 11
Thus, the required values of x are found to be 8 & 11.
(13) In the given fig. ∠A = 90° and AD⊥BC if BD = 2cm and CD = 8cm, find AD.
Solution:
Given that,
∠A = 90° and AD ⊥ or BC
Also, BD = 2cm and CD = 8cm.
From fig, ∠DCA + ∠DAC = 90° —— (i)
and ∠BAD + ∠DAC = 90° —– (ii)
From (i) & (ii), ∠BAD + ∠DAC = ∠DCA + ∠DAC
=>∠BAD = ∠DCA …….. (iii)
Now, In △BDA and △ADC,
∠BDA = ∠ADC (∵ each angle is of 90°)
∠BAD = ∠DCA (∵ from (iii)
=>△BDA ~ △ADC
Then, by similarity of triangles,
BD/AD = AD/DC = AB/AC
=> BA/AD = AD/DC
(AD)2 = BD × CD
(AD)2 = 2 × 8 = 16
AD = 4cm is the requited answer.
(14) A 15m high tower costs a shadow of 24 meters long at a certain time and at the same time, a telephone pole costs a shadow 16m long. Find the height in the telephone pole.
Solution:
Given that, height of tower AB = 15m and shadow of BC = 24m
At the same time & position, the height of a telephone pole is found to be DE = xm
And shadow of pole EF = 16m
=>△PQR ~ △MNO
Then, by similarity of triangles,
PQ/MN = ON/RQ
=> 15/x = 24/16
=> x = 15×16/24 = 240/24 = 10
Thus, the height of telephone pole is found to be 10m.
(15) A street light bulb is fixed on a pole 6m above the level of street. If a woman of height casts a shadow of 3m. Find the how far she is away from the base of the pole?
Solution:
Given that, A street light bulb is fixed on a pole 6m above the level of street.
Height of pole = AB = 6m
And height of woman = 1.5m.
And shadow of woman EF = 3m
As woman and pole are standing on the same line.
PM || MR =>△PRQ ~ △MNR
=> RQ/RN = PQ/MN ∵ by similarity of triangles
3+x/3 = 6/1.5
3+x/3 = 4
3 + x = 12
x = 9m
Thus, the woman is 9m far away from the base of the pole.
Exercise – 13.2
(1) (a) In fig. (i) Given below if DE || BG, AD = 3cm, BD = 4cm and BC = 5cm. Find (i) AE:EC (ii) DE
(b) In the fig. (ii) given below, PQ || AC, AP = 4cm, PB = 6cm and BC = 8cm, find CQ and BQ.
Solution:
(a) In fig. (i) DE || BG, AD = 3cm, BD = 4cm and BC = 5cm.
(i) AD/BD = AE/EC and AE/EC = AD/BD
AE/EC = ¾
=> AE:EC = 3:4
(ii) In △ADE and △ABC,
∠D = ∠B, ∠E = ∠C
=>△ADE ~ △ABC
Then, by similarity of triangles.
DE/BC = AD/AB
DE/5 = 3/(3+4)
DE = 15/7
(b) In fig. (ii) PQ || AC, AP = 4cm, PB = 6cm and BC = 8cm.
And ∠BQP = ∠BCA (∵ alternate angles are always equal)
∠B = ∠B (∵ common angle)
=>△ABC ~ △BPQ
Then, by similarity of triangles, BQ/BC = BF/AB = PQ/AC
BQ/BC = 6/(6+4) = PQ/AC
=> BQ/BC = 6/10 = PQ/AC
BQ/8 = 6/10 = PQ/AC
=> PQ/8 = 6/10 => BQ = 4.8 cm
Again, CQ = BC – BQ
CQ = (8 – 4.8)
CQ = 3.2cm
Thus, the required answer is
BQ = 4.8cm and CQ = 3.2cm.
(2) In given fig. DE || BC
(i) If AD = x, DB = x-2, AE = x + 2 and EC = x -1, find the value of x.
(ii) If DB = x – 3, AB = 2x, EC = x – 2 and AC = 2x + 3, find the value of x.
Solution:
Given that, DE || BC
(i) Let us consider △ABC,
AD/DB = AE/EC => x/x – 2 = (x + 2)/(x – 1)
=> x (x – 1) = (x – 2) (x + 2)
= x2 – x = x2 – 4
– x = – 4
x = 4 is the required value of x.
(ii) Given that, DB = x – 3, AB = 2x, EC = x – 2 and AC = 2x + 3 let us consider △ABC,
Then, AD/DB = AE/EC => 2x/(x – 2) = (2x + 3)/(x – 3)
2x (x – 3) = (2x + 3) (x – 2)
2x2 – 4x – 2x2 + 6x – 3x = – 9
– 7x + 6x = – 9
x = 9 is the required value of x.
(3) E and F are points on the sides PQ and PR respectively of a △PQR. For each of the following cases, states whether EF || QR.
(i) PE = 3.9cm, EQ = 3cm, PF = 8cm and RF = 9cm
(ii) PQ = 1.28cm, PR = 2.56cm, PE = 0.18cm and PF = 0.36cm.
Solution:
(i) Given that, E and F are points on the sides PQ & PR respectively of △PQR.
PE/EQ = 3.9/3 = 39/30 = 13/10
Then, PF/FR = 8/9
=> 13/10 ≠ 8/9
Thus, we conclude that PE/EQ ≠ PF/FR
Hence, EF is not parallel to QR.
(ii) Let us consider, △PQR
Here, PQ/PE = 1.28/0.18 = 128/18 = 64/9
and PR/PF = 2.56/0.36 = 256/36 = 64/9
=> 64/9 = 64/9
Thus, we conclude that PQ/PE = PR/PF
Hence, EF is parallel to QR.
(4) A and B are respectively the points on the sides PQ and PR of a triangle PQR so that PQ = 12.5cm, PA = 5cm, BR = 6cm, PB = 4cm. Is AB || QR? Give reason for your answer.
Solution:
Given that, A & B are respectively the points on the sides PQ and PR of a triangle PQR.
And PQ = 12.5cm, PA = 5cm, BR = 6cm, PB = 4cm.
Here, PQ/PA = 12.5/5 = 2.5/1 —– (i)
and PR/PB = (PB + BR)/PB = (4+6)/4 = 10/4 = 2.5 —– (ii)
From (i) & (ii) => PQ/PA = PR/PB
Thus, AB is parallel to QR. [Hence proved]
(5) (a) In the fig (i) given below, CD || LA and DE || AC. Find the length of CL if BE = 4cm and EC = 2cm.
(b) In the given fig. ∠D = ∠E and AD/BD = AE/EC. Prove that is an isosceles triangle.
Solution:
(a) From fig. (i)
CD || LA and DE || AC
In △BCA,
BE/BC = BD/BA => BE/(BE + EC) = BD/AB
4/(4+2) = BD/AB —- (i)
In △BLA, BC/BL = BD/AB => 6/(6+CL) = BD/AB …… (ii)
From (i) & (ii) => 6/(6+CL) = 4/6
=> 36 = 24 + 4CL
CL = 12/4
CL = 3cm is the required answer.
(b) In fig. ∠D = ∠E and AD/BD = AE/EC
In △ADE, ∠D = ∠E
AD = AE (∵ sides opposite to equal angles)
In △ABC, AD/DB = AE/EC
=> DE parallel to BC ∵ DB = FC
AD + DB = AE + EC
AB = AC
Thus, △ABC is an isosceles triangle. [Hence proved]
(6) In the fig below, A, B and C are points on OP, OQ & OR respectively such that AB || PQ and AC || PR. Show that BC || QR.
Solution:
Given that, A, B and C are points on OP, OQ & OR respectively so that AB || PQ and AC || PR.
In △ POQ,
AB || PQ
=> OA/AP = OB/BQ ——- (i)
In △OPR, AC || PR
Then, OA/AP = OC/CR —– (ii)
From (i) & (ii) => OB/BQ = OC/CR
Now, In △OQR
OB/BQ = OC/CR
=> BC || QR (Hence proved)
(7) ABCD is a trapezium in which AB || DC and its diagonals intersects each other at O. Using basic proportionality theorem prove that AO/BO = CO/DO.
Solution:
Given that, ABCD is a trapezium in which AB || DC of its diagonals intersects each other at O.
In △OAB and △OCD,
∠AOB = ∠COD (∵ vertically opposite angle are always equal)
∠OBA = ∠ODC (∵ alternate angles are always equal)
∠OAB = ∠OCD (∵ alternate angles are always equal)
=>△OAB ~ △OCD
Then, by similarity of two triangles,
OA/OC = OB/OD => AO/BO = CO/DO Hence proved.
(8) In the fig. below AD is bisector of ∠BAC. If AB = 6cm, AC = 4cm, and BD = 3cm. find BC.
Solution:
Given that,
In fig. AD is bisector of ∠BAC.
AB = 6cm, AC = 4cm & BD = 3cm
Here, we drawn a straight line from point C i.e. CE which is parallel to DA and joined AE.
∠1 = ∠2 —– (i)
CF || DE
=>∠2 = ∠4 (∵ alternate angles are always equal)
∠1 = ∠3 (∵ corresponding angles are always equal)
=>∠3 = ∠4 and AC = AE
In △ BCE, CE || DE
BD/DC = AB/AC => 3/DC = 6/4
DC = 12/6
DC = 2cm
Thus, BC = BD + DC
= 3 + 2
BC = 5cm.
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