Class 10 SELINA Solutions Maths Chapter 15 - Similarity (With Applications to Maps and Models)
Similarity (With Applications to Maps and Models) Exercise TEST YOURSELF
Solution 1(a)
Correct option: (iii) B and B’
Triangles ABC and A’B’C’ are similar to each other.
So, the corresponding vertices are:
A and A’
B and B’
C and C’
Solution 1(b)
Correct option: (i) AB and RS
Triangles ABC and RSP are similar to each other. So, their corresponding sides are:
AB and RS
BC and SP
AC and RP
Solution 1(c)
Correct option: (ii) A is false, B is true
Two congruent triangles are always similar
But, two similar triangles need not to be congruent.
Solution 1(d)
Correct option: (ii)
As triangles ABC and PQR are similar to each other.
Solution 1(e)
Correct option: (iii) 256 cm2
In ΔABC and ΔAEF,
∠A = ∠A … (Common angle)
∠AEF = ∠ABC … (Corresponding angles)
⇒ΔABC ∼ΔAEF … (By AA Postulate)
⇒ Area of ΔAEF = 256 cm2
Solution 1(f)
Correct option: (iii) Yes, AA
In ΔBAD and ΔACD,
∠D = ∠D … (Common angle)
∠ACD = ∠BAD … (Given)
⇒ΔBAD ∼ΔACD … (By AA Postulate)
Solution 1(g)
Correct option: (iv) 250 m
Scale = 3 : 500 … (Given)
⇒ Length of the ship = 250 m
Solution 2(i)
Solution 2(ii)
Solution 2(iii)
Given that XY || BC
So, △AXY ∼ △ABC
Solution 3(i)
Solution 3(ii)
Solution 3(iii)
Solution 4
Solution 5
Triangle ABC is similar to triangle PQR
Solution 6
Solution 7
Solution 8
Solution 9
Solution 10(i)
Solution 10(ii)
Solution 11(i)
Solution 11(ii)
Solution 11(iii)
Solution 12
BF + FE = 2BF
Hence,
FE = BF
(ii) In AFD, EG || FD. Using Basic Proportionality theorem,
… (1)
Now, AE = EB (as E is the mid-point of AB)
AE = 2EF (Since, EF = FB, by (i))
From (1),
Hence, AG: GD = 2: 1.
Solution 13
Let us assume two similar triangles as ABC PQR
Solution 14
The ratio between the altitudes of two similar triangles is same as the ratio between their sides.
(i) The ratio between the medians of two similar triangles is same as the ratio between their sides.
Required ratio = 3: 5
(ii) The ratio between the perimeters of two similar triangles is same as the ratio between their sides.
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.
Required ratio = (3)2 : (5)2 = 9: 25
Solution 15
The ratio between the areas of two similar triangles is same as the square of the ratio between their corresponding sides.
So, the ratio between the sides of the two triangles = 4: 5
(i) The ratio between the perimeters of two similar triangles is same as the ratio between their sides.
Required ratio = 4: 5
(ii) The ratio between the altitudes of two similar triangles is same as the ratio between their sides.
Required ratio = 4: 5
(iii) The ratio between the medians of two similar triangles is same as the ratio between their sides.
Required ratio = 4: 5
Solution 16
Scale :- 1 : 20000
1 cm represents 20000 cm= = 0.2 km
(i)
=
= 576 + 1024 = 1600
AC = 40 cm
Actual length of diagonal = 40 0.2 km = 8 km
(ii)
1 cm represents 0.2 km
1 cm2 represents 0.2 0.2
The area of the rectangle ABCD = AB BC
= 24 32 = 768
Actual area of the plot = 0.2 0.2 768 = 30.72 km2
Solution 17
The dimensions of the building are calculated as below.
Length = 1 50 m = 50 m
Breadth = 0.60 50 m = 30 m
Height = 1.20 50 m = 60 m
Thus, the actual dimensions of the building are 50 m 30 m 60 m.
(i)
Floor area of the room of the building
(ii)
Volume of the model of the building
Solution 18
(iii)
Solution 19
Triangle ABC is enlarged to DEF. So, the two triangles will be similar.
Longest side in ABC = BC = 6 cm
Corresponding longest side in DEF = EF = 9 cm
Scale factor = = 1.5
Solution 20
Let ABC and PQR be two isosceles triangles.
Then,
Also, A = P (Given)
Let AD and PS be the altitude in the respective triangles.
We know that the ratio of areas of two similar triangles is equal to the square of their corresponding altitudes.
Solution 21
In triangle ABC, PO || BC. Using Basic proportionality theorem,
(i)
(ii)
Solution 22
In ABC and EBD,
ACB = EDB (given)
ABC = EBD (common)
(by AA- similarity)
(i) We have,
(ii)
Solution 23
Solution 24(i)
Solution 24(ii)
Solution 24(iii)
Solution 25
(i) In AGB, DE || AB , by Basic proportionality theorem,
.... (1)
In GBC, EF || BC, by Basic proportionality theorem,
.... (2)
From (1) and (2), we get,
(ii)
From (i), we have:
Solution 26
i.
In ∆PQR and ∆SPR,
∠PSR = ∠QPR … given
∠PRQ = ∠PRS … common angle
⇒ ∆PQR ∼ ∆SPR (AA Test)
ii. Find the lengths of QR and PS.
Since ∆PQR ∼ ∆SPR … from (i)
iii.
Similarity (With Applications to Maps and Models) Exercise Ex. 15(A)
Solution 2(i)
Solution 2(ii)
Solution 3(i)
Solution 3(ii)
Solution 4(i)
Solution 4(ii)
Solution 5(i)
Solution 5(ii)
Solution 6
Solution 7(i)
Solution 7(ii)
In ΔADE and ΔABC
Solution 8
Solution 9
Solution 10
(i) False
(ii) True
(iii) True
(iv) False
(v) True
(vi) True
(vii) True
Solution 11
Solution 12
Solution 13
(i) In ∆ ABC and ∆ AMP,
BAC= PAM [Common]
ABC= PMA [Each = 90°]
∆ ABC ~ ∆ AMP [AA Similarity]
(ii)
Solution 14
(i) In PQM and PQR,
PMQ = PQR = 90o
QPM = RPQ (Common)
(ii) In QMR and PQR,
QMR = PQR = 90o
QRM = QRP (Common)
(iii) Adding the relations obtained in (i) and (ii), we get,
Solution 15
(i) In CDB,
1 + 2 +3 = 180o
1 + 3 = 90o ..... (1)(Since, 2 = 90o)
3 + 4 = 90o .....(2) (Since, ABC = 90o)
From (1) and (2),
1 + 3 = 3 + 4
1 = 4
Also, 2 = 5 = 90o
Hence, AD = 6.4 cm
(iii)
Solution 16
Solution 17
Given, AD2 = BD DC
So, these two triangles will be equiangular.
Solution 18
(i) The three pair of similar triangles are:
BEF and BDC
CEF and CAB
ABE and CDE
(ii) Since, ABE and CDE are similar,
Since, CEF and CAB are similar,
Solution 19
(i) Given, AP: PB = 4: 3.
Since, PQ || AC. Using Basic Proportionality theorem,
Now, PQB = ACB (Corresponding angles)
QPB = CAB (Corresponding angles)
(ii) ARC = QSP = 90o
ACR = SPQ (Alternate angles)
Solution 20
We have:
Solution 21
(i) Since, BD and CE are medians.
AD = DC
AE = BE
Hence, by converse of Basic Proportionality theorem,
ED || BC
In EGD and CGB,
(ii) Since,
In AED and ABC,
From (1),
Solution 1(a)
Correct option: (iii) AAA
In ΔABD and ΔCBA,
∠B = ∠B … (Common angle)
∠ADB = ∠CAB … (Each 90°)
Thus, ΔABD ∼ΔCBA … (By AA or AAA postulate)
Solution 1(b)
Correct option: (iii)
In ΔOCD and ΔOAB,
∠COD = ∠AOB … (Vertically opposite angles)
∠OCD = ∠OAB … (Internally opposite angles)
∠ODC = ∠OBA … (Internally opposite angles)
⇒ΔOCD ∼ΔOAB … AA postulate
Solution 1(c)
Correct option: (iv) 2.5 cm
In ΔACE and ΔBCD,
∠CAE = ∠CBD … (Each 90°)
∠C = ∠C … (Common angles)
⇒ΔACE ∼ΔBCD … AA postulate
Now, AB = AC – BC = 2.5 cm
Solution 1(d)
Correct option: (ii) ΔADE ∼ΔABC
In ΔADE and ΔABC
∠DAE = ∠BAC … (Given as angle A is common)
∠AED = ∠ACB … (Given)
⇒ΔADE ∼ΔABC … By AA postulate
∗∗ Answer in the book is given as (iii) ΔADE ∼ΔBAC, which is not correct.∗∗
Solution 1(e)
Correct option: (iv) SSS
Since, OD = 2 × OB, OC = 2 × OA and CD = 2 × AB
⇒ΔAOB ∼ΔCOD … By SSS postulate
Similarity (With Applications to Maps and Models) Exercise Ex. 15(B)
Solution 2(i)
Now, DE is parallel to BC.
Then, by Basic proportionality theorem, we have
Solution 2(ii)
Solution 2(iii)
Solution 2(iv)
Solution 2(v)
Solution 3(i)
Solution 3(ii)
Solution 3(iii)
Solution 4
Solution 5(i)
Solution 5(ii)
Solution 6
Solution 7
Solution 8
Similarly
Solution 9
Solution 10
Solution 11
Solution 1(a)
Correct option: (i) 3 : 8
Since, DE || BC, we have
… By Basic Proportionality Theorem
In ΔADE and ΔABC,
… From (I) and (II)
∠A = ∠A … (Common angle)
∴ΔADE ∼ΔABC … By SAS Postulate
DE : BC = 3 : 8
Solution 1(b)
Correct option: (iv) ΔADE ∼ΔABC
AB = AD + DB = AE + CE = AC … (I)
⇒ DE || BC … (By Converse of Basic Proportionality Theorem)
In ΔADE and ΔABC,
∠A = ∠A … (Common angle)
∠ADE = ∠ABC … (Corresponding angles)
⇒ΔADE ∼ΔABC … By AA Postulate
Solution 1(c)
Correct option: (iii) 6.4 cm
In ΔAOB and ΔCOD,
∠AOB = ∠COD … (Vertically opposite angles)
∠BAO = ∠DCO … (Alternate interior angles)
⇒ΔAOB ∼ΔCOD … By AA postulate
Solution 1(d)
Correct option: (iii) 8 cm
Since, DE || BC
… By Basic Proportionality Theorem
… (I)
In ΔADE and ΔABC,
∠A = ∠A … (Common angle)
∠ADE = ∠ABC … (Corresponding angles)
⇒ΔADE ∼ΔABC … By AA Postulate
Solution 1(e)
Correct option: (ii) similar
Two congruent triangles are similar.
Similarity (With Applications to Maps and Models) Exercise Ex. 15(C)
Solution 2
(i) AP =PB
(ii)
Solution 3
Let
Solution 4
Given,
(i)
(ii)
Solution 5
From the given information, we have:
Solution 6
(i)
(ii) Since LMN and MNR have common vertex at M and their bases LN and NR are along the same straight line
(iii) Since LQM and LQN have common vertex at L and their bases QM and QN are along the same straight line
Solution 7
Solution 8
Solution 9
(i) Given, DE || BC and
In ADE and ABC,
A = A(Corresponding Angles)
ADE = ABC(Corresponding Angles)
(By AA- similarity)
..........(1)
Now
Using (1), we get.........(2)
(ii) In DEF and CBF,
FDE = FCB(Alternate Angle)
DFE = BFC(Vertically Opposite Angle)
DEF CBF(By AA- similarity)
using (2)
.
(iii) Since the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides, therefore
Solution 10
Solution 1(a)
Correct option: (iii) 25 : 144
Since, DE || BC
… By Basic Proportionality Theorem
… (I)
In ΔADE and ΔABC,
∠ADE = ∠ABC … (Corresponding angles)
∠A = ∠A … (Common angles)
⇒ΔADE ∼ΔABC
Solution 1(b)
Correct option: (iv) 25 : 24
Since, DE || BC
… By Basic Proportionality Theorem
… (I)
In ΔADE and ΔABC,
∠ADE = ∠ABC … (Corresponding angles)
∠A = ∠A … (Common angles)
⇒ΔADE ∼ΔABC
Solution 1(c)
Correct option: (i) 4 : 49
Since, DE || BC
… By Basic Proportionality Theorem
… (I)
In ΔADE and ΔABC,
∠ADE = ∠ABC … (Corresponding angles)
∠A = ∠A … (Common angles)
⇒ΔADE ∼ΔABC … By AA Postulate
In ΔODE and ΔOCB,
∠DOE = ∠COB … Vertically opposite angles
∠ODE = ∠OCB … Alternate interior angles
⇒ΔODE ∼ΔOCB … By AA Postulate
Solution 1(d)
Correct option: (iv) 5 : 3
Since, DE || BC
… By Basic Proportionality Theorem
… (I)
In ΔADE and ΔABC,
∠ADE = ∠ABC … (Corresponding angles)
∠A = ∠A … (Common angles)
⇒ΔADE ∼ΔABC … By AA Postulate
Solution 1(e)
Correct option: (iii) 25 : 144
In right-angled triangle BAC.
AB2 + AC2 = BC2
⇒ AB = 12 cm … (I)
Let ∠C = x°
⇒∠B = (90 – x)° … (II)
In ΔADC and ΔBDA,
∠DAC = (90 – x)° = ∠DBA … From (II)
∠ADC = ∠BDA … (Each 90°)
⇒ΔADC ∼ΔBDA … By AA Postulate
Similarity (With Applications to Maps and Models) Exercise Ex. 15(D)
Solution 2(i)
Solution 2(ii)
Solution 3(i)
Solution 3(ii)
Solution 4(i)
Solution 4(ii)
Solution 5
Solution 6(i)
Solution 6(ii)
Solution 7
Solution 1(a)
Correct option: (i) 6.4 cm
Scale factor = 1:25
Solution 1(b)
Correct option: (ii) 2.4 m & 1.4 m
Length of the model of a rectangular object = 12 cm
Scale factor = 1 : 20
Solution 1(c)
Correct option: (iii) 1 : 15
Solution 1(d)
Correct option: (iv) 1 : 100
Let the scale factor = k
Area of the room = Length × breadth
⇒Area of the model = k2 × Area of the room
Therefore, Scale factor = 1 : 100
Solution 1(e)
Correct option: (ii) 5 : 3
Let the scale factor = k
Capacity of a container = Volume of the container = length × breadth × height