Class 10 SELINA Solutions Maths Chapter 15 - Similarity (With Applications to Maps and Models)
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),
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
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
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
Similarity (With Applications to Maps and Models) Exercise TEST YOURSELF
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.
