SELINA Solutions for Class 10 Maths Chapter 15 - Similarity (With Applications to Maps and Models)
Chapter 15 - Similarity (With Applications to Maps and Models) Exercise Ex. 15(A)
(i) False
(ii) True
(iii) True
(iv) False
(v) True
(vi) True
(vii) True
(i) In ∆ ABC and ∆ AMP,
BAC=
PAM [Common]
ABC=
PMA [Each = 90°]
∆ ABC ~ ∆ AMP [AA Similarity]
(ii)
(i)
(ii)
Since, triangles PQT and RQS are similar.
Hence, DP CR = DC
PR
(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,
(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)
Given, AE: EC = BE: ED
Draw EF || AB
In ABD, EF || AB
Using Basic Proportionality theorem,
Thus, in DCA, E and F are points on CA and DA
respectively such that
Thus, by converse of Basic proportionality theorem, FE || DC.
But, FE || AB.
Hence, AB || DC.
Thus, ABCD is a trapezium.
Given, AD2
= BD DC
So, these two triangles will be equiangular.
(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,
Given, QR is parallel to AB. Using Basic proportionality theorem,
Also, DR is parallel to QB. Using Basic proportionality theorem,
From (1) and (2), we get,
1 =
6 (Alternate
interior angles)
2 =
3 (Vertically
opposite angles)
DM = MC (M is the mid-point of CD)
So, DE = BC (Corresponding parts of congruent triangles)
Also, AD = BC (Opposite sides of a parallelogram)
AE = AD + DE = 2BC
Now, 1 =
6 and
4 =
5
(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)
We have:
(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),
Chapter 15 - Similarity (With Applications to Maps and Models) Exercise Ex. 15(B)
Now, DE is parallel to BC.
Then, by Basic proportionality theorem, we have
Chapter 15 - Similarity (With Applications to Maps and Models) Exercise Ex. 15(C)
We know that the ratio of the areas of two similar triangles is equal to the ratio of squares of their corresponding sides.
(i) Required ratio =
(ii) Required ratio =
(i) AP =PB
(ii)
Let
Given,
(i)
(ii)
From the given information, we have:
(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
(i)
(ii)
(i) Since APB and
CPB have common vertex at B and their bases AP and PC are along the same straight line
(ii) Since DPC and
BPA are similar
(iii) Since ADP and
APB have common vertex at A and their bases DP and PB are along the same straight line
(iv) Since APB and
ADB have common vertex at A and their bases BP and BD are along the same straight line
(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
Chapter 15 - Similarity (With Applications to Maps and Models) Exercise Ex. 15(D)
Given that triangle ABC is enlarged and the scale factor is m = 3 to the triangle A'B'C'.
OC = 21 cm
So, (OC)3 = OC'
i.e. 21 x 3 = OC'
i.e. OC' = 63 cm
Chapter 15 - Similarity (With Applications to Maps and Models) Exercise Ex. 15(E)
Given that XY || BC
So, △AXY ∼ △ABC
Join AR.
In ACR, BX || CR. By Basic Proportionality theorem,
In APR, XQ || AP. By Basic Proportionality theorem,
From (1) and (2), we get,
In ABC, PR || BC. By Basic proportionality theorem,
Also, in PAR and
ABC,
Similarly,
(i)
(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.
Let us assume two similar triangles as ABC
PQR
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
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
In PXY and
PQR, XY is parallel to QR, so corresponding angles are equal.
Hence, (By AA similarity criterion)
(i) We know that the ratio of areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.
(ii) Ar (trapezium XQRY) = Ar (PQR) - Ar (
PXY)
= (16x - x) cm2
= 15x cm2
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
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
(i)
(ii)
(iii)
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
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.
In triangle ABC, PO || BC. Using Basic proportionality theorem,
(i)
(ii)
In ABC and
EBD,
ACB =
EDB (given)
ABC =
EBD (common)
(by AA- similarity)
(i) We have,
(ii)
(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:
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.
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