Loading [MathJax]/jax/output/HTML-CSS/jax.js

S Chand Class 10 CHAPTER 12 Similar triangles Exercise 12 B

  Exercise 12 B


Question 1

Ans: 

(i) In Δ (a), third angle =180(75+55)=180130=50

So Three angles are 75,55,150
In (b) , third angle is = 180(55+80)=180135=45

Three angles are 45,80,55
We see that triangle (a) is different 

(ii) In Δ (a) third angle is = 180(100+45)
=180145=35

So Three angles will be = 100,35,45
In Δ (c) third angle is = 180(25+45)
=18070=110
So Three angles will be 110,25,45
 we see that triangle (c) is different

(iii) In Δ (a) two sides are equal
In Δ(b), no two sides are equal 
In Δ (c) two sides are equal 
soΔ (b)$ is different

Question 2

Ans:  (i) In the first pair of triangle 
The ratio in Corresponding sides are 
AB4z=BCxy=ACxz
69=4.56.75=710.5
23=450675=70105
23=23=23

Yes , it is the same ration 
So AByz=BCxy=ACxz=23

(ii) In the second pair of triangles
 The ratio in the corresponding sides are 
LmRP=mNRQ=LNPQ
162.4=81.2=142.1
16×1024=8×1012=14×1021
203=203=203=23
Yes , it is the same into 
So LmRP=mNRQ=LNPQ=23

Question 4

Ans: (i) In ABC,D and E are the points on the sides AB and AC respectively such that AD=3,BD=4.51AE =4 and CE = 6
(IMAGE TO BE ADDED)
 Now ADDB=34.5=23 and AEEC=46=23 uf ADDB=AEEC
So DE ||BC

(ii) In ABC,D and E are the points on the sides AB and AC respectively such that 
AB=  =12 cm,AD=8 cm,AE=12 cm and AC=18 cm

(IMAGE TO BE ADDED)
DB=ABAD=128=128=4 cm
EC=ACAE=1812=6 cm.
Now ADOB=84=21
and AEEC=126=21
if ADDB=AEEC
so DEDC

(iii) In ABC,D and E are the points on the sides' 
AB and AC respectively such that AB = 5.6cm
AD = 1.4 cm , AC = 7.2cm and AE = 1.8cm
(IMAGE TO BE ADDED)
So DB=ABAD=5.61.4=4.2 cm
EC=ACAE=7.21.8=5.4 cm Now ADDB=1.44.2=13AEEC=1.85.4=13
If ADDB=AEEC
So OEBC

Question 5

Ans: In the figure ABC and DEF are similar
AB=x,AC=7,BC=yDE=S,DF=3 and EF=8

If Δ are similarly
corresponding sides are proportional
 So ABDE=ACDF=BCEFx5=73=y8

x5=73
x=5×73=353=1123
and y8=73
y=8×73=563=1823

Question 6

Ans: if ABODCO
So Their Corresponding sides are Proportional 
So ABCD=AOOD=BOOC
But AB=3 cm,CD=2 cm,OC=3.2 cm and OD=2.4 cm

(IMAGE TO BE ADDED)
So 32=AO2.4=BO3.2
So A02.4=32
AO=3×2.42=3.6
and BO3.2=32
BO=3×3.22=4.8

Hence OA=3.6 cm and OB=4.8 cm

Question 7

Ans:  (IMAGE TO BE ADDED)

ABCADE
AD:DB=2:3
DE=5 cm,ALDE and AMBC
AL=x
If ABCADE
So ABAD=ACAE=BCDE
2+32=ACAE=BC5 (i) 52=BC3
BC=5×52=252 cm=12.5 cm

(ii) If ADEABC
So ALAM=ADAB
(In similar triangles altitudes are also proportional)
xAm=25AM=5x2 So AM=5x2

Question 8

Ans: In trapezium ABCD,ABDC
AB=4 cmBC=3 cm and, CD=6 cm
(IMAGE TO BE ADDED)

(i) In the figure
EDC and EAB
E=EEDC=EABECD=EBA SO EDCEAB

(ii) Let EB = x, then, EC = x+ 3
If EABEDC
soEBEC=ABDCxx+3=46
6x=4x+12
6x4x=122x=12
x=122=6
So EB=6 cm

Question 9

Ans: If the mean and the lamp post are perpendicular  On the ground 
Height of man PQ = 1.8m
And his shadow QN = 1.5m
(IMAGE TO BE ADDED)
Let LM be the lamp post and LM=xm
Now in △ L M N
So PNQLNM
 if PQLm=NQNm1.8x=1.51.5+51.8x=1.56.5
x=1.8×6.51.5=7.8

So height of the lamp post = 7.8m

Question 10

Ans: Two lines ACE and BCD intersect each other at C 
AC =4 cmCE=6 cm
BC=3 cm and CD=8 cm

(i) In ABC and CDE
BCAC=34 and CECD=68=34ACB=DCE

So ABCCDE because BCAC=CECD=34 and ACB=DCE
(ii) D=A=40
(iii) AB=x cm
and ACCD=ABOE=48=12
xDE=12DE=2x So DE=2x

Question 11

Ans: In right Angle ABC,C=90
 p is any point on AB
(IMAGE TO BE ADDED)

From P, PQ is drawn parallel to AC which meets BC at Q 
AC = 2.5cm, BC = 6cm, PQ=1cm
In ABC is PQAC

(i) So ABCPBQ
AC=BCBQ2.51=6BQ
BQ=6×12.5=C2.5=2.4 cm

(ii) In right PQB
BP2=PQ2+BQ2=(1)2+(2.412=1+5.76=6.75=(2.6)2 So DP=2.6 cm

Question 12
 
Ans: Let in one triangle ABC, 
A=72B=80
Then C=180(72+80=180152
=28
and in the other triangles PQB
P=28,Q=72
=P=28,Q=72
So R=180(28+72)=180100=80
if A=Q,B=R and C=P
So ADCPQR

Question 13
 
Ans: In the figure, AD=2.7 cm,AE=2.5 cmBE=1.1 cm and
BC=3.2 cm
(i) In ABC and ADE
A=AABC=ADE So ABCADE

(ii)  So BCPE=ABAD=ACAE
5.2DE=(1.1+2.5)2.7
5.2DE=3.62.7DE=5.2×2.73.6=3.9

Hence DE=3.9 cm

Question 14
 
Ans: ABC is a triangle AB. EF and CD are parallel line 
AB = 15cm, EG = 5cm. BC = 10cm and DC = 18cm
We have to find EF and AC 

(i) In EFG and CGD
EFG=GDC
So EFGCGD
So EGCh=EFCD
510=EF18EF=5×1010=9
So EF=9 cm
 
(ii)  similarly in ABC and ECF
C=CCAB=CEF So ABCECF SO ACEC=ABEFAC10+5=159AC15=159AC=15×159=2259=25SAC=25 cm

Question 15
 
Ans: ABCPQR and Let
In PQR,PQ=4 cmQK=7 cm and RP=8 cm
and let in ABC,AB=6 cm,BC=x and CA=y 
Then PQAB=QRBC=RPCA
46=7x=8y Now 7x=46x=7×64=212=10.5 and 8y=46
y=6×84=12

So other two corresponding sides are 10.5 cm and 12 cm 

Question 16
 
Ans: (a) In  ABC,DECB
so ABCADE
Now AD=3 cm,DB=4 cm,EC=12 cm
 let AE=x cm
if In ABC,DEBC
ADDB=AEEC34=AE1234=x12x=3×124=9
So AE=9 cm

(b) In ABC,DEBC
So ADDB=AEEC
2.4ED=2.74.5BD=2.4×4.52.7 So BD=4 cm

(c) In ABC,DEBC
 So ADDB=AEEC2.C6.5=3ECEC=6.5×32.6EC=152=7.5 So EC=7.5 cm

(d) In ABC,DEBC
SO ABC ADE
So ADAD=AEAC=26=AE9AE=2×96=3
So CE=ACAE=93=6 cm


















 



No comments:

Post a Comment

Contact Form

Name

Email *

Message *