Exercise 12 B
Question 1
Ans:
(i) In Δ (a), third angle =180∘−(75∘+55∘)=180∘−130∘=50∘
So Three angles are 75∘,55∘,150∘
In △(b) , third angle is = 180∘(55∘+80∘)=180∘−135∘=45∘
Three angles are 45∘,80∘,55∘
We see that triangle (a) is different
(ii) In Δ (a) third angle is = 180∘−(100∘+45∘)
=180∘−145=35∘
So Three angles will be = 100∘,35∘,45∘
In Δ (c) third angle is = 180∘−(25∘+45∘)
=180∘−70∘=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=AB−AD=12−8=12−8=4 cm
EC=AC−AE=18−12=6 cm.
Now ADOB=84=21
and AEEC=126=21
if ADDB=AEEC
so DE‖DC
(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=AB−AD=5.6−1.4=4.2 cm
EC=AC−AE=7.2−1.8=5.4 cm Now ADDB=1.44.2=13AEEC=1.85.4=13
If ADDB=AEEC
So OE‖BC
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=BCEF⇒x5=73=y8
x5=73
⇒x=5×73=353=1123
and y8=73
⇒y=8×73=563=1823
Question 6
Ans: if △ABO∼△DCO
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)
△ABC∽△ADE
AD:DB=2:3
DE=5 cm,AL⊥DE and AM⊥BC
AL=x
If △ABC∽△ADE
So ABAD=ACAE=BCDE
⇒2+32=ACAE=BC5 (i) 52=BC3
⇒BC=5×52=252 cm=12.5 cm
(ii) If △ADE∼△ABC
So ALAM=ADAB
(In similar triangles altitudes are also proportional)
⇒xAm=25⇒AM=5x2 So AM=5x2
Question 8
Ans: In trapezium ABCD,AB‖DC
AB=4 cmBC=3 cm and, CD=6 cm
(IMAGE TO BE ADDED)
(i) In the figure
△EDC and △EAB
∠E=∠E∠EDC=∠EAB∠ECD=∠EBA SO △EDC−△EAB
(ii) Let EB = x, then, EC = x+ 3
If △EAB∽△EDC
soEBEC=ABDC⇒xx+3=46
⇒6x=4x+12
⇒6x−4x=12⇒2x=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 △PNQ∽△LNM
if PQLm=NQNm⇒1.8x=1.51.5+5⇒1.8x=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 cm⋅CE=6 cm
BC=3 cm and CD=8 cm
(i) In △ABC and △CDE
BCAC=34 and CECD=68=34∠ACB=∠DCE
So △ABC∼△CDE because BCAC=CECD=34 and ∠ACB=∠DCE
(ii) ∠D=∠A=40∘
(iii) AB=x cm
and ACCD=ABOE=48=12
⇒xDE=12⇒DE=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 PQ‖AC
(i) So △ABC∽△PBQ
∴AC=BCBQ⇒2.51=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=72∘∠B=80∘
Then ∠C=180∘−(72∘+80∘=180∘−152∘
=28∘
and in the other triangles PQB
∠P=28∘,∠Q=72∘
=∠P=28∘,∠Q=72∘
So ∠R=180∘−(28∘+72∘)=180∘−100∘=80∘
if ∠A=∠Q,∠B=∠R and ∠C=∠P
So △ADC〜△PQR
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=∠A∠ABC=∠ADE So △ABC∼△ADE
(ii) So BCPE=ABAD=ACAE
⇒5.2DE=(1.1+2.5)2.7
⇒5.2DE=3.62.7⇒DE=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 △EFG∽△CGD
So EGCh=EFCD
⇒510=EF18⇒EF=5×1010=9
So EF=9 cm
(ii) similarly in △ABC and △ECF
∠C=∠C∠CAB=∠CEF So △ABC〜△ECF SO ACEC=ABEF⇒AC10+5=159⇒AC15=159⇒AC=15×159=2259=25S∘AC=25 cm
Question 15
Ans: △ABC∽△PQR and Let
In △PQR,PQ=4 cm⋅QK=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=46⇒x=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,DE‖CB
so △ABC∽△ADE
Now AD=3 cm,DB=4 cm,EC=12 cm
let AE=x cm
if In △ABC,DE‖BC
ADDB=AEEC⇒34=AE12⇒34=x12⇒x=3×124=9
So AE=9 cm
(b) In △ABC,DE‖BC
So ADDB=AEEC
⇒2.4ED=2.74.5⇒BD=2.4×4.52.7 So BD=4 cm
(c) In △ABC,DE‖BC
So ADDB=AEEC⇒2.C6.5=3EC⇒EC=6.5×32.6⇒EC=152=7.5 So EC=7.5 cm
(d) In △ABC,DE‖BC
SO △ABC △ADE
So ADAD=AEAC=26=AE9⇒AE=2×96=3
So CE=AC−AE=9−3=6 cm
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